output
stringlengths 52
181k
| instruction
stringlengths 296
182k
|
|---|---|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin >> n;
string s;
cin >> s;
int o = 0, c = 0, q = 0;
for (int i = 0; i < n; i++) {
if (s[i] == '?')
q++;
else if (s[i] == '(')
o++;
else
c++;
}
int to = 0, tc = 0;
for (int i = 0; i < n; i++) {
if (s[i] == '?') {
if (o < n / 2) {
s[i] = '(';
o++;
} else
s[i] = ')';
}
if (s[i] == '(')
to++;
else
tc++;
if (i < n - 1)
if (tc >= to) {
cout << ":(";
return 0;
}
if (tc > to) {
cout << ":(";
return 0;
}
if (to > n / 2) {
cout << ":(";
return 0;
}
}
cout << s;
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin >> n;
string s;
cin >> s;
int o = 0, c = 0, q = 0;
for (int i = 0; i < n; i++) {
if (s[i] == '?')
q++;
else if (s[i] == '(')
o++;
else
c++;
}
int to = 0, tc = 0;
for (int i = 0; i < n; i++) {
if (s[i] == '?') {
if (o < n / 2) {
s[i] = '(';
o++;
} else
s[i] = ')';
}
if (s[i] == '(')
to++;
else
tc++;
if (i < n - 1)
if (tc >= to) {
cout << ":(";
return 0;
}
if (tc > to) {
cout << ":(";
return 0;
}
if (to > n / 2) {
cout << ":(";
return 0;
}
}
cout << s;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 9, oo = 0x3f3f3f3f;
int n, m, h;
int ans[N][N], f[N], l[N], top[N][N], vis[N];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
string s;
cin >> n >> s;
if (n & 1) {
cout << ":(";
return 0;
}
stack<char> st;
bool f = 1;
int c1 = 0, c2 = 0;
for (int i = 0; i < n; ++i) {
c1 += (s[i] == '(');
c2 += (s[i] == ')');
}
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '(')
st.push(s[i]);
else if (s[i] == ')' && !st.empty())
st.pop();
else if (s[i] == '?') {
if (i < n - 1 && st.size() == 1) {
st.push('(');
s[i] = '(';
c1++;
} else if (i == n - 1 && st.size() == 1) {
s[i] = ')';
st.pop();
c2++;
} else if (c1 < n / 2) {
c1++;
s[i] = '(';
st.push('(');
} else if (c2 < n / 2) {
c2++;
s[i] = ')';
st.pop();
}
}
if (i < n - 1 && st.empty()) {
cout << ":(";
return 0;
}
}
stack<char> ch;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '(')
ch.push('(');
else if (s[i] == ')' && !ch.empty())
ch.pop();
else {
cout << ":(";
return 0;
}
}
if (ch.empty())
cout << s;
else
cout << ":(";
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 9, oo = 0x3f3f3f3f;
int n, m, h;
int ans[N][N], f[N], l[N], top[N][N], vis[N];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
string s;
cin >> n >> s;
if (n & 1) {
cout << ":(";
return 0;
}
stack<char> st;
bool f = 1;
int c1 = 0, c2 = 0;
for (int i = 0; i < n; ++i) {
c1 += (s[i] == '(');
c2 += (s[i] == ')');
}
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '(')
st.push(s[i]);
else if (s[i] == ')' && !st.empty())
st.pop();
else if (s[i] == '?') {
if (i < n - 1 && st.size() == 1) {
st.push('(');
s[i] = '(';
c1++;
} else if (i == n - 1 && st.size() == 1) {
s[i] = ')';
st.pop();
c2++;
} else if (c1 < n / 2) {
c1++;
s[i] = '(';
st.push('(');
} else if (c2 < n / 2) {
c2++;
s[i] = ')';
st.pop();
}
}
if (i < n - 1 && st.empty()) {
cout << ":(";
return 0;
}
}
stack<char> ch;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '(')
ch.push('(');
else if (s[i] == ')' && !ch.empty())
ch.pop();
else {
cout << ":(";
return 0;
}
}
if (ch.empty())
cout << s;
else
cout << ":(";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <class T>
using VV = vector<vector<T>>;
template <class T>
inline T SMIN(T& a, const T b) {
return a = (a > b) ? b : a;
}
template <class T>
inline T SMAX(T& a, const T b) {
return a = (a < b) ? b : a;
}
template <class Iter>
void __kumaerrc(Iter begin, Iter end) {
for (; begin != end; ++begin) {
cerr << *begin << ',';
}
cerr << endl;
}
void __kumaerr(istream_iterator<string> it) {
(void)it;
cerr << endl;
}
template <typename T, typename... Args>
void __kumaerr(istream_iterator<string> it, T a, Args... args) {
cerr << *it << "=" << a << ", ", __kumaerr(++it, args...);
}
template <typename S, typename T>
std::ostream& operator<<(std::ostream& _os, const std::pair<S, T>& _p) {
return _os << "{" << _p.first << ',' << _p.second << "}";
}
int N;
string SS;
void nope() {
cout << ":(" << endl;
exit(0);
}
void solve() {
string S = SS;
if (N % 2 == 1) nope();
if (S[0] == ')' || S[N - 1] == '(') nope();
int cnt = 0;
for (int i = 0; i < (N); ++i) cnt += S[i] == '(';
int b = 0;
for (int i = 0; i < (N); ++i) {
if (S[i] == '?') {
if (cnt < N / 2)
S[i] = '(', ++cnt;
else
S[i] = ')';
}
if (S[i] == '(')
++b;
else if (S[i] == ')')
--b;
if (i != N - 1 && b <= 0) nope();
}
if (b != 0) nope();
cout << S << endl;
}
void solve_wrong() {
string S = SS;
int b = 0;
for (int i = 0; i < (N); ++i) {
if (S[i] == '?') {
if (i == N - 1 || b > 0)
S[i] = ')';
else
S[i] = '(';
}
if (S[i] == '(') {
++b;
} else if (S[i] == ')') {
--b;
}
if (b < 0) nope();
}
if (b != 0) nope();
b = 0;
vector<int> X;
for (int i = 0; i < (N); ++i) {
if (S[i] == '(') {
++b;
} else if (S[i] == ')') {
--b;
}
if (i != N - 1 && b == 0) X.push_back(i);
}
{
string _s = "S";
replace(_s.begin(), _s.end(), ',', ' ');
stringstream _ss(_s);
istream_iterator<string> _it(_ss);
__kumaerr(_it, S);
};
{
cerr << "X"
<< ": ";
for (auto& x : (X)) {
cerr << x << ',';
}
cerr << endl;
};
for (int i = 0; i < (int((X).size())); ++i) {
int l = X[i], r = X[i] + 1;
int L = i == 0 ? 0 : X[i - 1], R = i == int((X).size()) - 1 ? N : X[i + 1];
{
string _s = "l, r";
replace(_s.begin(), _s.end(), ',', ' ');
stringstream _ss(_s);
istream_iterator<string> _it(_ss);
__kumaerr(_it, l, r);
};
while (l >= L && SS[l] != '?') --l;
while (r <= R && SS[r] != '?') ++r;
{
string _s = "l, r";
replace(_s.begin(), _s.end(), ',', ' ');
stringstream _ss(_s);
istream_iterator<string> _it(_ss);
__kumaerr(_it, l, r);
};
if (SS[l] == '?' && S[l] == ')' && SS[r] == '?' && S[r] == '(')
swap(S[l], S[r]), SS[l] = '#', SS[r] = '#';
else
nope();
}
cout << S << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout << setprecision(12) << fixed;
cin >> N >> SS;
solve();
return 0;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
using VV = vector<vector<T>>;
template <class T>
inline T SMIN(T& a, const T b) {
return a = (a > b) ? b : a;
}
template <class T>
inline T SMAX(T& a, const T b) {
return a = (a < b) ? b : a;
}
template <class Iter>
void __kumaerrc(Iter begin, Iter end) {
for (; begin != end; ++begin) {
cerr << *begin << ',';
}
cerr << endl;
}
void __kumaerr(istream_iterator<string> it) {
(void)it;
cerr << endl;
}
template <typename T, typename... Args>
void __kumaerr(istream_iterator<string> it, T a, Args... args) {
cerr << *it << "=" << a << ", ", __kumaerr(++it, args...);
}
template <typename S, typename T>
std::ostream& operator<<(std::ostream& _os, const std::pair<S, T>& _p) {
return _os << "{" << _p.first << ',' << _p.second << "}";
}
int N;
string SS;
void nope() {
cout << ":(" << endl;
exit(0);
}
void solve() {
string S = SS;
if (N % 2 == 1) nope();
if (S[0] == ')' || S[N - 1] == '(') nope();
int cnt = 0;
for (int i = 0; i < (N); ++i) cnt += S[i] == '(';
int b = 0;
for (int i = 0; i < (N); ++i) {
if (S[i] == '?') {
if (cnt < N / 2)
S[i] = '(', ++cnt;
else
S[i] = ')';
}
if (S[i] == '(')
++b;
else if (S[i] == ')')
--b;
if (i != N - 1 && b <= 0) nope();
}
if (b != 0) nope();
cout << S << endl;
}
void solve_wrong() {
string S = SS;
int b = 0;
for (int i = 0; i < (N); ++i) {
if (S[i] == '?') {
if (i == N - 1 || b > 0)
S[i] = ')';
else
S[i] = '(';
}
if (S[i] == '(') {
++b;
} else if (S[i] == ')') {
--b;
}
if (b < 0) nope();
}
if (b != 0) nope();
b = 0;
vector<int> X;
for (int i = 0; i < (N); ++i) {
if (S[i] == '(') {
++b;
} else if (S[i] == ')') {
--b;
}
if (i != N - 1 && b == 0) X.push_back(i);
}
{
string _s = "S";
replace(_s.begin(), _s.end(), ',', ' ');
stringstream _ss(_s);
istream_iterator<string> _it(_ss);
__kumaerr(_it, S);
};
{
cerr << "X"
<< ": ";
for (auto& x : (X)) {
cerr << x << ',';
}
cerr << endl;
};
for (int i = 0; i < (int((X).size())); ++i) {
int l = X[i], r = X[i] + 1;
int L = i == 0 ? 0 : X[i - 1], R = i == int((X).size()) - 1 ? N : X[i + 1];
{
string _s = "l, r";
replace(_s.begin(), _s.end(), ',', ' ');
stringstream _ss(_s);
istream_iterator<string> _it(_ss);
__kumaerr(_it, l, r);
};
while (l >= L && SS[l] != '?') --l;
while (r <= R && SS[r] != '?') ++r;
{
string _s = "l, r";
replace(_s.begin(), _s.end(), ',', ' ');
stringstream _ss(_s);
istream_iterator<string> _it(_ss);
__kumaerr(_it, l, r);
};
if (SS[l] == '?' && S[l] == ')' && SS[r] == '?' && S[r] == '(')
swap(S[l], S[r]), SS[l] = '#', SS[r] = '#';
else
nope();
}
cout << S << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout << setprecision(12) << fixed;
cin >> N >> SS;
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:247474112")
#pragma GCC optimize("Ofast")
using namespace std;
int32_t main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
long long n;
string s;
cin >> n >> s;
if (n & 1) {
return cout << ":(", 0;
}
long long sum = 0;
bool pos = 1;
long long j = 0;
long long q = 0;
long long tot(0);
for (long long i = 0; i < s.size(); ++i) {
if (s[i] == '?')
++q;
else if (s[i] == '(')
tot++;
else
tot--;
}
long long close = tot;
long long left = q - close;
long long open = left / 2;
close += left / 2;
for (long long i = 0; i < s.size(); ++i) {
if (s[i] == '?') {
if (open) {
s[i] = '(';
open--;
} else {
s[i] = ')';
close--;
}
}
}
for (long long i = 0; i < s.size() - 1; ++i) {
if (s[i] == '(') sum++;
if (s[i] == ')') sum--;
if (sum <= 0) pos = 0;
}
if (s.back() == '(')
sum++;
else if (s.back() == ')')
sum--;
else
s.back() = ')', sum--;
if (pos and sum == 0) {
cout << s;
} else {
cout << ":(";
}
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
### Response
```cpp
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:247474112")
#pragma GCC optimize("Ofast")
using namespace std;
int32_t main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
long long n;
string s;
cin >> n >> s;
if (n & 1) {
return cout << ":(", 0;
}
long long sum = 0;
bool pos = 1;
long long j = 0;
long long q = 0;
long long tot(0);
for (long long i = 0; i < s.size(); ++i) {
if (s[i] == '?')
++q;
else if (s[i] == '(')
tot++;
else
tot--;
}
long long close = tot;
long long left = q - close;
long long open = left / 2;
close += left / 2;
for (long long i = 0; i < s.size(); ++i) {
if (s[i] == '?') {
if (open) {
s[i] = '(';
open--;
} else {
s[i] = ')';
close--;
}
}
}
for (long long i = 0; i < s.size() - 1; ++i) {
if (s[i] == '(') sum++;
if (s[i] == ')') sum--;
if (sum <= 0) pos = 0;
}
if (s.back() == '(')
sum++;
else if (s.back() == ')')
sum--;
else
s.back() = ')', sum--;
if (pos and sum == 0) {
cout << s;
} else {
cout << ":(";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void wrong() { cout << ":("; }
int main() {
long long n, pre = 0, post = 0, i, x = 0;
string s;
cin >> n;
cin >> s;
if (n % 2) {
wrong();
return 0;
}
for (i = 0; i < s.length(); i++) {
if (s[i] == '(') {
x++;
}
}
for (i = 0; i < n; i++) {
if (s[i] == '(') {
pre++;
} else if (s[i] == ')') {
post++;
} else {
if (pre < n / 2 && x < n / 2) {
s[i] = '(';
pre++;
x++;
} else {
s[i] = ')';
post++;
}
}
if ((pre > n / 2 || post > n / 2 || pre == post || post > pre) &&
i != n - 1) {
wrong();
return 0;
}
}
if (pre == post) {
cout << s;
} else {
wrong();
return 0;
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void wrong() { cout << ":("; }
int main() {
long long n, pre = 0, post = 0, i, x = 0;
string s;
cin >> n;
cin >> s;
if (n % 2) {
wrong();
return 0;
}
for (i = 0; i < s.length(); i++) {
if (s[i] == '(') {
x++;
}
}
for (i = 0; i < n; i++) {
if (s[i] == '(') {
pre++;
} else if (s[i] == ')') {
post++;
} else {
if (pre < n / 2 && x < n / 2) {
s[i] = '(';
pre++;
x++;
} else {
s[i] = ')';
post++;
}
}
if ((pre > n / 2 || post > n / 2 || pre == post || post > pre) &&
i != n - 1) {
wrong();
return 0;
}
}
if (pre == post) {
cout << s;
} else {
wrong();
return 0;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const bool debug = false;
const bool INJUDGE = false;
const int MAXN = 100005;
const int MOD = 1000000007;
const int INF = 1000000007;
template <class T>
T emax(T& a, T b) {
return a = max(a, b);
}
template <class T>
T emin(T& a, T b) {
return a = min(a, b);
}
template <class T>
void byebye(T info) {
cout << info << endl;
exit(0);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> P) {
out << P.first << " " << P.second;
return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> V) {
for (int i = 0; i < V.size(); ++i) out << V[i] << " ";
return out;
}
char s[300005];
int len;
int main() {
scanf("%d", &len);
scanf("%s", s);
if (len & 1) byebye(":(");
int leftq = len / 2;
for (int i = 0; i < len; ++i) {
if (s[i] == '(') --leftq;
}
if (leftq < 0) byebye(":(");
for (int i = 0; i < len; ++i) {
if (s[i] == '?') {
if (leftq) {
--leftq;
s[i] = '(';
} else {
s[i] = ')';
}
}
}
int cnt = 0;
for (int i = 0; i < len; ++i) {
if (s[i] == '(')
++cnt;
else
--cnt;
if (i != len - 1 && !cnt) byebye(":(");
if (cnt < 0) byebye(":(");
}
if (cnt) {
byebye(":(");
}
puts(s);
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const bool debug = false;
const bool INJUDGE = false;
const int MAXN = 100005;
const int MOD = 1000000007;
const int INF = 1000000007;
template <class T>
T emax(T& a, T b) {
return a = max(a, b);
}
template <class T>
T emin(T& a, T b) {
return a = min(a, b);
}
template <class T>
void byebye(T info) {
cout << info << endl;
exit(0);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> P) {
out << P.first << " " << P.second;
return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> V) {
for (int i = 0; i < V.size(); ++i) out << V[i] << " ";
return out;
}
char s[300005];
int len;
int main() {
scanf("%d", &len);
scanf("%s", s);
if (len & 1) byebye(":(");
int leftq = len / 2;
for (int i = 0; i < len; ++i) {
if (s[i] == '(') --leftq;
}
if (leftq < 0) byebye(":(");
for (int i = 0; i < len; ++i) {
if (s[i] == '?') {
if (leftq) {
--leftq;
s[i] = '(';
} else {
s[i] = ')';
}
}
}
int cnt = 0;
for (int i = 0; i < len; ++i) {
if (s[i] == '(')
++cnt;
else
--cnt;
if (i != len - 1 && !cnt) byebye(":(");
if (cnt < 0) byebye(":(");
}
if (cnt) {
byebye(":(");
}
puts(s);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long n, k, t, m, q, x, flag = 0;
long long po(long long k, long long n, long long ans, long long temp,
long long ans1) {
if (n == 0) return ans;
while (temp <= n) {
ans *= ans1;
ans %= mod;
ans1 = ans1 * ans1;
ans1 %= mod;
n = n - temp;
temp *= 2;
}
return po(k, n, ans, 1, k) % mod;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
t = 1;
while (t--) {
cin >> n;
long long a = 0, b = 0, c = 0;
char x[n];
for (int i = 0; i < n; i++) {
cin >> x[i];
if (x[i] == '(') a++;
if (x[i] == ')') b++;
if (x[i] == '?') c++;
}
if (n % 2 == 1) {
cout << ":(";
return 0;
}
int temp = b;
for (int i = n - 1; i >= 0; i--) {
if (x[i] == '?' && temp < n / 2)
x[i] = ')', temp++;
else if (x[i] == '?')
x[i] = '(';
}
int c1 = 0, c2 = 0;
for (int i = 0; i < n; i++) {
if (x[i] == '(')
c1++;
else
c2++;
if (c2 >= c1 && i != n - 1) {
flag = 1;
break;
}
}
if (flag || c1 != c2) {
cout << ":(";
return 0;
} else
for (int i = 0; i < n; i++) cout << x[i];
}
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long n, k, t, m, q, x, flag = 0;
long long po(long long k, long long n, long long ans, long long temp,
long long ans1) {
if (n == 0) return ans;
while (temp <= n) {
ans *= ans1;
ans %= mod;
ans1 = ans1 * ans1;
ans1 %= mod;
n = n - temp;
temp *= 2;
}
return po(k, n, ans, 1, k) % mod;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
t = 1;
while (t--) {
cin >> n;
long long a = 0, b = 0, c = 0;
char x[n];
for (int i = 0; i < n; i++) {
cin >> x[i];
if (x[i] == '(') a++;
if (x[i] == ')') b++;
if (x[i] == '?') c++;
}
if (n % 2 == 1) {
cout << ":(";
return 0;
}
int temp = b;
for (int i = n - 1; i >= 0; i--) {
if (x[i] == '?' && temp < n / 2)
x[i] = ')', temp++;
else if (x[i] == '?')
x[i] = '(';
}
int c1 = 0, c2 = 0;
for (int i = 0; i < n; i++) {
if (x[i] == '(')
c1++;
else
c2++;
if (c2 >= c1 && i != n - 1) {
flag = 1;
break;
}
}
if (flag || c1 != c2) {
cout << ":(";
return 0;
} else
for (int i = 0; i < n; i++) cout << x[i];
}
return 0;
}
``` |
#include <bits/stdc++.h>
const unsigned int M = 1000000007;
using namespace std;
long long power(long long x, long long y) {
long long res = 1;
while (y > 0) {
if (y & 1) res = x * res;
y = y >> 1;
x = x * x;
}
return res;
}
void nik() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
signed main() {
nik();
long long t = 1;
while (t--) {
long long n, on = 0, ron, cnt = 0, off = 0;
cin >> n;
string s, ans = "";
cin >> s;
for (long long i = 0; i < (long long)s.size(); i++) {
if (s[i] == '(')
on++;
else if (s[i] == ')')
off++;
}
if (n % 2) return cout << ":(", 0;
ron = n / 2 - on;
for (long long i = 0; i < (long long)s.size(); i++) {
if (s[i] == '?') {
if (ron)
s[i] = '(', ron--;
else
s[i] = ')';
}
}
long long ok = 1;
for (long long i = 0; i < (long long)s.size(); i++) {
if (s[i] == '(')
cnt++;
else
cnt--;
if (cnt <= 0 && i != s.size() - 1) {
ok = 0;
break;
}
}
if (ok && cnt == 0)
cout << s;
else
cout << ":(";
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
### Response
```cpp
#include <bits/stdc++.h>
const unsigned int M = 1000000007;
using namespace std;
long long power(long long x, long long y) {
long long res = 1;
while (y > 0) {
if (y & 1) res = x * res;
y = y >> 1;
x = x * x;
}
return res;
}
void nik() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
}
signed main() {
nik();
long long t = 1;
while (t--) {
long long n, on = 0, ron, cnt = 0, off = 0;
cin >> n;
string s, ans = "";
cin >> s;
for (long long i = 0; i < (long long)s.size(); i++) {
if (s[i] == '(')
on++;
else if (s[i] == ')')
off++;
}
if (n % 2) return cout << ":(", 0;
ron = n / 2 - on;
for (long long i = 0; i < (long long)s.size(); i++) {
if (s[i] == '?') {
if (ron)
s[i] = '(', ron--;
else
s[i] = ')';
}
}
long long ok = 1;
for (long long i = 0; i < (long long)s.size(); i++) {
if (s[i] == '(')
cnt++;
else
cnt--;
if (cnt <= 0 && i != s.size() - 1) {
ok = 0;
break;
}
}
if (ok && cnt == 0)
cout << s;
else
cout << ":(";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void fail() {
cout << ":(" << endl;
exit(0);
}
int main() {
int N;
string S;
cin >> N >> S;
if (N % 2) fail();
int sum = 0, q = 0;
for (int i = 0; i < N; i++) {
if (S[i] == '?') {
q++;
} else {
sum += S[i] == '(' ? 1 : -1;
}
}
int L = (q - sum) / 2;
int R = q - L;
if (L < 0 || R < 0) fail();
for (int i = 0; i < N; i++) {
if (S[i] == '?') {
if (L > 0) {
L--;
S[i] = '(';
} else {
S[i] = ')';
}
}
}
sum = 0;
for (int i = 0; i < N; i++) {
sum += S[i] == '(' ? 1 : -1;
if (i < N - 1 && sum <= 0) fail();
}
if (sum != 0) fail();
cout << S << endl;
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.
In his favorite math class, the teacher taught him the following interesting definitions.
A parenthesis sequence is a string, containing only characters "(" and ")".
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.
We define that |s| as the length of string s. A strict prefix s[1... l] (1≤ l< |s|) of a string s = s_1s_2... s_{|s|} is string s_1s_2... s_l. Note that the empty string and the whole string are not strict prefixes of any string by the definition.
Having learned these definitions, he comes up with a new problem. He writes down a string s containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in s independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.
Input
The first line contains a single integer |s| (1≤ |s|≤ 3 ⋅ 10^5), the length of the string.
The second line contains a string s, containing only "(", ")" and "?".
Output
A single line contains a string representing the answer.
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
Examples
Input
6
(?????
Output
(()())
Input
10
(???(???(?
Output
:(
Note
It can be proved that there is no solution for the second sample, so print ":(".
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void fail() {
cout << ":(" << endl;
exit(0);
}
int main() {
int N;
string S;
cin >> N >> S;
if (N % 2) fail();
int sum = 0, q = 0;
for (int i = 0; i < N; i++) {
if (S[i] == '?') {
q++;
} else {
sum += S[i] == '(' ? 1 : -1;
}
}
int L = (q - sum) / 2;
int R = q - L;
if (L < 0 || R < 0) fail();
for (int i = 0; i < N; i++) {
if (S[i] == '?') {
if (L > 0) {
L--;
S[i] = '(';
} else {
S[i] = ')';
}
}
}
sum = 0;
for (int i = 0; i < N; i++) {
sum += S[i] == '(' ? 1 : -1;
if (i < N - 1 && sum <= 0) fail();
}
if (sum != 0) fail();
cout << S << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, re;
long long ans;
long long stck[100010];
bool chstck[100010];
bool over;
int main() {
scanf("%d", &n);
stck[0] = 1;
chstck[0] = true;
for (int i = 1; i <= n; i++) {
char in[10];
scanf("%s", in);
if (in[0] == 'a') {
ans += stck[re];
if (ans >= 4294967296 || !chstck[re]) {
over = true;
}
}
if (in[0] == 'f') {
int num;
scanf("%d", &num);
stck[re + 1] = stck[re] * num;
++re;
if (stck[re] < 4294967296 && chstck[re - 1])
chstck[re] = true;
else
chstck[re] = false;
}
if (in[0] == 'e') {
re--;
}
}
if (over) {
printf("OVERFLOW!!!");
return 0;
}
printf("%lld", ans);
}
| ### Prompt
Please create a solution in CPP to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, re;
long long ans;
long long stck[100010];
bool chstck[100010];
bool over;
int main() {
scanf("%d", &n);
stck[0] = 1;
chstck[0] = true;
for (int i = 1; i <= n; i++) {
char in[10];
scanf("%s", in);
if (in[0] == 'a') {
ans += stck[re];
if (ans >= 4294967296 || !chstck[re]) {
over = true;
}
}
if (in[0] == 'f') {
int num;
scanf("%d", &num);
stck[re + 1] = stck[re] * num;
++re;
if (stck[re] < 4294967296 && chstck[re - 1])
chstck[re] = true;
else
chstck[re] = false;
}
if (in[0] == 'e') {
re--;
}
}
if (over) {
printf("OVERFLOW!!!");
return 0;
}
printf("%lld", ans);
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long ans = pow(2.0, 32);
long long t, flag, loop = 1, n, sum = 0;
cin >> t;
string s;
stack<int> st;
flag = 0;
while (t--) {
cin >> s;
if (s == "add") {
sum += loop;
if (sum >= ans) flag = 1;
} else if (s == "end") {
if (!st.empty()) {
int k = st.top();
loop = loop / k;
st.pop();
}
} else {
cin >> n;
if (loop >= ans) {
st.push(1);
} else
st.push(n), loop = loop * n;
}
}
if (flag)
cout << "OVERFLOW!!!" << endl;
else
cout << sum << endl;
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long ans = pow(2.0, 32);
long long t, flag, loop = 1, n, sum = 0;
cin >> t;
string s;
stack<int> st;
flag = 0;
while (t--) {
cin >> s;
if (s == "add") {
sum += loop;
if (sum >= ans) flag = 1;
} else if (s == "end") {
if (!st.empty()) {
int k = st.top();
loop = loop / k;
st.pop();
}
} else {
cin >> n;
if (loop >= ans) {
st.push(1);
} else
st.push(n), loop = loop * n;
}
}
if (flag)
cout << "OVERFLOW!!!" << endl;
else
cout << sum << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
unsigned long long int x = 1, y = 0, lim = 0, prev = 1, ans = 0, t2;
lim = pow(2, 32) - 1;
cin >> n;
string s;
vector<int> v1, v2;
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "add") {
unsigned long long int t = 1;
for (int j = 0; j < v2.size(); j++) {
t *= v2[j];
if (t > lim) {
cout << "OVERFLOW!!!";
return 0;
}
}
prev = prev * t;
if (prev > lim) {
cout << "OVERFLOW!!!";
return 0;
}
for (int j = 0; j < v2.size(); j++) {
v1.push_back(v2[j]);
}
v2.clear();
ans += prev;
if (ans > lim) {
cout << "OVERFLOW!!!";
return 0;
}
} else if (s == "end") {
if (v2.size() == 0) {
t2 = v1[v1.size() - 1];
v1.pop_back();
prev /= t2;
} else {
v2.pop_back();
}
} else if (s == "for") {
int h;
cin >> h;
v2.push_back(h);
}
}
cout << ans;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
unsigned long long int x = 1, y = 0, lim = 0, prev = 1, ans = 0, t2;
lim = pow(2, 32) - 1;
cin >> n;
string s;
vector<int> v1, v2;
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "add") {
unsigned long long int t = 1;
for (int j = 0; j < v2.size(); j++) {
t *= v2[j];
if (t > lim) {
cout << "OVERFLOW!!!";
return 0;
}
}
prev = prev * t;
if (prev > lim) {
cout << "OVERFLOW!!!";
return 0;
}
for (int j = 0; j < v2.size(); j++) {
v1.push_back(v2[j]);
}
v2.clear();
ans += prev;
if (ans > lim) {
cout << "OVERFLOW!!!";
return 0;
}
} else if (s == "end") {
if (v2.size() == 0) {
t2 = v1[v1.size() - 1];
v1.pop_back();
prev /= t2;
} else {
v2.pop_back();
}
} else if (s == "for") {
int h;
cin >> h;
v2.push_back(h);
}
}
cout << ans;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
inline long long read() {
char ch = getchar();
long long x = 0, f = 0;
while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return f ? -x : x;
}
int l, stk[maxn], tp;
char op[maxn][10];
bool hhh[maxn];
long long x, y[maxn], a[maxn];
int main() {
l = read();
for (int i = (1); i <= (l); i++) {
scanf("%s", op[i] + 1);
if (op[i][1] == 'f') a[i] = read();
}
y[0] = 1;
for (int i = (1); i <= (l); i++) {
if (op[i][1] == 'f') {
stk[++tp] = i;
y[i] = y[i - 1] * a[i];
if (hhh[i - 1] || y[i] >= (1ll << 32)) hhh[i] = true;
} else {
if (op[i][1] == 'e')
y[i] = y[stk[tp] - 1], hhh[i] = hhh[stk[tp] - 1], tp--;
else {
y[i] = y[i - 1];
hhh[i] = hhh[i - 1];
if (hhh[i]) return puts("OVERFLOW!!!"), 0;
x += y[i];
if (x >= (1ll << 32)) return puts("OVERFLOW!!!"), 0;
}
}
}
cout << x << endl;
}
| ### Prompt
Generate a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
inline long long read() {
char ch = getchar();
long long x = 0, f = 0;
while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return f ? -x : x;
}
int l, stk[maxn], tp;
char op[maxn][10];
bool hhh[maxn];
long long x, y[maxn], a[maxn];
int main() {
l = read();
for (int i = (1); i <= (l); i++) {
scanf("%s", op[i] + 1);
if (op[i][1] == 'f') a[i] = read();
}
y[0] = 1;
for (int i = (1); i <= (l); i++) {
if (op[i][1] == 'f') {
stk[++tp] = i;
y[i] = y[i - 1] * a[i];
if (hhh[i - 1] || y[i] >= (1ll << 32)) hhh[i] = true;
} else {
if (op[i][1] == 'e')
y[i] = y[stk[tp] - 1], hhh[i] = hhh[stk[tp] - 1], tp--;
else {
y[i] = y[i - 1];
hhh[i] = hhh[i - 1];
if (hhh[i]) return puts("OVERFLOW!!!"), 0;
x += y[i];
if (x >= (1ll << 32)) return puts("OVERFLOW!!!"), 0;
}
}
}
cout << x << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int main() {
long long int ans = 0;
long long int m = (1LL << 32);
int l;
cin >> l;
vector<long long int> v = {1};
for (int i = 0; i < l; i++) {
string t;
cin >> t;
if (t == "add")
ans += v.back();
else if (t == "for") {
int num;
cin >> num;
v.push_back(min(m, v.back() * num));
} else
v.pop_back();
}
if (ans >= m)
cout << "OVERFLOW!!!";
else
cout << ans;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int main() {
long long int ans = 0;
long long int m = (1LL << 32);
int l;
cin >> l;
vector<long long int> v = {1};
for (int i = 0; i < l; i++) {
string t;
cin >> t;
if (t == "add")
ans += v.back();
else if (t == "for") {
int num;
cin >> num;
v.push_back(min(m, v.back() * num));
} else
v.pop_back();
}
if (ans >= m)
cout << "OVERFLOW!!!";
else
cout << ans;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long inf = 1e18;
const long double pai = acos(-1);
long long n, MX = 4294967295;
string s[100009];
long long N[100009];
long long en[100009];
long long num[109], here[100009];
long long val[109][100009];
long long mx[109];
void out() {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
bool OverFlow(long long x, long long y) { return x > inf / y; }
void fill() {
for (long long i = 1; i <= 100; i++) {
val[i][0] = 1;
long long j = 1;
while (j <= 1e5) {
val[i][j] = val[i][j - 1] * i;
if (val[i][j] > MX) {
mx[i] = j;
break;
}
j++;
mx[i] = j;
}
}
}
int main() {
fill();
cin >> n;
for (long long i = 0; i < n; i++) {
cin >> s[i];
if (s[i] == "for") cin >> N[i];
}
stack<long long> st;
for (long long i = 0; i < n; i++) {
if (s[i] == "add") continue;
if (s[i] == "for") {
st.push(i);
} else {
long long x = st.top();
st.pop();
en[x] = i;
}
}
for (long long i = 0; i < n; i++) {
if (s[i] != "for") continue;
here[i] = N[i];
here[en[i]] = -N[i];
}
long long crnt = 0;
for (long long i = 0; i < n; i++) {
if (here[i] > 0) num[here[i]]++;
if (here[i] < 0) num[-here[i]]--;
if (s[i] != "add") continue;
long long X = 1;
for (long long j = 1; j <= 100; j++) {
if (num[j] >= mx[j]) out();
if (OverFlow(X, val[j][num[j]])) out();
X *= val[j][num[j]];
if (X > MX) out();
}
crnt += X;
if (crnt > MX) out();
}
cout << crnt << endl;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long inf = 1e18;
const long double pai = acos(-1);
long long n, MX = 4294967295;
string s[100009];
long long N[100009];
long long en[100009];
long long num[109], here[100009];
long long val[109][100009];
long long mx[109];
void out() {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
bool OverFlow(long long x, long long y) { return x > inf / y; }
void fill() {
for (long long i = 1; i <= 100; i++) {
val[i][0] = 1;
long long j = 1;
while (j <= 1e5) {
val[i][j] = val[i][j - 1] * i;
if (val[i][j] > MX) {
mx[i] = j;
break;
}
j++;
mx[i] = j;
}
}
}
int main() {
fill();
cin >> n;
for (long long i = 0; i < n; i++) {
cin >> s[i];
if (s[i] == "for") cin >> N[i];
}
stack<long long> st;
for (long long i = 0; i < n; i++) {
if (s[i] == "add") continue;
if (s[i] == "for") {
st.push(i);
} else {
long long x = st.top();
st.pop();
en[x] = i;
}
}
for (long long i = 0; i < n; i++) {
if (s[i] != "for") continue;
here[i] = N[i];
here[en[i]] = -N[i];
}
long long crnt = 0;
for (long long i = 0; i < n; i++) {
if (here[i] > 0) num[here[i]]++;
if (here[i] < 0) num[-here[i]]--;
if (s[i] != "add") continue;
long long X = 1;
for (long long j = 1; j <= 100; j++) {
if (num[j] >= mx[j]) out();
if (OverFlow(X, val[j][num[j]])) out();
X *= val[j][num[j]];
if (X > MX) out();
}
crnt += X;
if (crnt > MX) out();
}
cout << crnt << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long p = 1000000007ll;
int main(void) {
long long n;
cin >> n;
vector<pair<long long, long long> > loops;
long long tmp = 0;
for (long long i = (0); i < (n); ++i) {
string com;
cin >> com;
if (com == "for") {
long long x;
cin >> x;
loops.emplace_back(x, tmp);
tmp = 0;
} else if (com == "end") {
tmp = tmp * loops.back().first + loops.back().second;
if (tmp >= (1ll << 32)) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
loops.pop_back();
} else {
++tmp;
}
}
if (tmp >= (1ll << 32)) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
cout << tmp << endl;
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long p = 1000000007ll;
int main(void) {
long long n;
cin >> n;
vector<pair<long long, long long> > loops;
long long tmp = 0;
for (long long i = (0); i < (n); ++i) {
string com;
cin >> com;
if (com == "for") {
long long x;
cin >> x;
loops.emplace_back(x, tmp);
tmp = 0;
} else if (com == "end") {
tmp = tmp * loops.back().first + loops.back().second;
if (tmp >= (1ll << 32)) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
loops.pop_back();
} else {
++tmp;
}
}
if (tmp >= (1ll << 32)) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
cout << tmp << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long MAX = 4294967295;
void run_prog(vector<pair<string, int> > prog, int n) {
unsigned long long pw = 1, num = 0, prev = 1, lvl = 0;
bool go_for = false, add = true;
map<int, unsigned long long> add_value, start_for;
for (int i(0); i < n; ++i) {
if (prog[i].first == "for") {
lvl++;
add_value[lvl] = 0;
if (lvl > 1) {
start_for[lvl] = prog[i].second * start_for[lvl - 1];
} else {
start_for[lvl] = prog[i].second;
}
} else if (prog[i].first == "add") {
add_value[lvl]++;
} else if (prog[i].first == "end") {
if (!start_for[lvl] and add_value[lvl]) {
num += MAX + 1;
break;
} else {
num += start_for[lvl] * add_value[lvl];
}
lvl--;
}
}
num += add_value[0];
if (num > MAX) {
cout << "OVERFLOW!!!\n";
} else {
cout << num << endl;
}
}
int main() {
int n;
cin >> n;
vector<pair<string, int> > prog(n);
string tmp;
int num;
for (int i(0); i < n; ++i) {
cin >> tmp;
if (tmp == "for") {
cin >> num;
prog[i].first = tmp;
prog[i].second = num;
} else {
prog[i].first = tmp;
}
}
run_prog(prog, n);
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long MAX = 4294967295;
void run_prog(vector<pair<string, int> > prog, int n) {
unsigned long long pw = 1, num = 0, prev = 1, lvl = 0;
bool go_for = false, add = true;
map<int, unsigned long long> add_value, start_for;
for (int i(0); i < n; ++i) {
if (prog[i].first == "for") {
lvl++;
add_value[lvl] = 0;
if (lvl > 1) {
start_for[lvl] = prog[i].second * start_for[lvl - 1];
} else {
start_for[lvl] = prog[i].second;
}
} else if (prog[i].first == "add") {
add_value[lvl]++;
} else if (prog[i].first == "end") {
if (!start_for[lvl] and add_value[lvl]) {
num += MAX + 1;
break;
} else {
num += start_for[lvl] * add_value[lvl];
}
lvl--;
}
}
num += add_value[0];
if (num > MAX) {
cout << "OVERFLOW!!!\n";
} else {
cout << num << endl;
}
}
int main() {
int n;
cin >> n;
vector<pair<string, int> > prog(n);
string tmp;
int num;
for (int i(0); i < n; ++i) {
cin >> tmp;
if (tmp == "for") {
cin >> num;
prog[i].first = tmp;
prog[i].second = num;
} else {
prog[i].first = tmp;
}
}
run_prog(prog, n);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long n, st[100005], tot = 0;
char s[100];
long long S = 1, ans = 0, u = ((1ll << 32) - 1ll);
inline void filehandling() {}
int main() {
filehandling();
scanf("%d", &n);
st[0] = 1;
for (int i = 1; i <= n; i++) {
int x;
scanf("%s", s);
if (s[0] == 'f') {
scanf("%d", &x);
S = min(u + 1, st[tot] * x);
st[++tot] = S;
} else if (s[0] == 'a') {
if (ans + st[tot] > u) {
printf("OVERFLOW!!!");
return 0;
} else
ans = ans + st[tot];
} else
tot--;
}
printf("%I64d", ans);
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, st[100005], tot = 0;
char s[100];
long long S = 1, ans = 0, u = ((1ll << 32) - 1ll);
inline void filehandling() {}
int main() {
filehandling();
scanf("%d", &n);
st[0] = 1;
for (int i = 1; i <= n; i++) {
int x;
scanf("%s", s);
if (s[0] == 'f') {
scanf("%d", &x);
S = min(u + 1, st[tot] * x);
st[++tot] = S;
} else if (s[0] == 'a') {
if (ans + st[tot] > u) {
printf("OVERFLOW!!!");
return 0;
} else
ans = ans + st[tot];
} else
tot--;
}
printf("%I64d", ans);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-9;
const long double PI = acos(-1.0);
const int MX = 6e5 + 5;
long long mod = 1e9 + 7;
int main() {
int tc;
cin >> tc;
stack<unsigned long long> f, v;
f.push(1);
v.push(0);
while (tc--) {
string s;
cin >> s;
if (s[0] == 'a') {
v.top()++;
if (v.top() > UINT_MAX) {
cout << "OVERFLOW!!!\n";
return 0;
}
} else if (s[0] == 'f') {
unsigned long long x;
cin >> x;
f.push(x);
v.push(0);
} else {
unsigned long long k = v.top() * f.top();
if (k > UINT_MAX) {
cout << "OVERFLOW!!!\n";
return 0;
}
v.pop();
f.pop();
v.top() += k;
if (v.top() > UINT_MAX) {
cout << "OVERFLOW!!!\n";
return 0;
}
}
}
cout << v.top() << '\n';
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-9;
const long double PI = acos(-1.0);
const int MX = 6e5 + 5;
long long mod = 1e9 + 7;
int main() {
int tc;
cin >> tc;
stack<unsigned long long> f, v;
f.push(1);
v.push(0);
while (tc--) {
string s;
cin >> s;
if (s[0] == 'a') {
v.top()++;
if (v.top() > UINT_MAX) {
cout << "OVERFLOW!!!\n";
return 0;
}
} else if (s[0] == 'f') {
unsigned long long x;
cin >> x;
f.push(x);
v.push(0);
} else {
unsigned long long k = v.top() * f.top();
if (k > UINT_MAX) {
cout << "OVERFLOW!!!\n";
return 0;
}
v.pop();
f.pop();
v.top() += k;
if (v.top() > UINT_MAX) {
cout << "OVERFLOW!!!\n";
return 0;
}
}
}
cout << v.top() << '\n';
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
const long long INF = 1ll << 32;
int t;
cin >> t;
long long add = 0;
long long x;
stack<long long> q;
q.push(1);
while (t--) {
string s;
int n;
cin >> s;
if (s == "for") {
cin >> n;
long long x = (q.top() * n);
q.push(min(INF, x));
} else if (s == "end") {
q.pop();
} else {
add += q.top();
}
}
if (add >= INF) {
cout << "OVERFLOW!!!";
} else
cout << add;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
const long long INF = 1ll << 32;
int t;
cin >> t;
long long add = 0;
long long x;
stack<long long> q;
q.push(1);
while (t--) {
string s;
int n;
cin >> s;
if (s == "for") {
cin >> n;
long long x = (q.top() * n);
q.push(min(INF, x));
} else if (s == "end") {
q.pop();
} else {
add += q.top();
}
}
if (add >= INF) {
cout << "OVERFLOW!!!";
} else
cout << add;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int l;
cin >> l;
long long int x = 0;
long long int depth = 1;
long long int tresh = pow(2, 32) - 1;
int st = 0;
int depthst = 0;
stack<long long int> nus;
nus.push(depth);
for (int i = 0; i < l; ++i) {
string cmd;
cin >> cmd;
if (cmd == "add") {
if (depthst == 1) {
st = 1;
}
x += depth;
} else if (cmd == "for") {
int nu;
cin >> nu;
if (depth * nu > tresh || depth == -1) {
depthst = 1;
depth = -1;
} else {
depth *= nu;
}
nus.push(depth);
} else if (cmd == "end") {
nus.pop();
depth = nus.top();
if (depth == -1)
depthst = 1;
else
depthst = 0;
}
if (x > tresh) st = 1;
}
if (st)
cout << "OVERFLOW!!!" << endl;
else
cout << x << endl;
return 0;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int l;
cin >> l;
long long int x = 0;
long long int depth = 1;
long long int tresh = pow(2, 32) - 1;
int st = 0;
int depthst = 0;
stack<long long int> nus;
nus.push(depth);
for (int i = 0; i < l; ++i) {
string cmd;
cin >> cmd;
if (cmd == "add") {
if (depthst == 1) {
st = 1;
}
x += depth;
} else if (cmd == "for") {
int nu;
cin >> nu;
if (depth * nu > tresh || depth == -1) {
depthst = 1;
depth = -1;
} else {
depth *= nu;
}
nus.push(depth);
} else if (cmd == "end") {
nus.pop();
depth = nus.top();
if (depth == -1)
depthst = 1;
else
depthst = 0;
}
if (x > tresh) st = 1;
}
if (st)
cout << "OVERFLOW!!!" << endl;
else
cout << x << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long l, x = 0, n, kt = 0;
stack<long long> s, tmp;
string cmd;
int main() {
cin >> l;
for (int i = 1; i <= l; i++) {
cin >> cmd;
if (cmd == "for") {
cin >> n;
s.push(n);
tmp.push(0);
}
if (cmd == "add") {
if (s.size() > 0) {
long long dm = tmp.top();
tmp.pop();
dm++;
tmp.push(dm);
} else {
x++;
}
}
if (cmd == "end") {
if (s.size() > 1) {
long long dm1 = tmp.top() * s.top();
tmp.pop();
s.pop();
long long dm2 = tmp.top();
tmp.pop();
dm2 = dm2 + dm1;
tmp.push(dm2);
if (tmp.top() * s.top() > (1LL << 32) - 1) {
kt = 1;
break;
}
} else {
x = x + tmp.top() * s.top();
s.pop();
tmp.pop();
}
}
if (x > (1LL << 32) - 1 || x < 0) {
kt = 1;
break;
}
}
if (kt == 0) {
cout << x;
} else {
cout << "OVERFLOW!!!";
}
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long l, x = 0, n, kt = 0;
stack<long long> s, tmp;
string cmd;
int main() {
cin >> l;
for (int i = 1; i <= l; i++) {
cin >> cmd;
if (cmd == "for") {
cin >> n;
s.push(n);
tmp.push(0);
}
if (cmd == "add") {
if (s.size() > 0) {
long long dm = tmp.top();
tmp.pop();
dm++;
tmp.push(dm);
} else {
x++;
}
}
if (cmd == "end") {
if (s.size() > 1) {
long long dm1 = tmp.top() * s.top();
tmp.pop();
s.pop();
long long dm2 = tmp.top();
tmp.pop();
dm2 = dm2 + dm1;
tmp.push(dm2);
if (tmp.top() * s.top() > (1LL << 32) - 1) {
kt = 1;
break;
}
} else {
x = x + tmp.top() * s.top();
s.pop();
tmp.pop();
}
}
if (x > (1LL << 32) - 1 || x < 0) {
kt = 1;
break;
}
}
if (kt == 0) {
cout << x;
} else {
cout << "OVERFLOW!!!";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a[100005], cor[100005];
int main() {
int n, t = 0;
long long int num = (1ll << 32) - 1;
cin >> n;
string st;
stack<long long int> s;
long long int x = 0, ct = 1;
s.push(ct);
while (n--) {
cin >> st;
if (st == "for") {
long long int e;
cin >> e;
if (s.top() > num) e = 1;
s.push(e * s.top());
} else if (st == "end") {
s.pop();
} else if (st == "add")
x += s.top();
if (x > num) {
cout << "OVERFLOW!!!";
return 0;
}
}
cout << x;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[100005], cor[100005];
int main() {
int n, t = 0;
long long int num = (1ll << 32) - 1;
cin >> n;
string st;
stack<long long int> s;
long long int x = 0, ct = 1;
s.push(ct);
while (n--) {
cin >> st;
if (st == "for") {
long long int e;
cin >> e;
if (s.top() > num) e = 1;
s.push(e * s.top());
} else if (st == "end") {
s.pop();
} else if (st == "add")
x += s.top();
if (x > num) {
cout << "OVERFLOW!!!";
return 0;
}
}
cout << x;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long k = 1;
long long subTotal[100005];
int main() {
deque<string> str;
string s, Sint = "";
long long total = 0, ch, ch2;
int n, m = 0, st;
cin >> n;
for (int i = 0; i < n + 1; i++) {
getline(cin, s);
str.push_back(s);
}
for (int i = 0; i < n + 1; i++) {
s = str.front();
str.pop_front();
if (s == "add" && m == 0) {
total++;
} else if (s[0] == 'f') {
m++;
for (int j = 4; j < s.size(); j++) {
Sint += s[j];
}
istringstream(Sint) >> st;
Sint = "";
if (k < 10000000000) {
subTotal[m] += k * st;
k = subTotal[m];
} else
subTotal[m] = k;
} else if (s == "add" && m > 0) {
total += subTotal[m];
} else if (s == "end") {
subTotal[m] = 0;
m--;
k = subTotal[m];
}
if (m == 0) {
k = 1;
Sint = "";
}
if (total > 100000000000) break;
}
if (total > pow(2, 32) - 1 || m > 1) {
cout << "OVERFLOW!!!" << endl;
} else
cout << total << endl;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long k = 1;
long long subTotal[100005];
int main() {
deque<string> str;
string s, Sint = "";
long long total = 0, ch, ch2;
int n, m = 0, st;
cin >> n;
for (int i = 0; i < n + 1; i++) {
getline(cin, s);
str.push_back(s);
}
for (int i = 0; i < n + 1; i++) {
s = str.front();
str.pop_front();
if (s == "add" && m == 0) {
total++;
} else if (s[0] == 'f') {
m++;
for (int j = 4; j < s.size(); j++) {
Sint += s[j];
}
istringstream(Sint) >> st;
Sint = "";
if (k < 10000000000) {
subTotal[m] += k * st;
k = subTotal[m];
} else
subTotal[m] = k;
} else if (s == "add" && m > 0) {
total += subTotal[m];
} else if (s == "end") {
subTotal[m] = 0;
m--;
k = subTotal[m];
}
if (m == 0) {
k = 1;
Sint = "";
}
if (total > 100000000000) break;
}
if (total > pow(2, 32) - 1 || m > 1) {
cout << "OVERFLOW!!!" << endl;
} else
cout << total << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, i;
long long maxim, val;
char a[100005][15];
long long solve() {
long long val = 0;
int x = 0, j = 4;
while (a[i][j] >= '0' && a[i][j] <= '9') {
x = x * 10 + a[i][j] - '0';
j++;
}
i++;
while (a[i][0] != 'e') {
if (a[i][0] == 'a') {
if (val != -1) {
val++;
}
i++;
} else {
long long val2 = solve();
if (val2 == -1) {
val = -1;
} else {
if (val != -1) {
val += val2;
}
}
}
if (val > maxim) {
val = -1;
}
}
i++;
if (val != -1) {
val *= x;
if (val > maxim) {
val = -1;
}
}
return val;
}
int main() {
cin >> n;
n++;
a[1][0] = 'f';
a[1][1] = 'o';
a[1][2] = 'r';
a[1][3] = ' ';
a[1][4] = '1';
for (i = 2; i <= n; i++) {
cin.get();
cin.get(a[i], 12);
}
n++;
a[n][0] = 'e';
a[n][1] = 'n';
a[n][2] = 'd';
i = 1;
maxim = (1LL << 32) - 1;
val = solve();
if (val == -1) {
cout << "OVERFLOW!!!";
} else {
cout << val;
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, i;
long long maxim, val;
char a[100005][15];
long long solve() {
long long val = 0;
int x = 0, j = 4;
while (a[i][j] >= '0' && a[i][j] <= '9') {
x = x * 10 + a[i][j] - '0';
j++;
}
i++;
while (a[i][0] != 'e') {
if (a[i][0] == 'a') {
if (val != -1) {
val++;
}
i++;
} else {
long long val2 = solve();
if (val2 == -1) {
val = -1;
} else {
if (val != -1) {
val += val2;
}
}
}
if (val > maxim) {
val = -1;
}
}
i++;
if (val != -1) {
val *= x;
if (val > maxim) {
val = -1;
}
}
return val;
}
int main() {
cin >> n;
n++;
a[1][0] = 'f';
a[1][1] = 'o';
a[1][2] = 'r';
a[1][3] = ' ';
a[1][4] = '1';
for (i = 2; i <= n; i++) {
cin.get();
cin.get(a[i], 12);
}
n++;
a[n][0] = 'e';
a[n][1] = 'n';
a[n][2] = 'd';
i = 1;
maxim = (1LL << 32) - 1;
val = solve();
if (val == -1) {
cout << "OVERFLOW!!!";
} else {
cout << val;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int q;
cin >> q;
stack<long long int> s;
s.push(1);
long long int w = pow(2, 32), sum = 0;
for (int i = 0; i < q; i++) {
string t;
cin >> t;
if (t == "for") {
long long int n;
cin >> n;
s.push(min(w, n * s.top()));
} else if (t == "end")
s.pop();
else {
sum += s.top();
}
}
if (sum < w)
cout << sum;
else
cout << "OVERFLOW!!!";
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int q;
cin >> q;
stack<long long int> s;
s.push(1);
long long int w = pow(2, 32), sum = 0;
for (int i = 0; i < q; i++) {
string t;
cin >> t;
if (t == "for") {
long long int n;
cin >> n;
s.push(min(w, n * s.top()));
} else if (t == "end")
s.pop();
else {
sum += s.top();
}
}
if (sum < w)
cout << sum;
else
cout << "OVERFLOW!!!";
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int md = 3e5 + 100;
vector<int> st;
vector<int> prp;
int main() {
int l;
cin >> l;
long long cur = 0;
while (l--) {
string ord;
cin >> ord;
if (ord == "for") {
long long x;
cin >> x;
prp.push_back(x);
if (x > 1) {
st.push_back(x);
}
} else if (ord == "end") {
long long x = prp[prp.size() - 1];
prp.pop_back();
if (x > 1) st.pop_back();
} else {
if ((int)st.size() > 32) {
puts("OVERFLOW!!!");
return 0;
} else {
long long prev = 1;
for (auto p : st) {
prev = prev * p;
if (prev >= (1ll << 32)) {
puts("OVERFLOW!!!");
return 0;
}
}
cur += prev;
}
}
if (cur >= (1ll << 32)) {
puts("OVERFLOW!!!");
return 0;
}
}
cout << cur;
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int md = 3e5 + 100;
vector<int> st;
vector<int> prp;
int main() {
int l;
cin >> l;
long long cur = 0;
while (l--) {
string ord;
cin >> ord;
if (ord == "for") {
long long x;
cin >> x;
prp.push_back(x);
if (x > 1) {
st.push_back(x);
}
} else if (ord == "end") {
long long x = prp[prp.size() - 1];
prp.pop_back();
if (x > 1) st.pop_back();
} else {
if ((int)st.size() > 32) {
puts("OVERFLOW!!!");
return 0;
} else {
long long prev = 1;
for (auto p : st) {
prev = prev * p;
if (prev >= (1ll << 32)) {
puts("OVERFLOW!!!");
return 0;
}
}
cur += prev;
}
}
if (cur >= (1ll << 32)) {
puts("OVERFLOW!!!");
return 0;
}
}
cout << cur;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long int m = (1LL << 32);
int main() {
int n;
cin >> n;
string s;
long long int x;
long long int ans = 0;
stack<long long int> st;
st.push(1);
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "for") {
cin >> x;
long long int t = st.top();
st.push(min(t * x, m));
} else if (s == "add") {
long long int t = st.top();
ans += t;
} else if (s == "end") {
st.pop();
}
}
if (ans >= m)
cout << "OVERFLOW!!!";
else
cout << ans;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int m = (1LL << 32);
int main() {
int n;
cin >> n;
string s;
long long int x;
long long int ans = 0;
stack<long long int> st;
st.push(1);
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "for") {
cin >> x;
long long int t = st.top();
st.push(min(t * x, m));
} else if (s == "add") {
long long int t = st.top();
ans += t;
} else if (s == "end") {
st.pop();
}
}
if (ans >= m)
cout << "OVERFLOW!!!";
else
cout << ans;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const double EPS = 1e-9;
const long long INF = 1e10;
long long addMod(long long a, long long b) {
if (a + b >= MOD) return a + b - MOD;
return a + b;
}
long long mulMod(long long a, long long b) { return (a * b) % MOD; }
long long fast(long long a, long long b) {
if (b == 0) return 1;
long long tmp = fast(a, b / 2);
tmp = mulMod(tmp, tmp);
if (b & 1) tmp = mulMod(tmp, a);
return tmp;
}
long long GCD(long long a, long long b) {
if (a > b) return GCD(b, a);
if (a == 0) return b;
return GCD(b % a, a);
}
stack<long long> lst;
int n;
string str;
int main() {
ios_base ::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long MX = 1;
for (int i = 0; i <= (int)32 - 1; i += 1) MX *= 2;
MX--;
cin >> n;
long long tot = 0, mul = 1, x, cnt = 0;
for (int i = 0; i <= (int)n - 1; i += 1) {
cin >> str;
tot = min(tot, INF);
if (str == "for") {
cin >> x;
if (mul <= INF) {
mul *= x;
lst.push(x);
} else
cnt++;
} else if (str == "end") {
if (cnt > 0) {
cnt--;
continue;
}
if (!lst.empty()) {
mul /= lst.top();
lst.pop();
}
} else {
tot += mul;
}
}
if (tot > MX)
cout << "OVERFLOW!!!" << endl;
else
cout << tot << endl;
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const double EPS = 1e-9;
const long long INF = 1e10;
long long addMod(long long a, long long b) {
if (a + b >= MOD) return a + b - MOD;
return a + b;
}
long long mulMod(long long a, long long b) { return (a * b) % MOD; }
long long fast(long long a, long long b) {
if (b == 0) return 1;
long long tmp = fast(a, b / 2);
tmp = mulMod(tmp, tmp);
if (b & 1) tmp = mulMod(tmp, a);
return tmp;
}
long long GCD(long long a, long long b) {
if (a > b) return GCD(b, a);
if (a == 0) return b;
return GCD(b % a, a);
}
stack<long long> lst;
int n;
string str;
int main() {
ios_base ::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long MX = 1;
for (int i = 0; i <= (int)32 - 1; i += 1) MX *= 2;
MX--;
cin >> n;
long long tot = 0, mul = 1, x, cnt = 0;
for (int i = 0; i <= (int)n - 1; i += 1) {
cin >> str;
tot = min(tot, INF);
if (str == "for") {
cin >> x;
if (mul <= INF) {
mul *= x;
lst.push(x);
} else
cnt++;
} else if (str == "end") {
if (cnt > 0) {
cnt--;
continue;
}
if (!lst.empty()) {
mul /= lst.top();
lst.pop();
}
} else {
tot += mul;
}
}
if (tot > MX)
cout << "OVERFLOW!!!" << endl;
else
cout << tot << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long N = (1ll << 32ll) - 1ll, M = 31;
long long l, r, i, j, t, n, m, k, f, ma = 0, d[410000], x, z, y, x2, y2, y3,
a[401000], b[401000], ans;
string s[300000];
int main() {
ios_base::sync_with_stdio(0);
cin >> n;
for (i = 1; i <= n; i++) {
cin >> s[i];
if (s[i] == "for") {
cin >> a[i];
b[i] = 1;
} else if (s[i] == "end") {
b[i] = 2;
} else
b[i] = 3;
}
t = 1;
l = 0;
for (i = 1; i <= n; i++) {
if (b[i] == 1) {
l++;
d[l] = a[i];
t *= a[i];
if (t > N) {
r = l;
for (j = i + 1; j <= n; j++) {
if (b[j] == 1) r++;
if (b[j] == 2) r--;
if (b[j] == 3) {
cout << "OVERFLOW!!!";
return 0;
}
if (r == l - 1) {
i = j - 1;
break;
}
}
}
} else if (b[i] == 2) {
l--;
t /= d[l + 1];
} else if (b[i] == 3)
ans += t;
}
if (ans > N)
cout << "OVERFLOW!!!";
else
cout << ans;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long N = (1ll << 32ll) - 1ll, M = 31;
long long l, r, i, j, t, n, m, k, f, ma = 0, d[410000], x, z, y, x2, y2, y3,
a[401000], b[401000], ans;
string s[300000];
int main() {
ios_base::sync_with_stdio(0);
cin >> n;
for (i = 1; i <= n; i++) {
cin >> s[i];
if (s[i] == "for") {
cin >> a[i];
b[i] = 1;
} else if (s[i] == "end") {
b[i] = 2;
} else
b[i] = 3;
}
t = 1;
l = 0;
for (i = 1; i <= n; i++) {
if (b[i] == 1) {
l++;
d[l] = a[i];
t *= a[i];
if (t > N) {
r = l;
for (j = i + 1; j <= n; j++) {
if (b[j] == 1) r++;
if (b[j] == 2) r--;
if (b[j] == 3) {
cout << "OVERFLOW!!!";
return 0;
}
if (r == l - 1) {
i = j - 1;
break;
}
}
}
} else if (b[i] == 2) {
l--;
t /= d[l + 1];
} else if (b[i] == 3)
ans += t;
}
if (ans > N)
cout << "OVERFLOW!!!";
else
cout << ans;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MAX = 4294967295;
int main() {
int l;
string in;
long long val = 0;
cin >> l;
long long x = 0;
stack<long long> fors;
for (int i = 0; i < l; i++) {
cin >> in;
if (in == "for") {
cin >> val;
long long curr = val;
if (!fors.empty()) {
if (fors.top() == -1) {
curr = -1;
} else {
curr *= fors.top();
}
}
if (curr > MAX) {
curr = -1;
}
fors.push(curr);
}
if (in == "end") {
fors.pop();
}
if (in == "add") {
if (fors.empty()) {
x++;
} else {
if (fors.top() == -1) {
cout << "OVERFLOW!!!";
return 0;
}
x += fors.top();
}
if (x > MAX) {
cout << "OVERFLOW!!!";
return 0;
}
}
}
cout << x;
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MAX = 4294967295;
int main() {
int l;
string in;
long long val = 0;
cin >> l;
long long x = 0;
stack<long long> fors;
for (int i = 0; i < l; i++) {
cin >> in;
if (in == "for") {
cin >> val;
long long curr = val;
if (!fors.empty()) {
if (fors.top() == -1) {
curr = -1;
} else {
curr *= fors.top();
}
}
if (curr > MAX) {
curr = -1;
}
fors.push(curr);
}
if (in == "end") {
fors.pop();
}
if (in == "add") {
if (fors.empty()) {
x++;
} else {
if (fors.top() == -1) {
cout << "OVERFLOW!!!";
return 0;
}
x += fors.top();
}
if (x > MAX) {
cout << "OVERFLOW!!!";
return 0;
}
}
}
cout << x;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int n;
cin >> n;
string str;
stack<long long int> s;
s.push(1);
long long int ans = 0;
long long int value = (long long int)(pow(2, 32));
while (n--) {
cin >> str;
if (str == "for") {
int x;
cin >> x;
s.push(min(value, s.top() * x));
} else if (str == "add") {
ans += s.top();
ans = min(ans, value);
} else
s.pop();
}
if (ans == value)
cout << "OVERFLOW!!!" << '\n';
else
cout << ans << '\n';
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int n;
cin >> n;
string str;
stack<long long int> s;
s.push(1);
long long int ans = 0;
long long int value = (long long int)(pow(2, 32));
while (n--) {
cin >> str;
if (str == "for") {
int x;
cin >> x;
s.push(min(value, s.top() * x));
} else if (str == "add") {
ans += s.top();
ans = min(ans, value);
} else
s.pop();
}
if (ans == value)
cout << "OVERFLOW!!!" << '\n';
else
cout << ans << '\n';
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = (int)2750131;
const int MOD = 1e9 + 7;
const int oo = (int)1e9;
const long long OO = (long long)1e18;
const double PI = acos(-1.0);
bool isprime(int n) {
if (n <= 1) return 0;
for (int i = 2; i * i <= n; i++)
if (n % i == 0) return false;
return 1;
}
long long intlog(long long base, long long x) { return (log(x) / log(base)); }
bool prime[N];
vector<int> next_primee;
void primee() {
memset(prime, 1, sizeof(prime));
for (int p = 2; p * p <= N; p++) {
if (prime[p] == true) {
for (int i = p * 2; i < N; i += p) prime[i] = false;
}
}
}
string decToBinary(long long n) {
string z = "";
int i = 0;
while (n > 0) {
z += (n % 2) + 48;
n = n / 2;
i++;
}
reverse(((z).begin()), ((z).end()));
return z;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long power(long long x, int y) {
long long temp;
if (y == 0) return 1;
temp = power(x, y / 2);
if (y % 2 == 0)
return temp * temp;
else
return x * temp * temp;
}
int get(int num) {
for (int i = 2; i * i <= num; i++) {
if (num % i == 0) return num / i;
}
return 0;
}
int main() {
int l, close = 0;
long long x = 0, val = 1;
bool overflow = 0;
vector<long long> v;
cin >> l;
for (int i = 0; i < l + 1; i++) {
string s;
getline(cin, s);
if (s[0] == 'f') {
if (close)
close++;
else if (val * stoll(s.substr(4, s.size() - 4)) >= (1ll << 32))
close++;
else
val *= stoll(s.substr(4, s.size() - 4)),
v.push_back(stoll(s.substr(4, s.size() - 4)));
} else if (s[0] == 'e') {
if (close)
close--;
else {
val /= v.back();
v.pop_back();
}
} else if (s[0] == 'a') {
x += val;
if ((x >= (1ll << 32)) || (close)) overflow = 1;
}
}
if (overflow)
puts("OVERFLOW!!!");
else
printf("%lld\n", x);
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = (int)2750131;
const int MOD = 1e9 + 7;
const int oo = (int)1e9;
const long long OO = (long long)1e18;
const double PI = acos(-1.0);
bool isprime(int n) {
if (n <= 1) return 0;
for (int i = 2; i * i <= n; i++)
if (n % i == 0) return false;
return 1;
}
long long intlog(long long base, long long x) { return (log(x) / log(base)); }
bool prime[N];
vector<int> next_primee;
void primee() {
memset(prime, 1, sizeof(prime));
for (int p = 2; p * p <= N; p++) {
if (prime[p] == true) {
for (int i = p * 2; i < N; i += p) prime[i] = false;
}
}
}
string decToBinary(long long n) {
string z = "";
int i = 0;
while (n > 0) {
z += (n % 2) + 48;
n = n / 2;
i++;
}
reverse(((z).begin()), ((z).end()));
return z;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long power(long long x, int y) {
long long temp;
if (y == 0) return 1;
temp = power(x, y / 2);
if (y % 2 == 0)
return temp * temp;
else
return x * temp * temp;
}
int get(int num) {
for (int i = 2; i * i <= num; i++) {
if (num % i == 0) return num / i;
}
return 0;
}
int main() {
int l, close = 0;
long long x = 0, val = 1;
bool overflow = 0;
vector<long long> v;
cin >> l;
for (int i = 0; i < l + 1; i++) {
string s;
getline(cin, s);
if (s[0] == 'f') {
if (close)
close++;
else if (val * stoll(s.substr(4, s.size() - 4)) >= (1ll << 32))
close++;
else
val *= stoll(s.substr(4, s.size() - 4)),
v.push_back(stoll(s.substr(4, s.size() - 4)));
} else if (s[0] == 'e') {
if (close)
close--;
else {
val /= v.back();
v.pop_back();
}
} else if (s[0] == 'a') {
x += val;
if ((x >= (1ll << 32)) || (close)) overflow = 1;
}
}
if (overflow)
puts("OVERFLOW!!!");
else
printf("%lld\n", x);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vll = vector<ll>;
const ll INF = 1LL << 32;
int main() {
cin.tie(NULL);
ios_base::sync_with_stdio(false);
ll res = 0;
int add_cnt = 0;
vll v;
int n;
cin >> n;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s == "add") {
if (v.empty())
res++;
else {
if (v.back() == 0) {
cout << "OVERFLOW!!!\n";
return 0;
}
res += v.back();
}
} else if (s == "end") {
v.pop_back();
} else {
ll x;
cin >> x;
if (v.empty())
v.push_back(x);
else {
v.push_back(v.back() * x);
if (v.back() >= INF) v.back() = 0;
}
}
}
if (res >= INF) {
cout << "OVERFLOW!!!\n";
return 0;
}
cout << res << '\n';
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vll = vector<ll>;
const ll INF = 1LL << 32;
int main() {
cin.tie(NULL);
ios_base::sync_with_stdio(false);
ll res = 0;
int add_cnt = 0;
vll v;
int n;
cin >> n;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s == "add") {
if (v.empty())
res++;
else {
if (v.back() == 0) {
cout << "OVERFLOW!!!\n";
return 0;
}
res += v.back();
}
} else if (s == "end") {
v.pop_back();
} else {
ll x;
cin >> x;
if (v.empty())
v.push_back(x);
else {
v.push_back(v.back() * x);
if (v.back() >= INF) v.back() = 0;
}
}
}
if (res >= INF) {
cout << "OVERFLOW!!!\n";
return 0;
}
cout << res << '\n';
}
``` |
#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXQ = 100000;
const long long INF = 1LL << 32;
int nq;
char qkind[MAXQ][10];
int qpar[MAXQ];
long long solve() {
long long ret = 0;
vector<long long> stck;
stck.push_back(1LL);
for (int i = (0); i < (nq); ++i) {
if (strcmp(qkind[i], "add") == 0) {
ret += stck.back();
if (ret >= INF) return -1;
}
if (strcmp(qkind[i], "for") == 0) {
long long nval = min(INF, stck.back() * qpar[i]);
stck.push_back(nval);
}
if (strcmp(qkind[i], "end") == 0) {
stck.pop_back();
}
}
return ret;
}
void run() {
scanf("%d", &nq);
for (int i = (0); i < (nq); ++i) {
scanf("%s", qkind[i]);
if (strcmp(qkind[i], "for") == 0) scanf("%d", &qpar[i]);
}
long long ans = solve();
if (ans == -1)
printf("OVERFLOW!!!\n");
else
printf("%lld\n", ans);
}
int main() {
run();
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXQ = 100000;
const long long INF = 1LL << 32;
int nq;
char qkind[MAXQ][10];
int qpar[MAXQ];
long long solve() {
long long ret = 0;
vector<long long> stck;
stck.push_back(1LL);
for (int i = (0); i < (nq); ++i) {
if (strcmp(qkind[i], "add") == 0) {
ret += stck.back();
if (ret >= INF) return -1;
}
if (strcmp(qkind[i], "for") == 0) {
long long nval = min(INF, stck.back() * qpar[i]);
stck.push_back(nval);
}
if (strcmp(qkind[i], "end") == 0) {
stck.pop_back();
}
}
return ret;
}
void run() {
scanf("%d", &nq);
for (int i = (0); i < (nq); ++i) {
scanf("%s", qkind[i]);
if (strcmp(qkind[i], "for") == 0) scanf("%d", &qpar[i]);
}
long long ans = solve();
if (ans == -1)
printf("OVERFLOW!!!\n");
else
printf("%lld\n", ans);
}
int main() {
run();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1ll << 32;
int main() {
int l;
cin >> l;
stack<long long> cur;
cur.push(1);
long long res = 0;
for (int _ = 0; _ < int(l); _++) {
string t;
cin >> t;
if (t == "for") {
int x;
cin >> x;
cur.push(min(INF, cur.top() * x));
} else if (t == "end") {
cur.pop();
} else {
res += cur.top();
}
}
if (res >= INF)
cout << "OVERFLOW!!!" << endl;
else
cout << res << endl;
}
| ### Prompt
Develop a solution in cpp to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1ll << 32;
int main() {
int l;
cin >> l;
stack<long long> cur;
cur.push(1);
long long res = 0;
for (int _ = 0; _ < int(l); _++) {
string t;
cin >> t;
if (t == "for") {
int x;
cin >> x;
cur.push(min(INF, cur.top() * x));
} else if (t == "end") {
cur.pop();
} else {
res += cur.top();
}
}
if (res >= INF)
cout << "OVERFLOW!!!" << endl;
else
cout << res << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d", &t);
stack<pair<long long int, long long int> > s;
long long int prod = 1;
long long int totalAdds = 0;
int flag = 0;
int overflow = 0;
for (int i = 0; i < t; i++) {
string str;
cin >> str;
if (flag == 1) continue;
if (str == "add") {
if (overflow || totalAdds + prod >= (1LL << 32)) {
flag = 1;
} else {
totalAdds += prod;
}
continue;
}
if (str == "for") {
long long int n;
scanf("%lld", &n);
if (!overflow && ((1LL << 32) - 1LL) / n >= prod) {
prod = prod * n;
} else {
overflow = 1;
}
s.push(make_pair(n, overflow));
continue;
}
if (str == "end") {
if (s.top().second == 0) {
prod = prod / s.top().first;
s.pop();
} else {
s.pop();
if (s.empty()) {
overflow = 0;
} else {
overflow = s.top().second;
}
}
}
}
if (flag || totalAdds >= (1LL << 32)) {
printf("OVERFLOW!!!\n");
} else {
printf("%lld\n", totalAdds);
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d", &t);
stack<pair<long long int, long long int> > s;
long long int prod = 1;
long long int totalAdds = 0;
int flag = 0;
int overflow = 0;
for (int i = 0; i < t; i++) {
string str;
cin >> str;
if (flag == 1) continue;
if (str == "add") {
if (overflow || totalAdds + prod >= (1LL << 32)) {
flag = 1;
} else {
totalAdds += prod;
}
continue;
}
if (str == "for") {
long long int n;
scanf("%lld", &n);
if (!overflow && ((1LL << 32) - 1LL) / n >= prod) {
prod = prod * n;
} else {
overflow = 1;
}
s.push(make_pair(n, overflow));
continue;
}
if (str == "end") {
if (s.top().second == 0) {
prod = prod / s.top().first;
s.pop();
} else {
s.pop();
if (s.empty()) {
overflow = 0;
} else {
overflow = s.top().second;
}
}
}
}
if (flag || totalAdds >= (1LL << 32)) {
printf("OVERFLOW!!!\n");
} else {
printf("%lld\n", totalAdds);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
const int mod = 1e9 + 7;
const ll inf = ll(1e18);
const double PI = acos(-1), eps = 1e-9;
const int mxn = 2e5, mxa = 1e5;
const ll mx = (1ll << 32) - 1;
int l;
string second;
void solve() {
cin >> l;
stack<ll> stk;
stk.push(1);
ll res = 0;
while (l--) {
cin >> second;
if (second == "add") {
if (res + stk.top() > mx) {
cout << "OVERFLOW!!!\n";
return;
}
res += stk.top();
} else if (second == "for") {
int n;
cin >> n;
stk.push(min(mx + 1, stk.top() * n));
} else
stk.pop();
}
cout << res << '\n';
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
t = 1;
while (t-- > 0) solve();
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
const int mod = 1e9 + 7;
const ll inf = ll(1e18);
const double PI = acos(-1), eps = 1e-9;
const int mxn = 2e5, mxa = 1e5;
const ll mx = (1ll << 32) - 1;
int l;
string second;
void solve() {
cin >> l;
stack<ll> stk;
stk.push(1);
ll res = 0;
while (l--) {
cin >> second;
if (second == "add") {
if (res + stk.top() > mx) {
cout << "OVERFLOW!!!\n";
return;
}
res += stk.top();
} else if (second == "for") {
int n;
cin >> n;
stk.push(min(mx + 1, stk.top() * n));
} else
stk.pop();
}
cout << res << '\n';
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int t;
t = 1;
while (t-- > 0) solve();
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int max = 4294967296, x = 0;
stack<long long int> s;
s.push(1);
int l, n = 0;
char command[4];
scanf("%d\n", &l);
while (l--) {
scanf("%s", command);
if (command[0] == 'f') {
scanf("%d", &n);
if (s.top() * n < max)
s.push(n * s.top());
else
s.push(max);
} else if (command[0] == 'a') {
x += s.top();
if (x >= max) break;
} else if (command[0] == 'e') {
s.pop();
}
}
if (x < max)
printf("%lld\n", x);
else
printf("OVERFLOW!!!\n");
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int max = 4294967296, x = 0;
stack<long long int> s;
s.push(1);
int l, n = 0;
char command[4];
scanf("%d\n", &l);
while (l--) {
scanf("%s", command);
if (command[0] == 'f') {
scanf("%d", &n);
if (s.top() * n < max)
s.push(n * s.top());
else
s.push(max);
} else if (command[0] == 'a') {
x += s.top();
if (x >= max) break;
} else if (command[0] == 'e') {
s.pop();
}
}
if (x < max)
printf("%lld\n", x);
else
printf("OVERFLOW!!!\n");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
long long maxim = (1LL << 32);
stack<long long> st;
long long prod = 1;
long long sum = 0;
bool a = false;
bool b = false;
int ct = 0;
for (int i = int(1); i <= int(n); i++) {
string s;
if (i == 1) cin.ignore();
getline(cin, s);
if (s[0] == 'f') {
long long val = 0;
long long u = 1;
for (int j = s.length() - 1; s[j] != ' '; j--) {
val += (s[j] - '0') * u;
u *= 10;
}
if (prod * val >= maxim || ct)
ct++;
else
prod *= val;
st.push(val);
} else if (s[0] == 'a') {
if (a)
sum = maxim;
else if (ct) {
sum = maxim;
a = true;
} else {
sum += prod;
if (sum >= maxim) a = true;
}
} else {
if (!ct)
prod /= (st.top());
else
ct--;
st.pop();
}
}
if (a)
cout << "OVERFLOW!!!";
else
cout << sum;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
long long maxim = (1LL << 32);
stack<long long> st;
long long prod = 1;
long long sum = 0;
bool a = false;
bool b = false;
int ct = 0;
for (int i = int(1); i <= int(n); i++) {
string s;
if (i == 1) cin.ignore();
getline(cin, s);
if (s[0] == 'f') {
long long val = 0;
long long u = 1;
for (int j = s.length() - 1; s[j] != ' '; j--) {
val += (s[j] - '0') * u;
u *= 10;
}
if (prod * val >= maxim || ct)
ct++;
else
prod *= val;
st.push(val);
} else if (s[0] == 'a') {
if (a)
sum = maxim;
else if (ct) {
sum = maxim;
a = true;
} else {
sum += prod;
if (sum >= maxim) a = true;
}
} else {
if (!ct)
prod /= (st.top());
else
ct--;
st.pop();
}
}
if (a)
cout << "OVERFLOW!!!";
else
cout << sum;
}
``` |
#include <bits/stdc++.h>
using namespace std;
unsigned long long use[105];
unsigned long long fastExpo(unsigned long long num, unsigned long long expo) {
if (expo == 0) return 1;
if (expo == 1) return num;
unsigned long long tmp = fastExpo(num, expo / 2);
if (tmp > 4294967295) return 0;
tmp *= tmp;
if (tmp > 4294967295) return 0;
if (expo % 2 == 1) tmp *= num;
if (tmp > 4294967295) return 0;
return tmp;
}
int main() {
std::ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int t;
stack<pair<string, unsigned long long>> prev;
unsigned long long add = 0;
unsigned long long x = 0;
cin >> t;
for (int i = 1; i <= t; i++) {
string cmd;
cin >> cmd;
if (cmd == "add") {
unsigned long long tmp = 1;
for (int j = 1; j <= 100; j++) {
tmp *= fastExpo(j, use[j]);
if (tmp == 0 || tmp > 4294967295) {
cout << "OVERFLOW!!!\n";
return 0;
}
}
x += tmp;
if (x > 4294967295) {
cout << "OVERFLOW!!!\n";
return 0;
}
} else if (cmd == "for") {
unsigned long long tmp;
cin >> tmp;
prev.push({cmd, tmp});
use[tmp]++;
} else {
while (prev.top().first != "for") {
prev.pop();
}
use[prev.top().second]--;
prev.pop();
}
}
cout << x << '\n';
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
unsigned long long use[105];
unsigned long long fastExpo(unsigned long long num, unsigned long long expo) {
if (expo == 0) return 1;
if (expo == 1) return num;
unsigned long long tmp = fastExpo(num, expo / 2);
if (tmp > 4294967295) return 0;
tmp *= tmp;
if (tmp > 4294967295) return 0;
if (expo % 2 == 1) tmp *= num;
if (tmp > 4294967295) return 0;
return tmp;
}
int main() {
std::ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int t;
stack<pair<string, unsigned long long>> prev;
unsigned long long add = 0;
unsigned long long x = 0;
cin >> t;
for (int i = 1; i <= t; i++) {
string cmd;
cin >> cmd;
if (cmd == "add") {
unsigned long long tmp = 1;
for (int j = 1; j <= 100; j++) {
tmp *= fastExpo(j, use[j]);
if (tmp == 0 || tmp > 4294967295) {
cout << "OVERFLOW!!!\n";
return 0;
}
}
x += tmp;
if (x > 4294967295) {
cout << "OVERFLOW!!!\n";
return 0;
}
} else if (cmd == "for") {
unsigned long long tmp;
cin >> tmp;
prev.push({cmd, tmp});
use[tmp]++;
} else {
while (prev.top().first != "for") {
prev.pop();
}
use[prev.top().second]--;
prev.pop();
}
}
cout << x << '\n';
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
stack<long long int> s;
long long int num(0);
long long int n;
cin >> n;
long long int lasttime(1);
long long int maxi = (1LL << 32);
while (n--) {
string in;
cin >> in;
if (in == "add" && s.empty())
num++;
else if (in == "add" && !s.empty()) {
num = num + s.top();
} else if (in == "for") {
long long int times;
cin >> times;
times = min(maxi, times * lasttime);
s.push(times);
lasttime = times;
} else {
s.pop();
lasttime = s.empty() ? 1 : s.top();
}
}
num >= maxi ? cout << "OVERFLOW!!!" : cout << num;
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
stack<long long int> s;
long long int num(0);
long long int n;
cin >> n;
long long int lasttime(1);
long long int maxi = (1LL << 32);
while (n--) {
string in;
cin >> in;
if (in == "add" && s.empty())
num++;
else if (in == "add" && !s.empty()) {
num = num + s.top();
} else if (in == "for") {
long long int times;
cin >> times;
times = min(maxi, times * lasttime);
s.push(times);
lasttime = times;
} else {
s.pop();
lasttime = s.empty() ? 1 : s.top();
}
}
num >= maxi ? cout << "OVERFLOW!!!" : cout << num;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long N = 4294967295;
stack<int> S;
char a[10];
int t;
long long sum, n, loop = 1;
bool ok = true;
int main() {
cin >> t;
while (t--) {
cin >> a;
if (a[0] == 'a') {
sum += loop;
if (sum > N) ok = false;
} else if (a[0] == 'f') {
cin >> n;
if (loop > N)
S.push(1);
else {
loop *= n;
S.push(n);
}
} else {
if (!S.empty()) {
loop /= S.top();
S.pop();
}
}
}
if (ok)
cout << sum << endl;
else
cout << "OVERFLOW!!!" << endl;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long N = 4294967295;
stack<int> S;
char a[10];
int t;
long long sum, n, loop = 1;
bool ok = true;
int main() {
cin >> t;
while (t--) {
cin >> a;
if (a[0] == 'a') {
sum += loop;
if (sum > N) ok = false;
} else if (a[0] == 'f') {
cin >> n;
if (loop > N)
S.push(1);
else {
loop *= n;
S.push(n);
}
} else {
if (!S.empty()) {
loop /= S.top();
S.pop();
}
}
}
if (ok)
cout << sum << endl;
else
cout << "OVERFLOW!!!" << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
long long T = 1, n, m, i, j, k;
while (T-- > 0) {
cin >> m;
long long stk[m + 1];
string second;
long long ans = 0;
k = 1;
stk[0] = 1;
for (i = 0; i < m; i++) {
cin >> second;
if (second == "for") {
cin >> n;
stk[k] = stk[k - 1] * n;
if (stk[k] > UINT32_MAX or stk[k] < 0) stk[k] = -1;
k++;
} else if (second == "end") {
k--;
} else {
if (stk[k - 1] < 0 or ans + stk[k - 1] > UINT_MAX) {
cout << "OVERFLOW!!!";
return 0;
}
ans += stk[k - 1];
}
}
cout << ans;
}
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
long long T = 1, n, m, i, j, k;
while (T-- > 0) {
cin >> m;
long long stk[m + 1];
string second;
long long ans = 0;
k = 1;
stk[0] = 1;
for (i = 0; i < m; i++) {
cin >> second;
if (second == "for") {
cin >> n;
stk[k] = stk[k - 1] * n;
if (stk[k] > UINT32_MAX or stk[k] < 0) stk[k] = -1;
k++;
} else if (second == "end") {
k--;
} else {
if (stk[k - 1] < 0 or ans + stk[k - 1] > UINT_MAX) {
cout << "OVERFLOW!!!";
return 0;
}
ans += stk[k - 1];
}
}
cout << ans;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const int inf = 2e9;
const long double eps = 1e-7;
const long double pi = acos(-1);
const int dx[4] = {0, 0, 1, -1};
const int dy[4] = {1, -1, 0, 0};
const int N = 1e5 + 11;
const int M = 1e3 + 11;
const int mod = 1e9 + 7;
string s[N];
long long a[N], p[N];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> s[i];
if (s[i] == "for") {
cin >> a[i];
}
}
stack<int> st;
long long ans = 0;
long long sum = 1;
int w = 0;
long long d = pow(2LL, 32) - 1;
for (int i = 1; i <= n; i++) {
if (s[i] == "add") {
if (w == 1) {
cout << "OVERFLOW!!!";
return 0;
}
ans += sum;
if (ans > d) {
cout << "OVERFLOW!!!";
return 0;
}
} else if (s[i] == "for") {
st.push(i);
if (w == 0) {
sum *= a[i];
if (sum > d) {
p[i] = 1;
w = 1;
}
}
} else {
if (w == 1) {
if (p[st.top()] == 1) sum /= a[st.top()], w = 0;
} else
sum /= a[st.top()];
st.pop();
}
}
cout << ans;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const int inf = 2e9;
const long double eps = 1e-7;
const long double pi = acos(-1);
const int dx[4] = {0, 0, 1, -1};
const int dy[4] = {1, -1, 0, 0};
const int N = 1e5 + 11;
const int M = 1e3 + 11;
const int mod = 1e9 + 7;
string s[N];
long long a[N], p[N];
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> s[i];
if (s[i] == "for") {
cin >> a[i];
}
}
stack<int> st;
long long ans = 0;
long long sum = 1;
int w = 0;
long long d = pow(2LL, 32) - 1;
for (int i = 1; i <= n; i++) {
if (s[i] == "add") {
if (w == 1) {
cout << "OVERFLOW!!!";
return 0;
}
ans += sum;
if (ans > d) {
cout << "OVERFLOW!!!";
return 0;
}
} else if (s[i] == "for") {
st.push(i);
if (w == 0) {
sum *= a[i];
if (sum > d) {
p[i] = 1;
w = 1;
}
}
} else {
if (w == 1) {
if (p[st.top()] == 1) sum /= a[st.top()], w = 0;
} else
sum /= a[st.top()];
st.pop();
}
}
cout << ans;
}
``` |
#include <bits/stdc++.h>
const int MOD = (int)1e9 + 7;
using namespace std;
stack<long long> st;
int main() {
long long n, num;
cin >> n;
long long sum = 0;
int over = 0;
st.push(1);
for (int i = 1; i <= n; i++) {
string s;
cin >> s;
if (over == 1) continue;
if (s[0] == 'a') {
sum += st.top();
if (sum > 4294967295) {
over = 1;
}
} else if (s[0] == 'f') {
cin >> num;
long long f = st.top();
if (st.top() > 4294967295)
st.push(st.top());
else
st.push(f * num);
} else {
st.pop();
}
}
if (over == 1)
cout << "OVERFLOW!!!" << endl;
else
cout << sum << endl;
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
const int MOD = (int)1e9 + 7;
using namespace std;
stack<long long> st;
int main() {
long long n, num;
cin >> n;
long long sum = 0;
int over = 0;
st.push(1);
for (int i = 1; i <= n; i++) {
string s;
cin >> s;
if (over == 1) continue;
if (s[0] == 'a') {
sum += st.top();
if (sum > 4294967295) {
over = 1;
}
} else if (s[0] == 'f') {
cin >> num;
long long f = st.top();
if (st.top() > 4294967295)
st.push(st.top());
else
st.push(f * num);
} else {
st.pop();
}
}
if (over == 1)
cout << "OVERFLOW!!!" << endl;
else
cout << sum << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
string S[100010];
int n[100010];
int L;
int now;
long long rec() {
long long res = 0;
while (now < L) {
if (res >= (1ll << 32)) {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
if (S[now] == "add") {
res++;
now++;
} else if (S[now] == "for") {
int pre = now;
now++;
long long z = rec();
res = res + z * n[pre];
} else {
now++;
return res;
}
}
if (res >= (1ll << 32)) {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
return res;
}
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
cin >> L;
for (int i = 0; i < (L); ++i) {
cin >> S[i];
if (S[i] == "for") cin >> n[i];
}
cout << rec() << endl;
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
string S[100010];
int n[100010];
int L;
int now;
long long rec() {
long long res = 0;
while (now < L) {
if (res >= (1ll << 32)) {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
if (S[now] == "add") {
res++;
now++;
} else if (S[now] == "for") {
int pre = now;
now++;
long long z = rec();
res = res + z * n[pre];
} else {
now++;
return res;
}
}
if (res >= (1ll << 32)) {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
return res;
}
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
cin >> L;
for (int i = 0; i < (L); ++i) {
cin >> S[i];
if (S[i] == "for") cin >> n[i];
}
cout << rec() << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using namespace std;
long long T, t;
void solve() {
long long int i, n, cnt = 0, a, sum2 = 0, ans = 1, x = 0, k, bound;
stack<long long int> s;
cin >> n;
ans = 1;
string st;
bound = (pow(2, 32) - 1);
for (i = 0; i < n; i++) {
cin >> st;
if (st == "for") {
cin >> a;
if (cnt > 0 || x + ans * a > bound) {
cnt++;
} else {
ans *= a;
s.push(a);
}
} else if (st == "end") {
if (cnt > 0)
cnt--;
else {
ans /= s.top();
s.pop();
}
} else {
if (x + ans > bound || cnt > 0) {
cout << "OVERFLOW!!!";
return;
}
x += ans;
}
}
if (x > bound)
cout << "OVERFLOW!!!";
else
cout << x;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
t = 1;
while (t--) solve();
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using namespace std;
long long T, t;
void solve() {
long long int i, n, cnt = 0, a, sum2 = 0, ans = 1, x = 0, k, bound;
stack<long long int> s;
cin >> n;
ans = 1;
string st;
bound = (pow(2, 32) - 1);
for (i = 0; i < n; i++) {
cin >> st;
if (st == "for") {
cin >> a;
if (cnt > 0 || x + ans * a > bound) {
cnt++;
} else {
ans *= a;
s.push(a);
}
} else if (st == "end") {
if (cnt > 0)
cnt--;
else {
ans /= s.top();
s.pop();
}
} else {
if (x + ans > bound || cnt > 0) {
cout << "OVERFLOW!!!";
return;
}
x += ans;
}
}
if (x > bound)
cout << "OVERFLOW!!!";
else
cout << x;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
t = 1;
while (t--) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long inf = 4294967296;
int main() {
int t, n, flag = 0;
scanf("%d", &t);
char ch[10];
long long loop = 1;
long long ans = 0;
stack<int> S;
while (t--) {
cin >> ch;
if (ch[0] == 'a') {
ans += loop;
if (ans >= inf) flag = 1;
} else if (ch[0] == 'f') {
cin >> n;
if (loop >= inf)
S.push(1);
else
S.push(n), loop = loop * n;
} else {
if (!S.empty()) {
int k = S.top();
loop /= k;
S.pop();
}
}
}
if (flag)
cout << "OVERFLOW!!!" << endl;
else
cout << ans << endl;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long inf = 4294967296;
int main() {
int t, n, flag = 0;
scanf("%d", &t);
char ch[10];
long long loop = 1;
long long ans = 0;
stack<int> S;
while (t--) {
cin >> ch;
if (ch[0] == 'a') {
ans += loop;
if (ans >= inf) flag = 1;
} else if (ch[0] == 'f') {
cin >> n;
if (loop >= inf)
S.push(1);
else
S.push(n), loop = loop * n;
} else {
if (!S.empty()) {
int k = S.top();
loop /= k;
S.pop();
}
}
}
if (flag)
cout << "OVERFLOW!!!" << endl;
else
cout << ans << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long over = ((long long)1 << 32) - 1;
int n, h;
long long x;
long long cnt = 0;
vector<long long> f;
int main(void) {
f.push_back(1);
cin >> n;
char t[4];
bool flag = 0;
for (int i = 1; i <= n; ++i) {
cin >> t;
if (t[0] == 'f') {
++cnt;
f.resize(cnt + 1);
cin >> h;
if (over / h < f[cnt - 1]) {
flag = 1;
}
f[cnt] = f[cnt - 1] * h;
} else if (t[0] == 'a') {
if (flag) {
printf("OVERFLOW!!!\n");
exit(0);
} else {
int tmp = x;
x += f[cnt];
if (x > over || x < tmp) {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
}
} else {
flag = 0;
--cnt;
}
}
cout << x << endl;
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long over = ((long long)1 << 32) - 1;
int n, h;
long long x;
long long cnt = 0;
vector<long long> f;
int main(void) {
f.push_back(1);
cin >> n;
char t[4];
bool flag = 0;
for (int i = 1; i <= n; ++i) {
cin >> t;
if (t[0] == 'f') {
++cnt;
f.resize(cnt + 1);
cin >> h;
if (over / h < f[cnt - 1]) {
flag = 1;
}
f[cnt] = f[cnt - 1] * h;
} else if (t[0] == 'a') {
if (flag) {
printf("OVERFLOW!!!\n");
exit(0);
} else {
int tmp = x;
x += f[cnt];
if (x > over || x < tmp) {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
}
} else {
flag = 0;
--cnt;
}
}
cout << x << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long MAX = pow(2, 32) - 1;
int n;
cin >> n;
stack<int> fors;
long long val = 0;
long long num = 1;
int i = 0;
while (i != n) {
string str;
cin >> str;
if (str == "add") {
val += num;
} else if (str == "for") {
int j;
cin >> j;
num *= j;
fors.push(j);
} else {
num /= fors.top();
fors.pop();
}
if (val > MAX) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
if (num > MAX) {
int end = 1;
while (end != 0) {
string s;
cin >> s;
if (s == "add") {
cout << "OVERFLOW!!!" << endl;
return 0;
} else if (s == "for") {
int k;
cin >> k;
end++;
} else {
end--;
}
i++;
}
num /= fors.top();
fors.pop();
}
i++;
}
cout << val << endl;
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long MAX = pow(2, 32) - 1;
int n;
cin >> n;
stack<int> fors;
long long val = 0;
long long num = 1;
int i = 0;
while (i != n) {
string str;
cin >> str;
if (str == "add") {
val += num;
} else if (str == "for") {
int j;
cin >> j;
num *= j;
fors.push(j);
} else {
num /= fors.top();
fors.pop();
}
if (val > MAX) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
if (num > MAX) {
int end = 1;
while (end != 0) {
string s;
cin >> s;
if (s == "add") {
cout << "OVERFLOW!!!" << endl;
return 0;
} else if (s == "for") {
int k;
cin >> k;
end++;
} else {
end--;
}
i++;
}
num /= fors.top();
fors.pop();
}
i++;
}
cout << val << endl;
}
``` |
#include <bits/stdc++.h>
const long double pie = 3.1415926535;
using namespace std;
long long fpow(long long n, long long p) {
long long m = 1;
while (p) {
if (p % 2) m *= n;
p >>= 1;
n *= n;
}
return m;
}
long long mfpow(long long n, long long p, long long M) {
long long m = 1;
n %= M;
while (p) {
if (p % 2) m = (m * n) % M;
n = (n * n) % M;
p >>= 1;
}
return m % M;
}
const vector<long long> days = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool br1, br2, br;
char ch, ch1, ch2;
long long n, k, x, y, z, mini, maxi, l, ind, ini, sum, t, len, r, w, imini, m;
string s, s1, s2;
stack<pair<long long, bool> > st;
void solve() {
long long maxim = fpow(2, 32) - 1;
cin >> x;
st.push(make_pair(1, 0));
for (long long i = 0; i < x; i++) {
cin >> s;
if (s == "for") {
cin >> m;
if (st.top().second == 1 || m * st.top().first > maxim)
st.push(make_pair(m * st.top().first, 1));
else
st.push(make_pair(m * st.top().first, 0));
}
if (s == "end") st.pop();
if (s == "add") {
n += st.top().first;
if (st.top().second == 1 || n > maxim) br = 1;
}
}
if (br)
cout << "OVERFLOW!!!" << endl;
else
cout << n << endl;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
solve();
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
const long double pie = 3.1415926535;
using namespace std;
long long fpow(long long n, long long p) {
long long m = 1;
while (p) {
if (p % 2) m *= n;
p >>= 1;
n *= n;
}
return m;
}
long long mfpow(long long n, long long p, long long M) {
long long m = 1;
n %= M;
while (p) {
if (p % 2) m = (m * n) % M;
n = (n * n) % M;
p >>= 1;
}
return m % M;
}
const vector<long long> days = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool br1, br2, br;
char ch, ch1, ch2;
long long n, k, x, y, z, mini, maxi, l, ind, ini, sum, t, len, r, w, imini, m;
string s, s1, s2;
stack<pair<long long, bool> > st;
void solve() {
long long maxim = fpow(2, 32) - 1;
cin >> x;
st.push(make_pair(1, 0));
for (long long i = 0; i < x; i++) {
cin >> s;
if (s == "for") {
cin >> m;
if (st.top().second == 1 || m * st.top().first > maxim)
st.push(make_pair(m * st.top().first, 1));
else
st.push(make_pair(m * st.top().first, 0));
}
if (s == "end") st.pop();
if (s == "add") {
n += st.top().first;
if (st.top().second == 1 || n > maxim) br = 1;
}
}
if (br)
cout << "OVERFLOW!!!" << endl;
else
cout << n << endl;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int l;
cin >> l;
std::stack<pair<string, long long> > cmds;
long long total = 0LL;
for (int i = 0; i < l; ++i) {
string cmd;
cin >> cmd;
if (cmd == "end") {
long long cnt = 0LL;
while (cmds.top().first != "for") {
cnt += cmds.top().second;
cmds.pop();
}
if (cnt == 0) {
cmds.pop();
continue;
}
if (INT_MAX * 2LL + 1 * 1LL < cnt ||
INT_MAX * 2LL + 1 * 1LL < cmds.top().second) {
cout << "OVERFLOW!!!";
return 0;
}
cnt *= cmds.top().second;
if (INT_MAX * 2LL + 1 * 1LL < cnt) {
cout << "OVERFLOW!!!";
return 0;
}
cmds.pop();
cmds.push(make_pair("add", cnt));
} else if (cmd == "add")
cmds.push(make_pair(cmd, 1LL));
else {
long long tmp;
cin >> tmp;
cmds.push(make_pair(cmd, tmp));
}
}
while (!cmds.empty()) {
total += cmds.top().second;
if (INT_MAX * 2LL + 1 * 1LL < total) {
cout << "OVERFLOW!!!";
return 0;
}
cmds.pop();
}
cout << total;
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int l;
cin >> l;
std::stack<pair<string, long long> > cmds;
long long total = 0LL;
for (int i = 0; i < l; ++i) {
string cmd;
cin >> cmd;
if (cmd == "end") {
long long cnt = 0LL;
while (cmds.top().first != "for") {
cnt += cmds.top().second;
cmds.pop();
}
if (cnt == 0) {
cmds.pop();
continue;
}
if (INT_MAX * 2LL + 1 * 1LL < cnt ||
INT_MAX * 2LL + 1 * 1LL < cmds.top().second) {
cout << "OVERFLOW!!!";
return 0;
}
cnt *= cmds.top().second;
if (INT_MAX * 2LL + 1 * 1LL < cnt) {
cout << "OVERFLOW!!!";
return 0;
}
cmds.pop();
cmds.push(make_pair("add", cnt));
} else if (cmd == "add")
cmds.push(make_pair(cmd, 1LL));
else {
long long tmp;
cin >> tmp;
cmds.push(make_pair(cmd, tmp));
}
}
while (!cmds.empty()) {
total += cmds.top().second;
if (INT_MAX * 2LL + 1 * 1LL < total) {
cout << "OVERFLOW!!!";
return 0;
}
cmds.pop();
}
cout << total;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int TESTS = 1;
while (TESTS--) {
stack<pair<string, long long>> st;
long long i, j, k, l = 0, m, n;
m = 4294967296;
cin >> n;
string s;
for (i = 0; i < n; i++) {
cin >> s;
if (s == "end") {
long long ans1 = 0;
while (st.top().first != "for") {
ans1 += st.top().second;
st.pop();
if (ans1 >= m) {
l = 0;
cout << "OVERFLOW!!!" << '\n';
return 0;
}
}
ans1 *= st.top().second;
if (ans1 >= m) {
l = 0;
cout << "OVERFLOW!!!" << '\n';
return 0;
}
st.pop();
st.push({"add", ans1});
continue;
}
if (s == "add") {
st.push({"add", 1});
continue;
}
cin >> l;
st.push({s, l});
}
long long ans = 0;
while (!st.empty()) {
ans += st.top().second;
st.pop();
if (ans >= m) {
l = 0;
cout << "OVERFLOW!!!" << '\n';
return 0;
}
}
if (ans >= m) {
l = 0;
cout << "OVERFLOW!!!" << '\n';
return 0;
}
cout << ans;
}
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int TESTS = 1;
while (TESTS--) {
stack<pair<string, long long>> st;
long long i, j, k, l = 0, m, n;
m = 4294967296;
cin >> n;
string s;
for (i = 0; i < n; i++) {
cin >> s;
if (s == "end") {
long long ans1 = 0;
while (st.top().first != "for") {
ans1 += st.top().second;
st.pop();
if (ans1 >= m) {
l = 0;
cout << "OVERFLOW!!!" << '\n';
return 0;
}
}
ans1 *= st.top().second;
if (ans1 >= m) {
l = 0;
cout << "OVERFLOW!!!" << '\n';
return 0;
}
st.pop();
st.push({"add", ans1});
continue;
}
if (s == "add") {
st.push({"add", 1});
continue;
}
cin >> l;
st.push({s, l});
}
long long ans = 0;
while (!st.empty()) {
ans += st.top().second;
st.pop();
if (ans >= m) {
l = 0;
cout << "OVERFLOW!!!" << '\n';
return 0;
}
}
if (ans >= m) {
l = 0;
cout << "OVERFLOW!!!" << '\n';
return 0;
}
cout << ans;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e10;
const long long N = 100001;
char en = '\n';
long long Y = 4294967296;
void solve() {
long long q, a = 1, ans = 0;
cin >> q;
stack<long long> w;
w.push(1);
bool first = 0;
while (q--) {
string second;
cin >> second;
if (second == "for") {
long long j;
cin >> j;
w.push(min(Y, w.top() * j));
}
if (second == "add") {
if (ans + w.top() < Y)
ans += w.top();
else
first = 1;
}
if (second == "end") {
long long r = w.top();
a /= r;
w.pop();
}
}
if (!first && ans < Y)
cout << ans;
else
cout << "OVERFLOW!!!";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
while (t--) {
solve();
cout << "\n";
}
cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e10;
const long long N = 100001;
char en = '\n';
long long Y = 4294967296;
void solve() {
long long q, a = 1, ans = 0;
cin >> q;
stack<long long> w;
w.push(1);
bool first = 0;
while (q--) {
string second;
cin >> second;
if (second == "for") {
long long j;
cin >> j;
w.push(min(Y, w.top() * j));
}
if (second == "add") {
if (ans + w.top() < Y)
ans += w.top();
else
first = 1;
}
if (second == "end") {
long long r = w.top();
a /= r;
w.pop();
}
}
if (!first && ans < Y)
cout << ans;
else
cout << "OVERFLOW!!!";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t = 1;
while (t--) {
solve();
cout << "\n";
}
cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll maxn = 1e5 + 5;
const ll limit = (1ll << 32) - 1;
ll vis[maxn];
int main() {
ll n;
scanf("%lld", &n);
stack<ll> st;
ll temp = 1;
int flag = 0;
ll ans = 0;
while (n--) {
char s[10];
cin >> s;
if (s[0] == 'f') {
ll m;
scanf("%lld", &m);
if (flag) {
continue;
}
st.push(m);
if (temp <= limit) {
vis[st.size()] = 1;
temp *= m;
}
} else if (s[0] == 'e') {
if (flag) {
continue;
}
int m = st.top();
if (vis[st.size()]) {
vis[st.size()] = 0;
temp /= m;
}
st.pop();
} else {
ans += temp;
if (ans > limit) {
flag = 1;
}
if (flag) {
continue;
}
}
}
if (!flag)
cout << ans << endl;
else
cout << "OVERFLOW!!!" << endl;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll maxn = 1e5 + 5;
const ll limit = (1ll << 32) - 1;
ll vis[maxn];
int main() {
ll n;
scanf("%lld", &n);
stack<ll> st;
ll temp = 1;
int flag = 0;
ll ans = 0;
while (n--) {
char s[10];
cin >> s;
if (s[0] == 'f') {
ll m;
scanf("%lld", &m);
if (flag) {
continue;
}
st.push(m);
if (temp <= limit) {
vis[st.size()] = 1;
temp *= m;
}
} else if (s[0] == 'e') {
if (flag) {
continue;
}
int m = st.top();
if (vis[st.size()]) {
vis[st.size()] = 0;
temp /= m;
}
st.pop();
} else {
ans += temp;
if (ans > limit) {
flag = 1;
}
if (flag) {
continue;
}
}
}
if (!flag)
cout << ans << endl;
else
cout << "OVERFLOW!!!" << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long ans = 0, tmp = 0;
int l, i, lp;
char data[100];
stack<pair<long long, long long> > stck;
pair<long long, long long> it;
scanf("%d", &l);
for (i = 0; i < l; i++) {
scanf("%s", data);
if (data[0] == 'a') {
if (stck.empty()) {
ans++;
} else {
it = stck.top();
it.second++;
stck.pop();
stck.push(it);
}
} else if (data[0] == 'f') {
scanf("%d", &lp);
stck.push(make_pair((long long)lp, 0ll));
} else {
if (!stck.empty()) {
it = stck.top();
tmp = it.first * it.second;
stck.pop();
if (stck.empty()) {
ans += tmp;
if (ans > 0xffffffffll) {
ans = -1;
break;
}
} else {
it = stck.top();
it.second += tmp;
if ((it.second * it.first) > 0xffffffffll) {
ans = -1;
break;
}
stck.pop();
stck.push(it);
}
}
}
}
if (ans > 0xffffffffll || ans == -1) {
printf("OVERFLOW!!!\n");
return 0;
}
printf("%lld\n", ans);
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long ans = 0, tmp = 0;
int l, i, lp;
char data[100];
stack<pair<long long, long long> > stck;
pair<long long, long long> it;
scanf("%d", &l);
for (i = 0; i < l; i++) {
scanf("%s", data);
if (data[0] == 'a') {
if (stck.empty()) {
ans++;
} else {
it = stck.top();
it.second++;
stck.pop();
stck.push(it);
}
} else if (data[0] == 'f') {
scanf("%d", &lp);
stck.push(make_pair((long long)lp, 0ll));
} else {
if (!stck.empty()) {
it = stck.top();
tmp = it.first * it.second;
stck.pop();
if (stck.empty()) {
ans += tmp;
if (ans > 0xffffffffll) {
ans = -1;
break;
}
} else {
it = stck.top();
it.second += tmp;
if ((it.second * it.first) > 0xffffffffll) {
ans = -1;
break;
}
stck.pop();
stck.push(it);
}
}
}
}
if (ans > 0xffffffffll || ans == -1) {
printf("OVERFLOW!!!\n");
return 0;
}
printf("%lld\n", ans);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1000000007;
const long long int big = 2999999999999999999;
const long long int small = -2999999999999999999;
void in2(long long int& a, long long int& b) {
scanf("%lld", &a);
scanf("%lld", &b);
}
void in3(long long int& a, long long int& b, long long int& c) {
scanf("%lld", &a);
scanf("%lld", &b);
scanf("%lld", &c);
}
void swap(long long int& a, long long int& b) {
a = a + b;
b = a - b;
a = a - b;
}
void arrayIn(vector<long long int>& a, long long int& n) {
for (long long int i = 0; i < n; i++) scanf("%lld", &a[i]);
}
const string add = "add";
const string loop = "for";
const string end = "end";
void solve() {
long long int x = 0;
long long int l;
cin >> l;
string s;
stack<long long int> fo;
long long int prod = 1;
vector<long long int> prods;
long long int p;
bool done = false;
long long int maxi = (1ll << 32) - 1;
for (long long int i = 0; i < l; i++) {
cin >> s;
if (s == loop) cin >> p;
if (!done) {
if (s == add) {
x += prod;
if (prods.size() > 0 || x > maxi) {
cout << "OVERFLOW!!!";
done = true;
}
} else if (s == loop) {
if (prod * p > maxi) {
prods.push_back(prod);
prod = p;
} else
prod *= p;
fo.push(p);
} else {
prod /= fo.top();
if (prod == 1 && prods.size() > 0) {
prod = prods.back();
prods.erase(prods.end() - 1);
}
fo.pop();
}
}
}
if (!done) cout << x;
}
int main() { solve(); }
| ### Prompt
In Cpp, your task is to solve the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1000000007;
const long long int big = 2999999999999999999;
const long long int small = -2999999999999999999;
void in2(long long int& a, long long int& b) {
scanf("%lld", &a);
scanf("%lld", &b);
}
void in3(long long int& a, long long int& b, long long int& c) {
scanf("%lld", &a);
scanf("%lld", &b);
scanf("%lld", &c);
}
void swap(long long int& a, long long int& b) {
a = a + b;
b = a - b;
a = a - b;
}
void arrayIn(vector<long long int>& a, long long int& n) {
for (long long int i = 0; i < n; i++) scanf("%lld", &a[i]);
}
const string add = "add";
const string loop = "for";
const string end = "end";
void solve() {
long long int x = 0;
long long int l;
cin >> l;
string s;
stack<long long int> fo;
long long int prod = 1;
vector<long long int> prods;
long long int p;
bool done = false;
long long int maxi = (1ll << 32) - 1;
for (long long int i = 0; i < l; i++) {
cin >> s;
if (s == loop) cin >> p;
if (!done) {
if (s == add) {
x += prod;
if (prods.size() > 0 || x > maxi) {
cout << "OVERFLOW!!!";
done = true;
}
} else if (s == loop) {
if (prod * p > maxi) {
prods.push_back(prod);
prod = p;
} else
prod *= p;
fo.push(p);
} else {
prod /= fo.top();
if (prod == 1 && prods.size() > 0) {
prod = prods.back();
prods.erase(prods.end() - 1);
}
fo.pop();
}
}
}
if (!done) cout << x;
}
int main() { solve(); }
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
bool isPowerOfTwo(long long int x) { return x && (!(x & (x - 1))); }
long long int Modular_Exponentiation(long long int x, long long int n,
long long int M) {
if (n == 0) return 1;
if (n % 2)
return (x * Modular_Exponentiation((x * x) % M, n / 2, M)) % M;
else
return (Modular_Exponentiation((x * x) % M, n / 2, M)) % M;
}
bool isPrime(long long int n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (long long int i = 5; i * i <= n; i += 6)
if (n % i == 0 || n % (i + 2) == 0) return false;
return true;
}
void solve() {
long long int c = (1LL << 32) - 1;
long long int l;
cin >> l;
stack<long long int> stk;
long long int ans = 0;
for (long long int i = 0; i < l; ++i) {
string str;
cin >> str;
if (str == "add") {
if (stk.empty())
ans++;
else
ans += (stk.top());
} else {
if (str == "for") {
long long int n;
cin >> n;
if (!stk.empty()) n = n * stk.top();
stk.push(min(c + 1, n));
} else {
if (stk.empty()) {
cout << "OVERFLOW!!!\n";
return;
}
stk.pop();
}
}
if (ans > c) {
cout << "OVERFLOW!!!\n";
return;
}
}
if (ans > c) {
cout << "OVERFLOW!!!\n";
return;
}
cout << ans;
}
int32_t main() {
std::ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int T = 1;
while (T--) solve();
cerr << "Time taken : " << fixed << setprecision(5)
<< ((float)clock() / CLOCKS_PER_SEC) * 1000 << " ms"
<< "\n";
cerr << "My CLOCKS_PER_SEC= " << CLOCKS_PER_SEC << "\n";
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
bool isPowerOfTwo(long long int x) { return x && (!(x & (x - 1))); }
long long int Modular_Exponentiation(long long int x, long long int n,
long long int M) {
if (n == 0) return 1;
if (n % 2)
return (x * Modular_Exponentiation((x * x) % M, n / 2, M)) % M;
else
return (Modular_Exponentiation((x * x) % M, n / 2, M)) % M;
}
bool isPrime(long long int n) {
if (n <= 1) return false;
if (n <= 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (long long int i = 5; i * i <= n; i += 6)
if (n % i == 0 || n % (i + 2) == 0) return false;
return true;
}
void solve() {
long long int c = (1LL << 32) - 1;
long long int l;
cin >> l;
stack<long long int> stk;
long long int ans = 0;
for (long long int i = 0; i < l; ++i) {
string str;
cin >> str;
if (str == "add") {
if (stk.empty())
ans++;
else
ans += (stk.top());
} else {
if (str == "for") {
long long int n;
cin >> n;
if (!stk.empty()) n = n * stk.top();
stk.push(min(c + 1, n));
} else {
if (stk.empty()) {
cout << "OVERFLOW!!!\n";
return;
}
stk.pop();
}
}
if (ans > c) {
cout << "OVERFLOW!!!\n";
return;
}
}
if (ans > c) {
cout << "OVERFLOW!!!\n";
return;
}
cout << ans;
}
int32_t main() {
std::ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int T = 1;
while (T--) solve();
cerr << "Time taken : " << fixed << setprecision(5)
<< ((float)clock() / CLOCKS_PER_SEC) * 1000 << " ms"
<< "\n";
cerr << "My CLOCKS_PER_SEC= " << CLOCKS_PER_SEC << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long l;
bool flag = false;
long long c = 1;
c <<= 32;
c -= 1;
cin >> l;
stack<long long> st;
st.push(1);
long long x = 0;
for (int i = 0; i < l; i++) {
string s;
cin >> s;
if (s == "for") {
long long t;
cin >> t;
if (st.top() > c) t = 1;
st.push(t * st.top());
} else if (s == "end")
st.pop();
else
x += st.top();
if (x > c) {
flag = true;
break;
}
}
if (flag)
cout << "OVERFLOW!!!";
else
cout << x;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long l;
bool flag = false;
long long c = 1;
c <<= 32;
c -= 1;
cin >> l;
stack<long long> st;
st.push(1);
long long x = 0;
for (int i = 0; i < l; i++) {
string s;
cin >> s;
if (s == "for") {
long long t;
cin >> t;
if (st.top() > c) t = 1;
st.push(t * st.top());
} else if (s == "end")
st.pop();
else
x += st.top();
if (x > c) {
flag = true;
break;
}
}
if (flag)
cout << "OVERFLOW!!!";
else
cout << x;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long num[100010];
int vis[100010] = {0};
char a[10], b[100010];
vector<long long> f;
vector<long long> e;
int flag = 0;
long long maxn = 0;
long long T;
long long res = 0;
long long solve(long long time) {
long long tt = 0;
long long add = 0;
long long sum = 0;
while (1) {
cin >> a;
T--;
if (a[0] == 'f') {
cin >> tt;
long long ttt = tt * time;
if (ttt > maxn) {
ttt = maxn + 1;
}
sum += solve(ttt);
} else if (a[0] == 'a') {
add++;
} else {
return sum + add * time;
}
if ((sum + add * time) > maxn) {
flag = 1;
return 0;
}
if (!T) {
res = sum + add * time;
return 0;
}
}
}
int main() {
maxn = pow(2, 32) - 1;
cin >> T;
solve(1);
if (flag)
cout << "OVERFLOW!!!" << endl;
else
cout << res << endl;
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long num[100010];
int vis[100010] = {0};
char a[10], b[100010];
vector<long long> f;
vector<long long> e;
int flag = 0;
long long maxn = 0;
long long T;
long long res = 0;
long long solve(long long time) {
long long tt = 0;
long long add = 0;
long long sum = 0;
while (1) {
cin >> a;
T--;
if (a[0] == 'f') {
cin >> tt;
long long ttt = tt * time;
if (ttt > maxn) {
ttt = maxn + 1;
}
sum += solve(ttt);
} else if (a[0] == 'a') {
add++;
} else {
return sum + add * time;
}
if ((sum + add * time) > maxn) {
flag = 1;
return 0;
}
if (!T) {
res = sum + add * time;
return 0;
}
}
}
int main() {
maxn = pow(2, 32) - 1;
cin >> T;
solve(1);
if (flag)
cout << "OVERFLOW!!!" << endl;
else
cout << res << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n;
int num[1000005], l[1000005];
long long ans, limit = 1;
string s[1000005];
void init() {
for (int i = 1; i <= 32; i++) limit *= 2;
limit--;
}
int main() {
init();
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> s[i];
if (s[i] == "for") cin >> num[i];
}
long long now = 1;
int tail = 0, yichu = 0;
for (int i = 1; i <= n; i++) {
if (s[i] == "for") {
if (now <= limit) {
now *= num[i];
l[++tail] = num[i];
} else
yichu++;
}
if (s[i] == "add") {
ans += now;
if (ans > limit) break;
}
if (s[i] == "end") {
if (!yichu)
now /= l[tail--];
else
yichu--;
}
}
if (ans > limit)
printf("OVERFLOW!!!");
else
printf("%lld", ans);
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n;
int num[1000005], l[1000005];
long long ans, limit = 1;
string s[1000005];
void init() {
for (int i = 1; i <= 32; i++) limit *= 2;
limit--;
}
int main() {
init();
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> s[i];
if (s[i] == "for") cin >> num[i];
}
long long now = 1;
int tail = 0, yichu = 0;
for (int i = 1; i <= n; i++) {
if (s[i] == "for") {
if (now <= limit) {
now *= num[i];
l[++tail] = num[i];
} else
yichu++;
}
if (s[i] == "add") {
ans += now;
if (ans > limit) break;
}
if (s[i] == "end") {
if (!yichu)
now /= l[tail--];
else
yichu--;
}
}
if (ans > limit)
printf("OVERFLOW!!!");
else
printf("%lld", ans);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, x, f = 1, es = 0, sum = 0;
string cad;
cin >> n;
stack<int> q;
while (n--) {
cin >> cad;
if (sum < (1LL << 32)) {
if (cad == "for") {
cin >> x;
if (es || f * x >= (1LL << 32))
es++;
else {
q.push(x);
f *= x;
}
}
if (cad == "add") {
if (sum + f < (1LL << 32) && es == 0)
sum += f;
else
sum = (1LL << 32);
}
if (cad == "end") {
if (es)
es--;
else {
if (!q.empty()) {
f /= q.top();
q.pop();
} else
f = 1;
}
}
}
}
if (sum < (1LL << 32))
cout << sum << "\n";
else
cout << "OVERFLOW!!!\n";
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, x, f = 1, es = 0, sum = 0;
string cad;
cin >> n;
stack<int> q;
while (n--) {
cin >> cad;
if (sum < (1LL << 32)) {
if (cad == "for") {
cin >> x;
if (es || f * x >= (1LL << 32))
es++;
else {
q.push(x);
f *= x;
}
}
if (cad == "add") {
if (sum + f < (1LL << 32) && es == 0)
sum += f;
else
sum = (1LL << 32);
}
if (cad == "end") {
if (es)
es--;
else {
if (!q.empty()) {
f /= q.top();
q.pop();
} else
f = 1;
}
}
}
}
if (sum < (1LL << 32))
cout << sum << "\n";
else
cout << "OVERFLOW!!!\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
stack<long long> sq;
int main() {
int n;
long long maxs = 1;
for (int i = 1; i <= 32; i++) maxs *= 2;
maxs--;
long long flag = 0;
long long unused = 0;
long long now = 0;
long long times = 1;
cin >> n;
while (n--) {
if (now > maxs) break;
string s;
cin >> s;
if (s == "add") {
if (flag == 1) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
now += times;
} else if (s == "for") {
long long d;
cin >> d;
if (flag == 1) {
unused++;
continue;
}
sq.push(d);
times *= d;
if (times > maxs) {
flag = 1;
unused++;
times /= sq.top();
sq.pop();
}
} else {
if (unused != 0) {
unused--;
if (unused == 0) {
flag = 0;
}
continue;
}
times /= sq.top();
sq.pop();
}
}
if (now > maxs) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << now << endl;
}
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
stack<long long> sq;
int main() {
int n;
long long maxs = 1;
for (int i = 1; i <= 32; i++) maxs *= 2;
maxs--;
long long flag = 0;
long long unused = 0;
long long now = 0;
long long times = 1;
cin >> n;
while (n--) {
if (now > maxs) break;
string s;
cin >> s;
if (s == "add") {
if (flag == 1) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
now += times;
} else if (s == "for") {
long long d;
cin >> d;
if (flag == 1) {
unused++;
continue;
}
sq.push(d);
times *= d;
if (times > maxs) {
flag = 1;
unused++;
times /= sq.top();
sq.pop();
}
} else {
if (unused != 0) {
unused--;
if (unused == 0) {
flag = 0;
}
continue;
}
times /= sq.top();
sq.pop();
}
}
if (now > maxs) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << now << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MOD = 20101009;
const int N = (int)1e5 + 5;
char op[N][10];
const long long lim = 1LL << 32;
int main() {
int m;
while (~scanf("%d", &m)) {
stack<long long> stk;
stack<bool> stkStatus;
getchar();
for (int i = 1; i <= m; i++) {
gets(op[i]);
}
stk.push(1);
stkStatus.push(true);
long long x = 0;
bool ok = true;
for (int i = 1; i <= m; i++) {
if (op[i][0] == 'a') {
x += stk.top();
if (stkStatus.top() == false || x >= lim) {
ok = false;
break;
}
} else if (op[i][0] == 'e') {
stk.pop();
stkStatus.pop();
} else {
long long val;
char uselessStr[10];
sscanf(op[i], "%s %lld", uselessStr, &val);
val *= stk.top();
stk.push(val);
stkStatus.push(stkStatus.top() & (val < lim));
}
}
if (ok) {
printf("%lld\n", x);
} else {
puts("OVERFLOW!!!");
}
}
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 20101009;
const int N = (int)1e5 + 5;
char op[N][10];
const long long lim = 1LL << 32;
int main() {
int m;
while (~scanf("%d", &m)) {
stack<long long> stk;
stack<bool> stkStatus;
getchar();
for (int i = 1; i <= m; i++) {
gets(op[i]);
}
stk.push(1);
stkStatus.push(true);
long long x = 0;
bool ok = true;
for (int i = 1; i <= m; i++) {
if (op[i][0] == 'a') {
x += stk.top();
if (stkStatus.top() == false || x >= lim) {
ok = false;
break;
}
} else if (op[i][0] == 'e') {
stk.pop();
stkStatus.pop();
} else {
long long val;
char uselessStr[10];
sscanf(op[i], "%s %lld", uselessStr, &val);
val *= stk.top();
stk.push(val);
stkStatus.push(stkStatus.top() & (val < lim));
}
}
if (ok) {
printf("%lld\n", x);
} else {
puts("OVERFLOW!!!");
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MAX = (1ll << 32) - 1;
int main() {
int l;
cin >> l;
stack<long long> nums;
stack<long long> times;
long long x = 0;
int flag = 0;
for (int i = 0; i < l; i++) {
string s;
cin >> s;
if (s == "add") {
x += 1;
if (x > MAX) {
flag = 1;
break;
}
} else if (s == "end") {
x *= times.top();
if (x > MAX) {
flag = 1;
break;
}
times.pop();
x += nums.top();
if (x > MAX) {
flag = 1;
break;
}
nums.pop();
} else {
long long k;
cin >> k;
times.push(k);
nums.push(x);
x = 0;
}
}
if (flag)
cout << "OVERFLOW!!!" << endl;
else
cout << x << endl;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MAX = (1ll << 32) - 1;
int main() {
int l;
cin >> l;
stack<long long> nums;
stack<long long> times;
long long x = 0;
int flag = 0;
for (int i = 0; i < l; i++) {
string s;
cin >> s;
if (s == "add") {
x += 1;
if (x > MAX) {
flag = 1;
break;
}
} else if (s == "end") {
x *= times.top();
if (x > MAX) {
flag = 1;
break;
}
times.pop();
x += nums.top();
if (x > MAX) {
flag = 1;
break;
}
nums.pop();
} else {
long long k;
cin >> k;
times.push(k);
nums.push(x);
x = 0;
}
}
if (flag)
cout << "OVERFLOW!!!" << endl;
else
cout << x << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
unsigned long long int N = 4294967295;
stack<pair<unsigned long long int, unsigned long long int>> buf;
unsigned long long int ans = 0;
void updater(unsigned long long int num) {
if (!buf.empty())
buf.top().first += num;
else
ans += num;
if ((!buf.empty() && buf.top().first > N) || ans > N) {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
unsigned long long int t;
cin >> t;
while (t--) {
string s;
cin >> s;
if (s[0] == 'f') {
unsigned long long int x;
cin >> x;
buf.push({0, x});
} else if (s[0] == 'a') {
updater(1);
} else if (s[0] == 'e') {
unsigned long long int temp = buf.top().first * buf.top().second;
if (temp > N) {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
buf.pop();
updater(temp);
}
}
cout << ans << endl;
}
| ### Prompt
Develop a solution in cpp to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
unsigned long long int N = 4294967295;
stack<pair<unsigned long long int, unsigned long long int>> buf;
unsigned long long int ans = 0;
void updater(unsigned long long int num) {
if (!buf.empty())
buf.top().first += num;
else
ans += num;
if ((!buf.empty() && buf.top().first > N) || ans > N) {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
unsigned long long int t;
cin >> t;
while (t--) {
string s;
cin >> s;
if (s[0] == 'f') {
unsigned long long int x;
cin >> x;
buf.push({0, x});
} else if (s[0] == 'a') {
updater(1);
} else if (s[0] == 'e') {
unsigned long long int temp = buf.top().first * buf.top().second;
if (temp > N) {
cout << "OVERFLOW!!!" << endl;
exit(0);
}
buf.pop();
updater(temp);
}
}
cout << ans << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline long long sbt(long long x) { return __builtin_popcount(x); }
inline long long mul(long long a, long long b, long long m = 1000000007) {
return (a * b) % m;
}
inline long long add(long long a, long long b, long long m = 1000000007) {
return (a + b) % m;
}
inline long long sub(long long a, long long b, long long m = 1000000007) {
return (a - b + m) % m;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t;
cin >> t;
long long val = 0;
long long n = 1;
long long gg = (1LL << 32);
stack<long long> st;
st.push(1LL);
while (t--) {
string s;
cin >> s;
if (s == "add") {
val = val + st.top();
} else if (s == "for") {
long long m;
cin >> m;
st.push(min(gg, st.top() * m));
} else if (s == "end") {
st.pop();
}
}
if (val >= gg) {
cout << "OVERFLOW!!!";
} else {
cout << val;
}
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline long long sbt(long long x) { return __builtin_popcount(x); }
inline long long mul(long long a, long long b, long long m = 1000000007) {
return (a * b) % m;
}
inline long long add(long long a, long long b, long long m = 1000000007) {
return (a + b) % m;
}
inline long long sub(long long a, long long b, long long m = 1000000007) {
return (a - b + m) % m;
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t;
cin >> t;
long long val = 0;
long long n = 1;
long long gg = (1LL << 32);
stack<long long> st;
st.push(1LL);
while (t--) {
string s;
cin >> s;
if (s == "add") {
val = val + st.top();
} else if (s == "for") {
long long m;
cin >> m;
st.push(min(gg, st.top() * m));
} else if (s == "end") {
st.pop();
}
}
if (val >= gg) {
cout << "OVERFLOW!!!";
} else {
cout << val;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long getMaxL() {
long long ans = 1;
int n = 32;
for (int i = 0; i < n; ++i) {
ans *= 2;
}
return ans - 1;
}
int main() {
long long maxL = getMaxL();
int n;
scanf("%d", &n);
stack<long long> S;
long long ans = 0;
bool flag = false;
S.push(1);
while (n--) {
char com[10];
scanf("%s", com);
if (com[0] == 'a') {
ans += S.top();
if (ans > maxL) {
flag = true;
break;
}
} else if (com[0] == 'e') {
S.pop();
} else {
long long v;
scanf("%I64d", &v);
S.push(min(maxL + 1, S.top() * v));
}
}
if (flag) {
puts("OVERFLOW!!!");
} else {
printf("%I64d\n", ans);
}
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long getMaxL() {
long long ans = 1;
int n = 32;
for (int i = 0; i < n; ++i) {
ans *= 2;
}
return ans - 1;
}
int main() {
long long maxL = getMaxL();
int n;
scanf("%d", &n);
stack<long long> S;
long long ans = 0;
bool flag = false;
S.push(1);
while (n--) {
char com[10];
scanf("%s", com);
if (com[0] == 'a') {
ans += S.top();
if (ans > maxL) {
flag = true;
break;
}
} else if (com[0] == 'e') {
S.pop();
} else {
long long v;
scanf("%I64d", &v);
S.push(min(maxL + 1, S.top() * v));
}
}
if (flag) {
puts("OVERFLOW!!!");
} else {
printf("%I64d\n", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long int p2[63] = {1,
2,
4,
8,
16,
32,
64,
128,
256,
512,
1024,
2048,
4096,
8192,
16384,
32768,
65536,
131072,
262144,
524288,
1048576,
2097152,
4194304,
8388608,
16777216,
33554432,
67108864,
134217728,
268435456,
536870912,
1073741824,
2147483648,
4294967296,
8589934592,
17179869184,
34359738368,
68719476736,
137438953472,
274877906944,
549755813888,
1099511627776,
2199023255552,
4398046511104,
8796093022208,
17592186044416,
35184372088832,
70368744177664,
140737488355328,
281474976710656,
562949953421312,
1125899906842624,
2251799813685248,
4503599627370496,
9007199254740992,
18014398509481984,
36028797018963968,
72057594037927936,
144115188075855872,
288230376151711744,
576460752303423488,
1152921504606846976,
2305843009213693952,
4611686018427387904};
const long long int maxN = 2e5 + 10;
const long long int maxB = 320;
const long long int INF = 1e14;
map<char, long long int> mp;
map<long long int, char> rmp;
enum operators {
left_parenthesis = -1,
right_parenthesis = -2,
addition = -3,
multiplication = -4
};
deque<long long int> infix_notation, postfix_notation;
void infix_to_postfix() {
infix_notation.push_back(right_parenthesis);
stack<long long int> curr;
curr.push(left_parenthesis);
for (auto x : infix_notation) {
if (x >= 0)
postfix_notation.push_back(x);
else if (x == left_parenthesis)
curr.push(x);
else if (x == addition) {
while (curr.top() == multiplication) {
postfix_notation.push_back(curr.top());
curr.pop();
}
curr.push(x);
} else if (x == multiplication)
curr.push(x);
else if (x == right_parenthesis) {
while (curr.top() != left_parenthesis) {
postfix_notation.push_back(curr.top());
curr.pop();
}
curr.pop();
}
}
return;
}
long long int calculate() {
stack<long long int> curr;
long long int a, b, result;
for (auto x : postfix_notation) {
if (x < 0) {
a = curr.top();
curr.pop();
b = curr.top();
curr.pop();
if (x == addition)
result = a + b;
else
result = a * b;
curr.push(result);
if (result >= p2[32]) return -1;
} else
curr.push(x);
}
return curr.top();
}
int main() {
long long int number_of_test_case = 1, test_case;
for (test_case = 1; test_case <= number_of_test_case; test_case++) {
long long int i, j, k, m, n, query, q;
mp['('] = -1;
mp[')'] = -2;
mp['+'] = -3;
mp['*'] = -4;
rmp[-1] = '(';
rmp[-2] = ')';
rmp[-3] = '+';
rmp[-4] = '*';
string tmp;
cin >> n;
for (i = 0; i < (n); ++i) {
cin >> tmp;
if (tmp == "for") {
cin >> m;
infix_notation.push_back(m);
infix_notation.push_back(multiplication);
infix_notation.push_back(left_parenthesis);
} else if (tmp == "add") {
infix_notation.push_back(1);
infix_notation.push_back(addition);
} else {
infix_notation.push_back(0);
infix_notation.push_back(right_parenthesis);
infix_notation.push_back(addition);
}
}
infix_notation.pop_back();
infix_to_postfix();
long long int ans = calculate();
if (ans == -1)
cout << "OVERFLOW!!!\n";
else
cout << ans << "\n";
}
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int p2[63] = {1,
2,
4,
8,
16,
32,
64,
128,
256,
512,
1024,
2048,
4096,
8192,
16384,
32768,
65536,
131072,
262144,
524288,
1048576,
2097152,
4194304,
8388608,
16777216,
33554432,
67108864,
134217728,
268435456,
536870912,
1073741824,
2147483648,
4294967296,
8589934592,
17179869184,
34359738368,
68719476736,
137438953472,
274877906944,
549755813888,
1099511627776,
2199023255552,
4398046511104,
8796093022208,
17592186044416,
35184372088832,
70368744177664,
140737488355328,
281474976710656,
562949953421312,
1125899906842624,
2251799813685248,
4503599627370496,
9007199254740992,
18014398509481984,
36028797018963968,
72057594037927936,
144115188075855872,
288230376151711744,
576460752303423488,
1152921504606846976,
2305843009213693952,
4611686018427387904};
const long long int maxN = 2e5 + 10;
const long long int maxB = 320;
const long long int INF = 1e14;
map<char, long long int> mp;
map<long long int, char> rmp;
enum operators {
left_parenthesis = -1,
right_parenthesis = -2,
addition = -3,
multiplication = -4
};
deque<long long int> infix_notation, postfix_notation;
void infix_to_postfix() {
infix_notation.push_back(right_parenthesis);
stack<long long int> curr;
curr.push(left_parenthesis);
for (auto x : infix_notation) {
if (x >= 0)
postfix_notation.push_back(x);
else if (x == left_parenthesis)
curr.push(x);
else if (x == addition) {
while (curr.top() == multiplication) {
postfix_notation.push_back(curr.top());
curr.pop();
}
curr.push(x);
} else if (x == multiplication)
curr.push(x);
else if (x == right_parenthesis) {
while (curr.top() != left_parenthesis) {
postfix_notation.push_back(curr.top());
curr.pop();
}
curr.pop();
}
}
return;
}
long long int calculate() {
stack<long long int> curr;
long long int a, b, result;
for (auto x : postfix_notation) {
if (x < 0) {
a = curr.top();
curr.pop();
b = curr.top();
curr.pop();
if (x == addition)
result = a + b;
else
result = a * b;
curr.push(result);
if (result >= p2[32]) return -1;
} else
curr.push(x);
}
return curr.top();
}
int main() {
long long int number_of_test_case = 1, test_case;
for (test_case = 1; test_case <= number_of_test_case; test_case++) {
long long int i, j, k, m, n, query, q;
mp['('] = -1;
mp[')'] = -2;
mp['+'] = -3;
mp['*'] = -4;
rmp[-1] = '(';
rmp[-2] = ')';
rmp[-3] = '+';
rmp[-4] = '*';
string tmp;
cin >> n;
for (i = 0; i < (n); ++i) {
cin >> tmp;
if (tmp == "for") {
cin >> m;
infix_notation.push_back(m);
infix_notation.push_back(multiplication);
infix_notation.push_back(left_parenthesis);
} else if (tmp == "add") {
infix_notation.push_back(1);
infix_notation.push_back(addition);
} else {
infix_notation.push_back(0);
infix_notation.push_back(right_parenthesis);
infix_notation.push_back(addition);
}
}
infix_notation.pop_back();
infix_to_postfix();
long long int ans = calculate();
if (ans == -1)
cout << "OVERFLOW!!!\n";
else
cout << ans << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, i, j, T, ti;
long long MX = 4294967295;
long long temp;
long long loop(int n) {
long long ret = 0;
string str;
int rn;
cin >> str;
m--;
while (str != "end") {
if (str == "add") {
ret++;
} else if (str == "for") {
cin >> rn;
temp = loop(rn);
if (temp == -1) break;
ret += temp;
} else {
cerr << "WOW " << str << endl;
}
if (m == 0) break;
cin >> str;
m--;
}
if (temp == -1) return -1;
if (ret * n > MX) {
return -1;
}
return ret * n;
}
int main() {
ios::sync_with_stdio(false);
cin >> m;
temp = 0;
temp = loop(1);
if (temp == -1) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << temp << endl;
}
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, i, j, T, ti;
long long MX = 4294967295;
long long temp;
long long loop(int n) {
long long ret = 0;
string str;
int rn;
cin >> str;
m--;
while (str != "end") {
if (str == "add") {
ret++;
} else if (str == "for") {
cin >> rn;
temp = loop(rn);
if (temp == -1) break;
ret += temp;
} else {
cerr << "WOW " << str << endl;
}
if (m == 0) break;
cin >> str;
m--;
}
if (temp == -1) return -1;
if (ret * n > MX) {
return -1;
}
return ret * n;
}
int main() {
ios::sync_with_stdio(false);
cin >> m;
temp = 0;
temp = loop(1);
if (temp == -1) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << temp << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int l;
scanf("%d", &l);
long long add = 0, loop = 1;
long long res = 0;
stack<int> loops, bad;
bool over = false;
while (l--) {
string s;
cin >> s;
if (s == "add") {
res += loop;
if (bad.size() > 0) return cout << "OVERFLOW!!!" << endl, 0;
} else if (s == "end") {
if (bad.size() > 0) {
bad.pop();
} else if (loops.size() > 0) {
loop /= loops.top();
loops.pop();
}
} else if (s == "for") {
long long n;
cin >> n;
if (bad.size() > 0)
bad.push(n);
else if (loop * n < 4294967296) {
loops.push(n);
loop *= n;
} else {
bad.push(n);
}
}
if (res >= 4294967296) return cout << "OVERFLOW!!!" << endl, 0;
}
if (res >= 4294967296) return cout << "OVERFLOW!!!" << endl, 0;
cout << res << endl;
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int l;
scanf("%d", &l);
long long add = 0, loop = 1;
long long res = 0;
stack<int> loops, bad;
bool over = false;
while (l--) {
string s;
cin >> s;
if (s == "add") {
res += loop;
if (bad.size() > 0) return cout << "OVERFLOW!!!" << endl, 0;
} else if (s == "end") {
if (bad.size() > 0) {
bad.pop();
} else if (loops.size() > 0) {
loop /= loops.top();
loops.pop();
}
} else if (s == "for") {
long long n;
cin >> n;
if (bad.size() > 0)
bad.push(n);
else if (loop * n < 4294967296) {
loops.push(n);
loop *= n;
} else {
bad.push(n);
}
}
if (res >= 4294967296) return cout << "OVERFLOW!!!" << endl, 0;
}
if (res >= 4294967296) return cout << "OVERFLOW!!!" << endl, 0;
cout << res << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x7fffffff;
const int maxn = 2e6 + 7;
int main() {
int l;
while (cin >> l) {
long long ans = 0;
stack<long long> FOR, NUM;
NUM.push(0);
int flag = 0;
for (int i = 0; i < l; i++) {
string op;
cin >> op;
if (op[0] == 'a') {
long long a = NUM.top();
a++;
if (a >= (1ll << 32)) {
flag = 1;
a = 0;
}
NUM.pop();
NUM.push(a);
} else if (op[0] == 'f') {
long long for_n;
cin >> for_n;
FOR.push(for_n);
NUM.push(0);
} else {
int for_n = FOR.top();
FOR.pop();
long long t_num = for_n * NUM.top();
NUM.pop();
t_num += NUM.top();
if (t_num >= (1ll << 32)) {
flag = 1;
t_num = 0;
}
NUM.pop();
NUM.push(t_num);
}
}
if (flag) {
cout << "OVERFLOW!!!" << endl;
continue;
}
cout << NUM.top() << endl;
}
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x7fffffff;
const int maxn = 2e6 + 7;
int main() {
int l;
while (cin >> l) {
long long ans = 0;
stack<long long> FOR, NUM;
NUM.push(0);
int flag = 0;
for (int i = 0; i < l; i++) {
string op;
cin >> op;
if (op[0] == 'a') {
long long a = NUM.top();
a++;
if (a >= (1ll << 32)) {
flag = 1;
a = 0;
}
NUM.pop();
NUM.push(a);
} else if (op[0] == 'f') {
long long for_n;
cin >> for_n;
FOR.push(for_n);
NUM.push(0);
} else {
int for_n = FOR.top();
FOR.pop();
long long t_num = for_n * NUM.top();
NUM.pop();
t_num += NUM.top();
if (t_num >= (1ll << 32)) {
flag = 1;
t_num = 0;
}
NUM.pop();
NUM.push(t_num);
}
}
if (flag) {
cout << "OVERFLOW!!!" << endl;
continue;
}
cout << NUM.top() << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1LL << 32;
long long ans = 0;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int l;
cin >> l;
stack<long long> cur;
cur.push(1);
while (l--) {
string s;
cin >> s;
if (s == "for") {
int n;
cin >> n;
cur.push(min(INF, cur.top() * n));
} else if (s == "end")
cur.pop();
else
ans += cur.top();
}
if (ans >= INF)
cout << "OVERFLOW!!!"
<< "\n";
else
cout << ans << "\n";
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1LL << 32;
long long ans = 0;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int l;
cin >> l;
stack<long long> cur;
cur.push(1);
while (l--) {
string s;
cin >> s;
if (s == "for") {
int n;
cin >> n;
cur.push(min(INF, cur.top() * n));
} else if (s == "end")
cur.pop();
else
ans += cur.top();
}
if (ans >= INF)
cout << "OVERFLOW!!!"
<< "\n";
else
cout << ans << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long a, n, w, h, m, k, b;
string s, sl, sr;
void solve() {
long long ans = 0;
vector<long long> vec;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "for") {
cin >> a;
if (vec.size() == 0)
vec.push_back(a);
else
vec.push_back(a * vec.back());
if (vec.back() > UINT_MAX || vec.back() < 0) vec[vec.size() - 1] = -1;
}
if (s == "end") {
vec.pop_back();
}
if (s == "add") {
if (vec.size() == 0) {
ans++;
} else {
if (vec.back() == -1) {
cout << "OVERFLOW!!!";
return;
}
ans += vec.back();
}
if (ans > UINT_MAX) {
cout << "OVERFLOW!!!";
return;
}
}
}
cout << ans;
}
int main() {
ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
cout << setprecision(10);
int t = 1;
while (t--) {
solve();
cout << '\n';
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long a, n, w, h, m, k, b;
string s, sl, sr;
void solve() {
long long ans = 0;
vector<long long> vec;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "for") {
cin >> a;
if (vec.size() == 0)
vec.push_back(a);
else
vec.push_back(a * vec.back());
if (vec.back() > UINT_MAX || vec.back() < 0) vec[vec.size() - 1] = -1;
}
if (s == "end") {
vec.pop_back();
}
if (s == "add") {
if (vec.size() == 0) {
ans++;
} else {
if (vec.back() == -1) {
cout << "OVERFLOW!!!";
return;
}
ans += vec.back();
}
if (ans > UINT_MAX) {
cout << "OVERFLOW!!!";
return;
}
}
}
cout << ans;
}
int main() {
ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
cout << setprecision(10);
int t = 1;
while (t--) {
solve();
cout << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int ar[100005];
int N;
int main() {
cin >> N;
vector<long long> stack;
vector<long long> test;
string str;
long long ans = 0ll;
long long over = (1ll << 32ll) - 1ll;
for (int i = 0; i < N; i++) {
cin >> str;
if (str[0] == 'a') {
if (stack.size() >= 32) {
cout << "OVERFLOW!!!" << endl;
return (0);
}
long long cur_mult = 1ll;
for (int j = 0; j < stack.size(); j++) {
cur_mult *= stack[j];
if (cur_mult > over - ans) {
cout << "OVERFLOW!!!" << endl;
return (0);
}
}
if (ans > over - cur_mult) {
cout << "OVERFLOW!!!" << endl;
return (0);
}
ans += (cur_mult);
} else if (str[0] == 'f') {
long long x;
cin >> x;
test.push_back(x);
if (x != 1) stack.push_back(x);
} else {
if (test[test.size() - 1] != 1) stack.pop_back();
test.pop_back();
}
}
cout << ans << endl;
return (0);
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int ar[100005];
int N;
int main() {
cin >> N;
vector<long long> stack;
vector<long long> test;
string str;
long long ans = 0ll;
long long over = (1ll << 32ll) - 1ll;
for (int i = 0; i < N; i++) {
cin >> str;
if (str[0] == 'a') {
if (stack.size() >= 32) {
cout << "OVERFLOW!!!" << endl;
return (0);
}
long long cur_mult = 1ll;
for (int j = 0; j < stack.size(); j++) {
cur_mult *= stack[j];
if (cur_mult > over - ans) {
cout << "OVERFLOW!!!" << endl;
return (0);
}
}
if (ans > over - cur_mult) {
cout << "OVERFLOW!!!" << endl;
return (0);
}
ans += (cur_mult);
} else if (str[0] == 'f') {
long long x;
cin >> x;
test.push_back(x);
if (x != 1) stack.push_back(x);
} else {
if (test[test.size() - 1] != 1) stack.pop_back();
test.pop_back();
}
}
cout << ans << endl;
return (0);
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 * 1 + 5, inf = 1ll << 32;
int l;
long long ans;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> l;
stack<long long> loop;
loop.push(1);
for (string input; l > 0; l--) {
cin >> input;
if (input == "for") {
long long inputi;
cin >> inputi;
loop.push(min(inputi * loop.top(), inf));
} else if (input == "end")
loop.pop();
else {
if (ans >= inf - loop.top()) return cout << "OVERFLOW!!!", 0;
ans += loop.top();
}
}
cout << ans;
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 * 1 + 5, inf = 1ll << 32;
int l;
long long ans;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> l;
stack<long long> loop;
loop.push(1);
for (string input; l > 0; l--) {
cin >> input;
if (input == "for") {
long long inputi;
cin >> inputi;
loop.push(min(inputi * loop.top(), inf));
} else if (input == "end")
loop.pop();
else {
if (ans >= inf - loop.top()) return cout << "OVERFLOW!!!", 0;
ans += loop.top();
}
}
cout << ans;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long int m = (1LL << 32);
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
string s;
long long int x;
long long int ans = 0;
stack<long long int> st;
st.push(1);
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "for") {
cin >> x;
long long int t = st.top();
st.push(min(t * x, m));
} else if (s == "add") {
long long int t = st.top();
ans += t;
} else if (s == "end") {
st.pop();
}
}
if (ans >= m)
cout << "OVERFLOW!!!";
else
cout << ans;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long int m = (1LL << 32);
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
string s;
long long int x;
long long int ans = 0;
stack<long long int> st;
st.push(1);
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "for") {
cin >> x;
long long int t = st.top();
st.push(min(t * x, m));
} else if (s == "add") {
long long int t = st.top();
ans += t;
} else if (s == "end") {
st.pop();
}
}
if (ans >= m)
cout << "OVERFLOW!!!";
else
cout << ans;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int l;
cin >> l;
long long cur = 1;
long long x = 0;
bool over = false;
stack<long long> s;
for (int i = int(1); i <= int(l); i++) {
string thing;
cin >> thing;
if (thing == "for") {
int len;
cin >> len;
if (!over) {
cur *= len;
s.push(len);
if (cur > 5e9) over = true;
} else {
s.push(-1);
}
}
if (thing == "end") {
int backy = s.top();
s.pop();
if (backy != -1) {
if (!over)
cur /= backy;
else {
over = false;
cur /= backy;
}
}
}
if (thing == "add") {
if (over) {
cout << "OVERFLOW!!!" << endl;
return 0;
} else
x += cur;
}
}
if (x <= 4294967295)
cout << x << endl;
else
cout << "OVERFLOW!!!" << endl;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int l;
cin >> l;
long long cur = 1;
long long x = 0;
bool over = false;
stack<long long> s;
for (int i = int(1); i <= int(l); i++) {
string thing;
cin >> thing;
if (thing == "for") {
int len;
cin >> len;
if (!over) {
cur *= len;
s.push(len);
if (cur > 5e9) over = true;
} else {
s.push(-1);
}
}
if (thing == "end") {
int backy = s.top();
s.pop();
if (backy != -1) {
if (!over)
cur /= backy;
else {
over = false;
cur /= backy;
}
}
}
if (thing == "add") {
if (over) {
cout << "OVERFLOW!!!" << endl;
return 0;
} else
x += cur;
}
}
if (x <= 4294967295)
cout << x << endl;
else
cout << "OVERFLOW!!!" << endl;
}
``` |
#include <bits/stdc++.h>
template <typename T>
inline void read(T& x) {
T res = 0, flag = 1;
char ch;
ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') flag = -1;
ch = getchar();
}
while (isdigit(ch)) {
res = res * 10 + ch - 48;
ch = getchar();
}
x = res * flag;
}
const long long INF = 1LL << 32;
int l;
long long ans, x;
char buf[10];
std::stack<long long> s;
int main() {
read(l);
s.push(1);
for (int i = 1; i <= l; ++i) {
scanf("%s", buf);
if (buf[0] == 'a')
ans += s.top();
else if (buf[0] == 'e')
s.pop();
else {
read(x);
s.push(std::min(INF, x * s.top()));
}
}
if (ans >= INF)
puts("OVERFLOW!!!");
else
printf("%lld\n", ans);
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
template <typename T>
inline void read(T& x) {
T res = 0, flag = 1;
char ch;
ch = getchar();
while (!isdigit(ch)) {
if (ch == '-') flag = -1;
ch = getchar();
}
while (isdigit(ch)) {
res = res * 10 + ch - 48;
ch = getchar();
}
x = res * flag;
}
const long long INF = 1LL << 32;
int l;
long long ans, x;
char buf[10];
std::stack<long long> s;
int main() {
read(l);
s.push(1);
for (int i = 1; i <= l; ++i) {
scanf("%s", buf);
if (buf[0] == 'a')
ans += s.top();
else if (buf[0] == 'e')
s.pop();
else {
read(x);
s.push(std::min(INF, x * s.top()));
}
}
if (ans >= INF)
puts("OVERFLOW!!!");
else
printf("%lld\n", ans);
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
const ll inf = (1e18);
const ll mod = 1000000007;
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
struct __INIT {
__INIT() {
cin.tie(0);
ios::sync_with_stdio(false);
cout << fixed << setprecision(15);
}
} __init;
template <class T>
bool chmax(T &a, const T &b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
if (a > b) {
a = b;
return 1;
}
return 0;
}
int main() {
int n;
cin >> n;
stack<P> q;
ll ans = 0;
bool over = false;
bool ok = true;
int memo = 0;
ll now = 1;
ll ma = (1ll << 32) - 1;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s == "for") {
ll t;
cin >> t;
q.emplace(t, i);
if (over || now * t > ma) {
if (!over) memo = i;
over = true;
} else
now *= t;
}
if (s == "add") {
if (over) ok = false;
if (!ok) continue;
if (ans + now > ma)
ok = false;
else
ans += now;
}
if (s == "end") {
if (!ok) continue;
if (q.size()) {
P p = q.top();
if (over) {
if (memo == p.second) over = false;
} else {
now /= p.first;
}
q.pop();
}
}
}
if (!ok)
cout << "OVERFLOW!!!"
<< "\n";
else
cout << ans << "\n";
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
const ll inf = (1e18);
const ll mod = 1000000007;
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
struct __INIT {
__INIT() {
cin.tie(0);
ios::sync_with_stdio(false);
cout << fixed << setprecision(15);
}
} __init;
template <class T>
bool chmax(T &a, const T &b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
if (a > b) {
a = b;
return 1;
}
return 0;
}
int main() {
int n;
cin >> n;
stack<P> q;
ll ans = 0;
bool over = false;
bool ok = true;
int memo = 0;
ll now = 1;
ll ma = (1ll << 32) - 1;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s == "for") {
ll t;
cin >> t;
q.emplace(t, i);
if (over || now * t > ma) {
if (!over) memo = i;
over = true;
} else
now *= t;
}
if (s == "add") {
if (over) ok = false;
if (!ok) continue;
if (ans + now > ma)
ok = false;
else
ans += now;
}
if (s == "end") {
if (!ok) continue;
if (q.size()) {
P p = q.top();
if (over) {
if (memo == p.second) over = false;
} else {
now /= p.first;
}
q.pop();
}
}
}
if (!ok)
cout << "OVERFLOW!!!"
<< "\n";
else
cout << ans << "\n";
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
int n;
int main() {
cin >> n;
const long long LLIMIT = (long long)UINT32_MAX;
long long res = 0;
v.push_back(1);
string s;
getline(cin, s);
for (int i = 0; i < n; ++i) {
getline(cin, s);
if (s[0] == 'a') {
if (LLIMIT - res < v.back()) {
cout << "OVERFLOW!!!";
return 0;
}
res += v.back();
} else if (s[0] == 'f') {
long long t = 0;
for (int j = 4; j < s.size(); ++j) {
t *= 10;
t += (int)s[j] - 48;
}
if (LLIMIT < v.back() * t)
v.push_back(LLIMIT + 1);
else
v.push_back(v.back() * t);
} else if (s[0] == 'e') {
v.pop_back();
}
}
if (res > LLIMIT)
cout << "OVERFLOW!!!";
else
cout << res;
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
int n;
int main() {
cin >> n;
const long long LLIMIT = (long long)UINT32_MAX;
long long res = 0;
v.push_back(1);
string s;
getline(cin, s);
for (int i = 0; i < n; ++i) {
getline(cin, s);
if (s[0] == 'a') {
if (LLIMIT - res < v.back()) {
cout << "OVERFLOW!!!";
return 0;
}
res += v.back();
} else if (s[0] == 'f') {
long long t = 0;
for (int j = 4; j < s.size(); ++j) {
t *= 10;
t += (int)s[j] - 48;
}
if (LLIMIT < v.back() * t)
v.push_back(LLIMIT + 1);
else
v.push_back(v.back() * t);
} else if (s[0] == 'e') {
v.pop_back();
}
}
if (res > LLIMIT)
cout << "OVERFLOW!!!";
else
cout << res;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long int l, ans = 0, multi = 1, flag = 0;
cin >> l;
stack<long long int> S;
while (l--) {
string s;
cin >> s;
if (s == "for") {
long long int n;
cin >> n;
if (multi * n > pow(2, 32) - 1 || flag < 0) {
flag--;
} else {
multi *= n;
S.push(n);
}
} else if (s == "end") {
if (flag < 0) {
flag++;
} else {
multi /= S.top();
S.pop();
}
} else {
if (S.size() == 0) {
ans++;
} else {
if (flag < 0) {
cout << "OVERFLOW!!!";
return 0;
}
ans += multi;
}
if (ans > pow(2, 32) - 1) {
cout << "OVERFLOW!!!";
return 0;
}
}
}
if (ans >= 0 && ans <= pow(2, 32) - 1) {
cout << ans;
} else {
cout << "OVERFLOW!!!";
}
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long int l, ans = 0, multi = 1, flag = 0;
cin >> l;
stack<long long int> S;
while (l--) {
string s;
cin >> s;
if (s == "for") {
long long int n;
cin >> n;
if (multi * n > pow(2, 32) - 1 || flag < 0) {
flag--;
} else {
multi *= n;
S.push(n);
}
} else if (s == "end") {
if (flag < 0) {
flag++;
} else {
multi /= S.top();
S.pop();
}
} else {
if (S.size() == 0) {
ans++;
} else {
if (flag < 0) {
cout << "OVERFLOW!!!";
return 0;
}
ans += multi;
}
if (ans > pow(2, 32) - 1) {
cout << "OVERFLOW!!!";
return 0;
}
}
}
if (ans >= 0 && ans <= pow(2, 32) - 1) {
cout << ans;
} else {
cout << "OVERFLOW!!!";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long l, n, now, res = 1, O, a[100001], b[100001], ans, INF = pow(2, 32);
char s[5];
int main() {
scanf("%I64d", &l), a[0] = 1;
while (l--) {
scanf("%s", s);
if (s[0] == 'f')
scanf("%I64d", &n), (res >= INF) ? O++ : res *= n, a[++now] = n;
if (s[0] == 'e') O ? O-- : res /= a[now], --now;
if (s[0] == 'a') {
if (O || ans + res >= INF) {
printf("OVERFLOW!!!");
return 0;
}
ans += res;
}
}
printf("%I64d", ans);
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long l, n, now, res = 1, O, a[100001], b[100001], ans, INF = pow(2, 32);
char s[5];
int main() {
scanf("%I64d", &l), a[0] = 1;
while (l--) {
scanf("%s", s);
if (s[0] == 'f')
scanf("%I64d", &n), (res >= INF) ? O++ : res *= n, a[++now] = n;
if (s[0] == 'e') O ? O-- : res /= a[now], --now;
if (s[0] == 'a') {
if (O || ans + res >= INF) {
printf("OVERFLOW!!!");
return 0;
}
ans += res;
}
}
printf("%I64d", ans);
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long INF = (1ll << 32) - 1;
long long dfs() {
string tmp;
long long res = 0, ti;
while (cin >> tmp) {
if (tmp[0] == 'a') {
if (res != -1 && res >= 0 && res <= INF)
res++;
else
res = -1;
} else if (tmp[0] == 'f') {
cin >> ti;
long long tmpans = dfs() * ti;
if (res == -1 || res < 0 || tmpans < 0 || tmpans == -1 ||
res + tmpans > INF)
res = -1;
else
res += tmpans;
} else if (tmp[0] == 'e') {
if (res > INF || res == -1 || res < 0) return -1;
return res;
}
}
return res;
}
int main(int argc, char** argv) {
int n;
scanf("%d", &n);
long long ans = dfs();
if (ans == -1 || ans > INF)
printf("OVERFLOW!!!\n");
else
printf("%lld\n", ans);
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long INF = (1ll << 32) - 1;
long long dfs() {
string tmp;
long long res = 0, ti;
while (cin >> tmp) {
if (tmp[0] == 'a') {
if (res != -1 && res >= 0 && res <= INF)
res++;
else
res = -1;
} else if (tmp[0] == 'f') {
cin >> ti;
long long tmpans = dfs() * ti;
if (res == -1 || res < 0 || tmpans < 0 || tmpans == -1 ||
res + tmpans > INF)
res = -1;
else
res += tmpans;
} else if (tmp[0] == 'e') {
if (res > INF || res == -1 || res < 0) return -1;
return res;
}
}
return res;
}
int main(int argc, char** argv) {
int n;
scanf("%d", &n);
long long ans = dfs();
if (ans == -1 || ans > INF)
printf("OVERFLOW!!!\n");
else
printf("%lld\n", ans);
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")
using namespace std;
int main() {
int l;
cin >> l;
long long lim = 4294967295LL, n, x = 0;
string str;
stack<pair<int, int> > s;
while (l--) {
cin >> str;
if (str == "for") {
cin >> n;
s.push(make_pair(x, n));
} else if (str == "add") {
x++;
} else {
pair<int, int> p = s.top();
long long change = (x - p.first);
x = p.first + p.second * change;
s.pop();
}
if (x > lim) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
}
cout << x << endl;
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")
using namespace std;
int main() {
int l;
cin >> l;
long long lim = 4294967295LL, n, x = 0;
string str;
stack<pair<int, int> > s;
while (l--) {
cin >> str;
if (str == "for") {
cin >> n;
s.push(make_pair(x, n));
} else if (str == "add") {
x++;
} else {
pair<int, int> p = s.top();
long long change = (x - p.first);
x = p.first + p.second * change;
s.pop();
}
if (x > lim) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
}
cout << x << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long maxn = (1LL << 32) - 1;
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
long long x = 0, k;
cin >> k;
int n;
string str;
stack<int> st;
stack<long long> add;
long long mul = 0;
long long t = 0;
add.push(0);
bool over = false;
while (k) {
k--;
cin >> str;
if (str == "for") {
cin >> n;
if (over) continue;
add.push(0);
st.push(n);
} else if (str == "add") {
if (over) continue;
if (st.empty()) {
add.top() += 1;
} else {
add.top() = add.top() + 1;
}
} else {
if (over) continue;
mul = st.top();
st.pop();
t = add.top();
add.pop();
t *= mul;
if (t > maxn) over = true;
add.top() += t;
}
}
if (over || add.top() > maxn) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
cout << add.top() << endl;
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long maxn = (1LL << 32) - 1;
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(0);
long long x = 0, k;
cin >> k;
int n;
string str;
stack<int> st;
stack<long long> add;
long long mul = 0;
long long t = 0;
add.push(0);
bool over = false;
while (k) {
k--;
cin >> str;
if (str == "for") {
cin >> n;
if (over) continue;
add.push(0);
st.push(n);
} else if (str == "add") {
if (over) continue;
if (st.empty()) {
add.top() += 1;
} else {
add.top() = add.top() + 1;
}
} else {
if (over) continue;
mul = st.top();
st.pop();
t = add.top();
add.pop();
t *= mul;
if (t > maxn) over = true;
add.top() += t;
}
}
if (over || add.top() > maxn) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
cout << add.top() << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long int N = 1e5 + 5, inf = 1e18;
long long int expo(long long int b, long long int e, long long int m) {
long long int a = 1;
while (e != 0) {
if ((e & 1) == 1) a = (a * b) % m;
b = (b * b) % m;
e >>= 1;
}
return a % m;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int n;
cin >> n;
long long int mx = (1LL << 32) - 1;
long long int ans = 0;
long long int mul = 1, of = 0, ex = 0, f = 0;
vector<long long int> v;
while (n--) {
string s;
cin >> s;
if (s[0] == 'a') {
if (of == 1) {
f = 1;
} else {
ans += mul;
}
} else if (s[0] == 'e') {
if (ex > 0) {
ex--;
if (ex == 0) {
of = 0;
}
} else {
mul /= v[v.size() - 1];
}
v.pop_back();
} else {
long long int a;
cin >> a;
v.push_back(a);
if ((of == 1) || (mul * a > mx)) {
of = 1;
ex++;
} else {
mul *= a;
}
}
}
if ((ans > mx) || (f == 1)) {
cout << "OVERFLOW!!!"
<< "\n";
} else {
cout << ans << "\n";
}
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int N = 1e5 + 5, inf = 1e18;
long long int expo(long long int b, long long int e, long long int m) {
long long int a = 1;
while (e != 0) {
if ((e & 1) == 1) a = (a * b) % m;
b = (b * b) % m;
e >>= 1;
}
return a % m;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int n;
cin >> n;
long long int mx = (1LL << 32) - 1;
long long int ans = 0;
long long int mul = 1, of = 0, ex = 0, f = 0;
vector<long long int> v;
while (n--) {
string s;
cin >> s;
if (s[0] == 'a') {
if (of == 1) {
f = 1;
} else {
ans += mul;
}
} else if (s[0] == 'e') {
if (ex > 0) {
ex--;
if (ex == 0) {
of = 0;
}
} else {
mul /= v[v.size() - 1];
}
v.pop_back();
} else {
long long int a;
cin >> a;
v.push_back(a);
if ((of == 1) || (mul * a > mx)) {
of = 1;
ex++;
} else {
mul *= a;
}
}
}
if ((ans > mx) || (f == 1)) {
cout << "OVERFLOW!!!"
<< "\n";
} else {
cout << ans << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int l;
cin >> l;
int er = 0;
long long int ok = 1;
vector<long long int> v1;
long long int add = 0;
bool f = true;
long long int ans = 0;
while (l--) {
string a;
cin >> a;
if (a == "add") {
if (v1.size() == 0)
ans++;
else {
if (v1[v1.size() - 1] == 0) {
f = false;
} else {
ans += v1[v1.size() - 1];
}
}
} else if (a == "for") {
if (add) {
ans += add;
add = 0;
}
long long int number;
cin >> number;
if (v1.size() == 0) {
v1.push_back(number);
continue;
}
if (number * v1[v1.size() - 1] > 4294967295) {
v1.push_back(0);
continue;
}
v1.push_back(number * v1[v1.size() - 1]);
} else {
v1.erase(v1.begin() + v1.size() - 1);
}
}
ans += add;
if (ans > 4294967295) f = false;
if (!f) {
cout << "OVERFLOW!!!" << endl;
} else
cout << ans << endl;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int l;
cin >> l;
int er = 0;
long long int ok = 1;
vector<long long int> v1;
long long int add = 0;
bool f = true;
long long int ans = 0;
while (l--) {
string a;
cin >> a;
if (a == "add") {
if (v1.size() == 0)
ans++;
else {
if (v1[v1.size() - 1] == 0) {
f = false;
} else {
ans += v1[v1.size() - 1];
}
}
} else if (a == "for") {
if (add) {
ans += add;
add = 0;
}
long long int number;
cin >> number;
if (v1.size() == 0) {
v1.push_back(number);
continue;
}
if (number * v1[v1.size() - 1] > 4294967295) {
v1.push_back(0);
continue;
}
v1.push_back(number * v1[v1.size() - 1]);
} else {
v1.erase(v1.begin() + v1.size() - 1);
}
}
ans += add;
if (ans > 4294967295) f = false;
if (!f) {
cout << "OVERFLOW!!!" << endl;
} else
cout << ans << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
string instru;
long long int x = 0, ciclo, inf = 1LL << 32;
vector<long long int> pila;
cin >> n;
pila.push_back(1);
for (int i = 0; i < n; i++) {
cin >> instru;
if (instru == "add") {
x += pila.back();
} else if (instru == "for") {
cin >> ciclo;
pila.push_back(min(inf, ciclo * pila.back()));
} else if (instru == "end") {
pila.pop_back();
}
}
if (x < inf)
cout << x << "\n";
else
cout << "OVERFLOW!!!\n";
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
string instru;
long long int x = 0, ciclo, inf = 1LL << 32;
vector<long long int> pila;
cin >> n;
pila.push_back(1);
for (int i = 0; i < n; i++) {
cin >> instru;
if (instru == "add") {
x += pila.back();
} else if (instru == "for") {
cin >> ciclo;
pila.push_back(min(inf, ciclo * pila.back()));
} else if (instru == "end") {
pila.pop_back();
}
}
if (x < inf)
cout << x << "\n";
else
cout << "OVERFLOW!!!\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
bool OF = 0;
stack<long long> st;
char s[14];
long long INF = (1ll << 32) - 1;
int main() {
int n, t;
scanf("%d", &n);
for (int i = 0; i < n && !OF; i++) {
scanf("%s", s);
if (s[0] == 'f') {
scanf("%d", &t);
st.push(t);
} else if (s[0] == 'a')
st.push(-1);
else {
long long tp = 0;
while (!st.empty() && st.top() <= 0) {
tp += st.top();
if (tp < -INF) {
OF = 1;
break;
}
st.pop();
}
long long x = st.top();
st.pop();
st.push(tp * x);
}
}
if (OF)
puts("OVERFLOW!!!");
else {
long long ans = 0;
while (!st.empty()) {
ans += st.top();
st.pop();
}
ans = -ans;
if (ans <= INF)
printf("%lld\n", ans);
else
puts("OVERFLOW!!!");
}
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
bool OF = 0;
stack<long long> st;
char s[14];
long long INF = (1ll << 32) - 1;
int main() {
int n, t;
scanf("%d", &n);
for (int i = 0; i < n && !OF; i++) {
scanf("%s", s);
if (s[0] == 'f') {
scanf("%d", &t);
st.push(t);
} else if (s[0] == 'a')
st.push(-1);
else {
long long tp = 0;
while (!st.empty() && st.top() <= 0) {
tp += st.top();
if (tp < -INF) {
OF = 1;
break;
}
st.pop();
}
long long x = st.top();
st.pop();
st.push(tp * x);
}
}
if (OF)
puts("OVERFLOW!!!");
else {
long long ans = 0;
while (!st.empty()) {
ans += st.top();
st.pop();
}
ans = -ans;
if (ans <= INF)
printf("%lld\n", ans);
else
puts("OVERFLOW!!!");
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
long long t, q = 1;
int main() {
long long x = 0, ov = 0;
cin >> t;
while (t--) {
string s;
cin >> s;
if (s[0] == 'f') {
long long w;
cin >> w;
if (q * w >= 4294967296 || ov > 0) {
ov++;
} else {
q *= w;
v.push_back(w);
}
} else if (s[0] == 'e') {
if (ov > 0) {
ov--;
} else {
q = q / v[v.size() - 1];
v.pop_back();
}
} else {
if (ov > 0 || x + q >= 4294967296) {
cout << "OVERFLOW!!!";
return 0;
}
x += q;
}
}
cout << x;
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<long long> v;
long long t, q = 1;
int main() {
long long x = 0, ov = 0;
cin >> t;
while (t--) {
string s;
cin >> s;
if (s[0] == 'f') {
long long w;
cin >> w;
if (q * w >= 4294967296 || ov > 0) {
ov++;
} else {
q *= w;
v.push_back(w);
}
} else if (s[0] == 'e') {
if (ov > 0) {
ov--;
} else {
q = q / v[v.size() - 1];
v.pop_back();
}
} else {
if (ov > 0 || x + q >= 4294967296) {
cout << "OVERFLOW!!!";
return 0;
}
x += q;
}
}
cout << x;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct S {
int a, b;
S() {}
S(int _a, int _b) {
a = _a;
b = _b;
}
const bool operator<(const S &o) const { return a < o.a; }
};
priority_queue<int, vector<int>, greater<int>> mpq;
const int SPACE = 0, NL = 1;
string exm;
inline void showAll(vector<int> &v, int sep) {
for (int &here : v) printf("%d%c", here, (sep) ? '\n' : ' ');
}
inline void exf(void) {
cout << exm << "\n";
exit(0);
}
const int SIZE = 100009;
int main() {
exm = "OVERFLOW!!!";
int l;
scanf("%d", &l);
long long ans = 0;
vector<long long> st;
st.push_back(1);
while (l--) {
char cmd[10];
scanf("%s", cmd + 1);
if (cmd[1] == 'a') {
long long back = st.back();
if (back == -1) exf();
ans += back;
} else if (cmd[1] == 'f') {
int n;
scanf("%d", &n);
long long back = st.back();
if (back != -1) back = back * n;
if (back >= (1LL << 32))
st.push_back(-1);
else
st.push_back(back);
} else {
st.pop_back();
}
if (ans >= (1LL << 32)) exf();
}
printf("%lld\n", ans);
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct S {
int a, b;
S() {}
S(int _a, int _b) {
a = _a;
b = _b;
}
const bool operator<(const S &o) const { return a < o.a; }
};
priority_queue<int, vector<int>, greater<int>> mpq;
const int SPACE = 0, NL = 1;
string exm;
inline void showAll(vector<int> &v, int sep) {
for (int &here : v) printf("%d%c", here, (sep) ? '\n' : ' ');
}
inline void exf(void) {
cout << exm << "\n";
exit(0);
}
const int SIZE = 100009;
int main() {
exm = "OVERFLOW!!!";
int l;
scanf("%d", &l);
long long ans = 0;
vector<long long> st;
st.push_back(1);
while (l--) {
char cmd[10];
scanf("%s", cmd + 1);
if (cmd[1] == 'a') {
long long back = st.back();
if (back == -1) exf();
ans += back;
} else if (cmd[1] == 'f') {
int n;
scanf("%d", &n);
long long back = st.back();
if (back != -1) back = back * n;
if (back >= (1LL << 32))
st.push_back(-1);
else
st.push_back(back);
} else {
st.pop_back();
}
if (ans >= (1LL << 32)) exf();
}
printf("%lld\n", ans);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int L, n;
unsigned long long val = 0, outer = 0;
string str;
bool ended = false;
stack<int> Loop;
vector<unsigned long long> Value;
cin >> L;
while (L-- && !ended) {
cin >> str;
if (str[0] == 'a') {
if (Loop.size() > 0) {
(Value)[((Value).size()) - 1]++;
if (outer + (Value)[((Value).size()) - 1] > 4294967295) {
ended = true;
break;
}
} else {
outer++;
if (outer > 4294967295) {
ended = true;
break;
}
}
}
if (str[0] == 'f') {
cin >> n;
Loop.push(n);
(Value).push_back((0));
}
if (str[0] == 'e') {
n = Loop.top();
Loop.pop();
val = (Value)[((Value).size()) - 1];
Value.pop_back();
if (Value.size() > 0) {
(Value)[((Value).size()) - 1] += n * val;
if (outer + (Value)[((Value).size()) - 1] > 4294967295) {
ended = true;
break;
}
} else {
outer += n * val;
if (outer > 4294967295) {
ended = true;
break;
}
}
}
}
if (ended) {
cout << "OVERFLOW!!!";
} else {
cout << outer;
}
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int L, n;
unsigned long long val = 0, outer = 0;
string str;
bool ended = false;
stack<int> Loop;
vector<unsigned long long> Value;
cin >> L;
while (L-- && !ended) {
cin >> str;
if (str[0] == 'a') {
if (Loop.size() > 0) {
(Value)[((Value).size()) - 1]++;
if (outer + (Value)[((Value).size()) - 1] > 4294967295) {
ended = true;
break;
}
} else {
outer++;
if (outer > 4294967295) {
ended = true;
break;
}
}
}
if (str[0] == 'f') {
cin >> n;
Loop.push(n);
(Value).push_back((0));
}
if (str[0] == 'e') {
n = Loop.top();
Loop.pop();
val = (Value)[((Value).size()) - 1];
Value.pop_back();
if (Value.size() > 0) {
(Value)[((Value).size()) - 1] += n * val;
if (outer + (Value)[((Value).size()) - 1] > 4294967295) {
ended = true;
break;
}
} else {
outer += n * val;
if (outer > 4294967295) {
ended = true;
break;
}
}
}
}
if (ended) {
cout << "OVERFLOW!!!";
} else {
cout << outer;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct Point {
int x;
int y;
};
struct PPoint {
Point x;
Point y;
};
long long gcd(long long a, long long b) {
if (a < b) swap(a, b);
if (b <= 0) return a;
return gcd(b, a % b);
}
long long l;
long long x;
int main(void) {
cin.tie(NULL);
ios_base::sync_with_stdio(false);
cin >> l;
stack<long long> stk;
stk.push(1);
const long long limit = (1LL << 32);
while (l--) {
string s;
cin >> s;
if (s == "add") {
x += stk.top();
if (x >= limit) {
cout << "OVERFLOW!!!" << '\n';
exit(0);
}
} else if (s == "end") {
stk.pop();
} else {
int n;
cin >> n;
stk.push(min(limit, stk.top() * n));
}
}
cout << x << '\n';
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct Point {
int x;
int y;
};
struct PPoint {
Point x;
Point y;
};
long long gcd(long long a, long long b) {
if (a < b) swap(a, b);
if (b <= 0) return a;
return gcd(b, a % b);
}
long long l;
long long x;
int main(void) {
cin.tie(NULL);
ios_base::sync_with_stdio(false);
cin >> l;
stack<long long> stk;
stk.push(1);
const long long limit = (1LL << 32);
while (l--) {
string s;
cin >> s;
if (s == "add") {
x += stk.top();
if (x >= limit) {
cout << "OVERFLOW!!!" << '\n';
exit(0);
}
} else if (s == "end") {
stk.pop();
} else {
int n;
cin >> n;
stk.push(min(limit, stk.top() * n));
}
}
cout << x << '\n';
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T bigmod(T p, T e, T M) {
long long int ret = 1;
for (; e > 0; e >>= 1) {
if (e & 1) ret = (ret * p) % M;
p = (p * p) % M;
}
return (T)ret;
}
template <class T>
inline T modinverse(T a, T M) {
return bigmod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long int x, n, a, l;
char s[10];
int main() {
scanf("%lld", &l);
long long int lim = pow(2, 32);
stack<long long int> st;
while (l--) {
scanf("%s", s);
if (s[0] == 'f') {
scanf("%lld", &a);
if (st.size() == 0)
n = 1;
else
n = st.top();
n = min(n * a, lim);
st.push(n);
} else if (s[0] == 'a') {
if (st.size() == 0)
x++;
else
x += st.top();
} else
st.pop();
if (x >= lim) return printf("OVERFLOW!!!\n"), 0;
}
printf("%lld\n", x);
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T bigmod(T p, T e, T M) {
long long int ret = 1;
for (; e > 0; e >>= 1) {
if (e & 1) ret = (ret * p) % M;
p = (p * p) % M;
}
return (T)ret;
}
template <class T>
inline T modinverse(T a, T M) {
return bigmod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long int x, n, a, l;
char s[10];
int main() {
scanf("%lld", &l);
long long int lim = pow(2, 32);
stack<long long int> st;
while (l--) {
scanf("%s", s);
if (s[0] == 'f') {
scanf("%lld", &a);
if (st.size() == 0)
n = 1;
else
n = st.top();
n = min(n * a, lim);
st.push(n);
} else if (s[0] == 'a') {
if (st.size() == 0)
x++;
else
x += st.top();
} else
st.pop();
if (x >= lim) return printf("OVERFLOW!!!\n"), 0;
}
printf("%lld\n", x);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
stack<long long> s;
char p[5];
const long long inf = pow(2, 32);
int main() {
ios::sync_with_stdio(false);
long long n, ans = 0, i = 1;
for (cin >> n, s.push(1); i <= n; i++) {
int x;
cin >> p;
if (p[0] == 'a')
ans += s.top();
else if (p[0] == 'f') {
cin >> x;
s.push(min(inf, s.top() * x));
} else
s.pop();
}
if (ans < inf)
cout << ans << endl;
else
cout << "OVERFLOW!!!" << endl;
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
stack<long long> s;
char p[5];
const long long inf = pow(2, 32);
int main() {
ios::sync_with_stdio(false);
long long n, ans = 0, i = 1;
for (cin >> n, s.push(1); i <= n; i++) {
int x;
cin >> p;
if (p[0] == 'a')
ans += s.top();
else if (p[0] == 'f') {
cin >> x;
s.push(min(inf, s.top() * x));
} else
s.pop();
}
if (ans < inf)
cout << ans << endl;
else
cout << "OVERFLOW!!!" << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct Node {
int t;
int v;
};
int l;
Node comm[500005];
const long long INF = (1LL << 32) - 1;
int main() {
char s[100];
while (scanf("%d", &l) != EOF) {
for (int i = 0; i < l; ++i) {
scanf("%s", s);
if ('a' == s[0]) {
comm[i].t = 1;
} else if ('f' == s[0]) {
comm[i].t = 2;
scanf("%d", &(comm[i].v));
} else if ('e' == s[0]) {
comm[i].t = 3;
} else {
printf("ERROR\n");
}
}
long long x = 0;
int p = 0;
stack<int> q;
stack<long long> value;
stack<long long> yv;
value.push(0);
while (p < l) {
long long y = value.top();
value.pop();
while (p < l && 1 == comm[p].t) {
++y;
++p;
}
if (p == l) value.push(y);
if (2 == comm[p].t) {
yv.push(y);
value.push(0);
q.push(comm[p].v);
++p;
} else if (3 == comm[p].t) {
long long vt = y * q.top();
q.pop();
vt += yv.top();
yv.pop();
value.push(vt);
++p;
}
if (value.top() > INF) {
x = -1;
break;
}
}
if (value.top() > INF) x = -1;
if (-1 == x) {
printf("OVERFLOW!!!\n");
} else {
x = value.top();
printf("%I64d\n", x);
}
}
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct Node {
int t;
int v;
};
int l;
Node comm[500005];
const long long INF = (1LL << 32) - 1;
int main() {
char s[100];
while (scanf("%d", &l) != EOF) {
for (int i = 0; i < l; ++i) {
scanf("%s", s);
if ('a' == s[0]) {
comm[i].t = 1;
} else if ('f' == s[0]) {
comm[i].t = 2;
scanf("%d", &(comm[i].v));
} else if ('e' == s[0]) {
comm[i].t = 3;
} else {
printf("ERROR\n");
}
}
long long x = 0;
int p = 0;
stack<int> q;
stack<long long> value;
stack<long long> yv;
value.push(0);
while (p < l) {
long long y = value.top();
value.pop();
while (p < l && 1 == comm[p].t) {
++y;
++p;
}
if (p == l) value.push(y);
if (2 == comm[p].t) {
yv.push(y);
value.push(0);
q.push(comm[p].v);
++p;
} else if (3 == comm[p].t) {
long long vt = y * q.top();
q.pop();
vt += yv.top();
yv.pop();
value.push(vt);
++p;
}
if (value.top() > INF) {
x = -1;
break;
}
}
if (value.top() > INF) x = -1;
if (-1 == x) {
printf("OVERFLOW!!!\n");
} else {
x = value.top();
printf("%I64d\n", x);
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1ll << 32;
int main() {
long long t;
cin >> t;
int flag = 0;
long long ans = 0;
long long multi = 1;
stack<long long> ss;
ss.push(1);
for (int i = 0; i < t; ++i) {
string s;
cin >> s;
if (s.compare("add") == 0) {
ans += min(ss.top(), INF);
}
if (s.compare("for") == 0) {
long long a;
cin >> a;
ss.push(min(a * ss.top(), INF));
}
if (s.compare("end") == 0) {
ss.pop();
}
if (ans >= INF) {
flag = 1;
}
}
if (ans >= INF || flag == 1) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << ans << endl;
}
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1ll << 32;
int main() {
long long t;
cin >> t;
int flag = 0;
long long ans = 0;
long long multi = 1;
stack<long long> ss;
ss.push(1);
for (int i = 0; i < t; ++i) {
string s;
cin >> s;
if (s.compare("add") == 0) {
ans += min(ss.top(), INF);
}
if (s.compare("for") == 0) {
long long a;
cin >> a;
ss.push(min(a * ss.top(), INF));
}
if (s.compare("end") == 0) {
ss.pop();
}
if (ans >= INF) {
flag = 1;
}
}
if (ans >= INF || flag == 1) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
stack<long long> khara;
long long ans = 0LL;
int main() {
int n;
cin >> n;
khara.push(1LL);
bool fucked = false;
while (n--) {
string s;
cin >> s;
if (s == "for") {
int x;
cin >> x;
khara.push(min(khara.top() * x, (1LL << 32)));
}
if (fucked) continue;
if (s == "add") ans += khara.top();
if (s == "end") khara.pop();
if (ans >= (1ll << 32)) fucked = true;
}
if (fucked)
cout << "OVERFLOW!!!" << endl;
else
cout << ans << endl;
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
You are given a function f written in some basic language. The function accepts an integer value, which is immediately written into some variable x. x is an integer variable and can be assigned values from 0 to 2^{32}-1. The function contains three types of commands:
* for n — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
After the execution of these commands, value of x is returned.
Every "for n" is matched with "end", thus the function is guaranteed to be valid. "for n" can be immediately followed by "end"."add" command can be outside of any for loops.
Notice that "add" commands might overflow the value of x! It means that the value of x becomes greater than 2^{32}-1 after some "add" command.
Now you run f(0) and wonder if the resulting value of x is correct or some overflow made it incorrect.
If overflow happened then output "OVERFLOW!!!", otherwise print the resulting value of x.
Input
The first line contains a single integer l (1 ≤ l ≤ 10^5) — the number of lines in the function.
Each of the next l lines contains a single command of one of three types:
* for n (1 ≤ n ≤ 100) — for loop;
* end — every command between "for n" and corresponding "end" is executed n times;
* add — adds 1 to x.
Output
If overflow happened during execution of f(0), then output "OVERFLOW!!!", otherwise print the resulting value of x.
Examples
Input
9
add
for 43
end
for 10
for 15
add
end
add
end
Output
161
Input
2
for 62
end
Output
0
Input
11
for 100
for 100
for 100
for 100
for 100
add
end
end
end
end
end
Output
OVERFLOW!!!
Note
In the first example the first "add" is executed 1 time, the second "add" is executed 150 times and the last "add" is executed 10 times. Note that "for n" can be immediately followed by "end" and that "add" can be outside of any for loops.
In the second example there are no commands "add", thus the returning value is 0.
In the third example "add" command is executed too many times, which causes x to go over 2^{32}-1.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
stack<long long> khara;
long long ans = 0LL;
int main() {
int n;
cin >> n;
khara.push(1LL);
bool fucked = false;
while (n--) {
string s;
cin >> s;
if (s == "for") {
int x;
cin >> x;
khara.push(min(khara.top() * x, (1LL << 32)));
}
if (fucked) continue;
if (s == "add") ans += khara.top();
if (s == "end") khara.pop();
if (ans >= (1ll << 32)) fucked = true;
}
if (fucked)
cout << "OVERFLOW!!!" << endl;
else
cout << ans << endl;
return 0;
}
``` |
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