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stringlengths 52
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#include <bits/stdc++.h>
using namespace std;
int main() {
deque<pair<long long, long long> > st;
int n;
long long val;
long long t;
cin >> n;
for (int i = 1; i <= n; i++) {
val = 0;
string s;
cin >> s;
if (s == "add")
if (st.size() && st.back().first == 1)
st.back().second++;
else
st.push_back({1, 1});
if (s == "for") {
cin >> t;
st.push_back({2, t});
}
if (s == "end") {
while (st.back().first != 2) {
val += st.back().second;
st.pop_back();
}
val *= st.back().second;
st.pop_back();
if (val >= (1LL << 32)) {
st.push_back({1, val});
break;
}
if (st.size() && st.back().first == 1)
st.back().second += val;
else
st.push_back({1, val});
}
}
if (st.back().second >= (1LL << 32))
cout << "OVERFLOW!!!";
else
cout << st.back().second;
}
| ### 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 main() {
deque<pair<long long, long long> > st;
int n;
long long val;
long long t;
cin >> n;
for (int i = 1; i <= n; i++) {
val = 0;
string s;
cin >> s;
if (s == "add")
if (st.size() && st.back().first == 1)
st.back().second++;
else
st.push_back({1, 1});
if (s == "for") {
cin >> t;
st.push_back({2, t});
}
if (s == "end") {
while (st.back().first != 2) {
val += st.back().second;
st.pop_back();
}
val *= st.back().second;
st.pop_back();
if (val >= (1LL << 32)) {
st.push_back({1, val});
break;
}
if (st.size() && st.back().first == 1)
st.back().second += val;
else
st.push_back({1, val});
}
}
if (st.back().second >= (1LL << 32))
cout << "OVERFLOW!!!";
else
cout << st.back().second;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const char ch[3][4] = {"for", "end", "add"};
stack<int> s;
char c[4];
int n;
long long p, f[100001], fa[100001], tp[100001];
int main() {
scanf("%d", &n);
while (n--) {
scanf("%s", c);
if (!strcmp(c, ch[0])) {
scanf("%lld", &tp[++p]);
if (s.empty())
fa[p] = 0;
else
fa[p] = s.top();
s.push(p);
} else if (!strcmp(c, ch[1])) {
f[fa[s.top()]] += f[s.top()] * tp[s.top()];
if (f[fa[s.top()]] >= 1ll << 32) {
puts("OVERFLOW!!!");
return 0;
}
s.pop();
} else if (s.empty())
f[0]++;
else
f[s.top()]++;
}
if (f[0] >= 1ll << 32)
puts("OVERFLOW!!!");
else
printf("%lld", f[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;
const char ch[3][4] = {"for", "end", "add"};
stack<int> s;
char c[4];
int n;
long long p, f[100001], fa[100001], tp[100001];
int main() {
scanf("%d", &n);
while (n--) {
scanf("%s", c);
if (!strcmp(c, ch[0])) {
scanf("%lld", &tp[++p]);
if (s.empty())
fa[p] = 0;
else
fa[p] = s.top();
s.push(p);
} else if (!strcmp(c, ch[1])) {
f[fa[s.top()]] += f[s.top()] * tp[s.top()];
if (f[fa[s.top()]] >= 1ll << 32) {
puts("OVERFLOW!!!");
return 0;
}
s.pop();
} else if (s.empty())
f[0]++;
else
f[s.top()]++;
}
if (f[0] >= 1ll << 32)
puts("OVERFLOW!!!");
else
printf("%lld", f[0]);
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long ans = 0;
long long l = 0;
long long f(long long n) {
long long lans = 0;
string s;
cin >> s;
l--;
while (!(s[0] == 'e')) {
if (s[0] == 'a') {
lans++;
}
if (s[0] == 'f') {
long long ii;
cin >> ii;
lans += f(ii);
if (lans >= 4294967296) {
cout << "OVERFLOW!!!";
exit(0);
}
}
cin >> s;
l--;
}
lans *= n;
return (lans);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int TESTS = 1;
while (TESTS--) {
cin >> l;
long long i, j, k, m, n;
while (l) {
l--;
string s;
cin >> s;
if (s[0] == 'a') {
ans++;
}
if (s[0] == 'f') {
cin >> i;
long long ii;
ii = f(i);
ans += ii;
if (ans >= 4294967296) {
cout << "OVERFLOW!!!";
exit(0);
}
}
}
if (ans >= 4294967296 || ans < 0) {
cout << "OVERFLOW!!!";
exit(0);
}
cout << ans;
}
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;
long long ans = 0;
long long l = 0;
long long f(long long n) {
long long lans = 0;
string s;
cin >> s;
l--;
while (!(s[0] == 'e')) {
if (s[0] == 'a') {
lans++;
}
if (s[0] == 'f') {
long long ii;
cin >> ii;
lans += f(ii);
if (lans >= 4294967296) {
cout << "OVERFLOW!!!";
exit(0);
}
}
cin >> s;
l--;
}
lans *= n;
return (lans);
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int TESTS = 1;
while (TESTS--) {
cin >> l;
long long i, j, k, m, n;
while (l) {
l--;
string s;
cin >> s;
if (s[0] == 'a') {
ans++;
}
if (s[0] == 'f') {
cin >> i;
long long ii;
ii = f(i);
ans += ii;
if (ans >= 4294967296) {
cout << "OVERFLOW!!!";
exit(0);
}
}
}
if (ans >= 4294967296 || ans < 0) {
cout << "OVERFLOW!!!";
exit(0);
}
cout << ans;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long cnd = (1ll << 32) - 1;
long long x = 0;
void check() {
if (x > cnd) {
cout << "OVERFLOW!!!\n";
exit(0);
}
}
int main() {
long long i, j, k, l, n, test_case, m, r;
test_case = 1;
cin >> test_case;
stack<long long> STA;
long long cur = 1;
string str;
int cnt = 0;
bool fl = 0;
while (test_case--) {
cin >> str;
if (str == "add") {
x += cur;
if (fl) {
cout << "OVERFLOW!!!";
exit(0);
}
check();
} else if (str == "for") {
cin >> j;
if (!fl) {
STA.push(j);
cur = cur * j;
} else {
cnt++;
}
if (cur > cnd) {
fl = 1;
}
} else {
if (cnt) {
cnt--;
} else {
cur = cur / (STA.top());
STA.pop();
}
if (cnt == 0) {
fl = 0;
}
}
}
cout << x << "\n";
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 cnd = (1ll << 32) - 1;
long long x = 0;
void check() {
if (x > cnd) {
cout << "OVERFLOW!!!\n";
exit(0);
}
}
int main() {
long long i, j, k, l, n, test_case, m, r;
test_case = 1;
cin >> test_case;
stack<long long> STA;
long long cur = 1;
string str;
int cnt = 0;
bool fl = 0;
while (test_case--) {
cin >> str;
if (str == "add") {
x += cur;
if (fl) {
cout << "OVERFLOW!!!";
exit(0);
}
check();
} else if (str == "for") {
cin >> j;
if (!fl) {
STA.push(j);
cur = cur * j;
} else {
cnt++;
}
if (cur > cnd) {
fl = 1;
}
} else {
if (cnt) {
cnt--;
} else {
cur = cur / (STA.top());
STA.pop();
}
if (cnt == 0) {
fl = 0;
}
}
}
cout << x << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
x = 0;
int flag = 1;
char c = getchar();
while (!isdigit(c)) {
if (c == '-') flag = -1;
c = getchar();
}
while (isdigit(c)) {
x = x * 10 + c - '0';
c = getchar();
}
x = x * flag;
}
const int maxn = 1e5 + 10;
int n, r;
string s;
long long x;
stack<int> q;
bool inq[maxn];
long long dfs() {
if (r >= n) return 1LL;
long long ret = 0;
if (ret >= (1LL << 32)) {
printf("OVERFLOW!!!\n");
exit(0);
}
long long a;
while (r < n) {
r++;
cin >> s;
if (ret >= (1LL << 32)) {
printf("OVERFLOW!!!\n");
exit(0);
}
if (s == "add") ret++;
if (s == "end") {
inq[q.top()] = 0;
q.pop();
return ret;
}
if (s == "for") {
int now = r;
read(a);
q.push(r);
inq[now] = 1;
ret = ret + a * dfs();
while (inq[now]) {
ret = ret + a * dfs();
if (ret >= (1LL << 32)) {
printf("OVERFLOW!!!\n");
exit(0);
}
}
return ret;
}
}
return ret;
}
int main() {
read(n);
while (r < n) {
x += dfs();
if (x >= (1LL << 32)) {
printf("OVERFLOW!!!\n");
return 0;
}
}
if (x >= 0)
printf("%lld\n", x);
else
printf("OVERFLOW!!!\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;
template <typename T>
void read(T &x) {
x = 0;
int flag = 1;
char c = getchar();
while (!isdigit(c)) {
if (c == '-') flag = -1;
c = getchar();
}
while (isdigit(c)) {
x = x * 10 + c - '0';
c = getchar();
}
x = x * flag;
}
const int maxn = 1e5 + 10;
int n, r;
string s;
long long x;
stack<int> q;
bool inq[maxn];
long long dfs() {
if (r >= n) return 1LL;
long long ret = 0;
if (ret >= (1LL << 32)) {
printf("OVERFLOW!!!\n");
exit(0);
}
long long a;
while (r < n) {
r++;
cin >> s;
if (ret >= (1LL << 32)) {
printf("OVERFLOW!!!\n");
exit(0);
}
if (s == "add") ret++;
if (s == "end") {
inq[q.top()] = 0;
q.pop();
return ret;
}
if (s == "for") {
int now = r;
read(a);
q.push(r);
inq[now] = 1;
ret = ret + a * dfs();
while (inq[now]) {
ret = ret + a * dfs();
if (ret >= (1LL << 32)) {
printf("OVERFLOW!!!\n");
exit(0);
}
}
return ret;
}
}
return ret;
}
int main() {
read(n);
while (r < n) {
x += dfs();
if (x >= (1LL << 32)) {
printf("OVERFLOW!!!\n");
return 0;
}
}
if (x >= 0)
printf("%lld\n", x);
else
printf("OVERFLOW!!!\n");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const long long inf = (1ll << 32);
stack<long long> s;
long long ans, l, tp;
string s1;
int main() {
cin >> l;
s.push(1ll);
for (int i = 1; i <= l; i++) {
cin >> s1;
if (s1 == "end") s.pop();
if (s1 == "add") ans += s.top();
if (s1 == "for") {
cin >> tp;
s.push(min(inf, tp * s.top()));
}
}
if (ans < inf)
cout << ans << endl;
else
cout << "OVERFLOW!!!" << 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 int maxn = 1e5 + 5;
const long long inf = (1ll << 32);
stack<long long> s;
long long ans, l, tp;
string s1;
int main() {
cin >> l;
s.push(1ll);
for (int i = 1; i <= l; i++) {
cin >> s1;
if (s1 == "end") s.pop();
if (s1 == "add") ans += s.top();
if (s1 == "for") {
cin >> tp;
s.push(min(inf, tp * s.top()));
}
}
if (ans < inf)
cout << ans << endl;
else
cout << "OVERFLOW!!!" << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long inf = 4294967296;
int main() {
int t, n, flag = 0;
while (~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;
while (~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 int N = 1e5 + 3;
const long long Mxn = (1ll << 32) - 1;
int l;
long long res;
int main() {
scanf("%d", &l);
static long long stk[N];
stk[0] = 1ll;
while (l--) {
static char ch[21];
static int h, n;
scanf("%s", ch + 1);
switch (ch[1]) {
case 'f': {
scanf("%lld", &stk[++h]);
if (stk[h - 1] == -1)
stk[h] = -1;
else if (1ll * stk[h] * stk[h - 1] > Mxn)
stk[h] = -1;
else
stk[h] = stk[h] * stk[h - 1];
break;
}
case 'a': {
if (res != -1) {
res += stk[h];
if (res > Mxn || stk[h] == -1) res = -1;
}
break;
}
case 'e': {
h--;
break;
}
}
}
if (res == -1)
puts("OVERFLOW!!!");
else
printf("%lld\n", res);
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 int N = 1e5 + 3;
const long long Mxn = (1ll << 32) - 1;
int l;
long long res;
int main() {
scanf("%d", &l);
static long long stk[N];
stk[0] = 1ll;
while (l--) {
static char ch[21];
static int h, n;
scanf("%s", ch + 1);
switch (ch[1]) {
case 'f': {
scanf("%lld", &stk[++h]);
if (stk[h - 1] == -1)
stk[h] = -1;
else if (1ll * stk[h] * stk[h - 1] > Mxn)
stk[h] = -1;
else
stk[h] = stk[h] * stk[h - 1];
break;
}
case 'a': {
if (res != -1) {
res += stk[h];
if (res > Mxn || stk[h] == -1) res = -1;
}
break;
}
case 'e': {
h--;
break;
}
}
}
if (res == -1)
puts("OVERFLOW!!!");
else
printf("%lld\n", res);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long l, i, j, k, c, cur[100010], a[100010], m, MAX;
string s;
int main() {
cur[0] = m = 0;
MAX = (1ll << 32);
for (cin >> l; l > 0; l--) {
cin >> s;
if (s[0] == 'a') {
cur[m]++;
} else if (s[0] == 'e') {
m--;
c = cur[m + 1] * a[m];
if (c >= MAX) {
printf("OVERFLOW!!!\n");
return 0;
}
cur[m] += c;
} else if (s[0] == 'f') {
scanf("%I64d", &a[m++]);
cur[m] = 0;
}
if (cur[m] >= MAX) {
printf("OVERFLOW!!!\n");
return 0;
}
}
cout << cur[0] << endl;
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 l, i, j, k, c, cur[100010], a[100010], m, MAX;
string s;
int main() {
cur[0] = m = 0;
MAX = (1ll << 32);
for (cin >> l; l > 0; l--) {
cin >> s;
if (s[0] == 'a') {
cur[m]++;
} else if (s[0] == 'e') {
m--;
c = cur[m + 1] * a[m];
if (c >= MAX) {
printf("OVERFLOW!!!\n");
return 0;
}
cur[m] += c;
} else if (s[0] == 'f') {
scanf("%I64d", &a[m++]);
cur[m] = 0;
}
if (cur[m] >= MAX) {
printf("OVERFLOW!!!\n");
return 0;
}
}
cout << cur[0] << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <class T>
bool mini(T &a, T b) {
return a > b ? (a = b, true) : false;
}
template <class T>
bool maxi(T &a, T b) {
return a < b ? (a = b, true) : false;
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.precision(10);
cout << fixed;
int n;
cin >> n;
bool ok = 1;
vector<long long> st;
long long ans = 0;
long long ax = (1ll << 32) - 1;
for (int i = 0; i < (int)(n); i++) {
string s;
cin >> s;
if (s == "for") {
int x;
cin >> x;
st.push_back(min(x * (st.size() ? st.back() : 1), ax + 1));
} else if (s == "end")
st.pop_back();
else {
ans += st.size() ? st.back() : 1;
if (ans > ax) {
ok = 0;
break;
}
}
}
if (ok)
cout << ans << "\n";
else
cout << "OVERFLOW!!!"
<< "\n";
}
| ### 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;
template <class T>
bool mini(T &a, T b) {
return a > b ? (a = b, true) : false;
}
template <class T>
bool maxi(T &a, T b) {
return a < b ? (a = b, true) : false;
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.precision(10);
cout << fixed;
int n;
cin >> n;
bool ok = 1;
vector<long long> st;
long long ans = 0;
long long ax = (1ll << 32) - 1;
for (int i = 0; i < (int)(n); i++) {
string s;
cin >> s;
if (s == "for") {
int x;
cin >> x;
st.push_back(min(x * (st.size() ? st.back() : 1), ax + 1));
} else if (s == "end")
st.pop_back();
else {
ans += st.size() ? st.back() : 1;
if (ans > ax) {
ok = 0;
break;
}
}
}
if (ok)
cout << ans << "\n";
else
cout << "OVERFLOW!!!"
<< "\n";
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline int get() {
char ch = getchar();
for (; ch < 'a' || ch > 'z'; ch = getchar())
;
getchar(), getchar();
if (ch == 'a') return 0;
if (ch == 'e') return 2;
return 1;
}
inline int read() {
int x = 0;
char ch = getchar();
for (; ch < '0' || ch > '9'; ch = getchar())
;
for (; ch >= '0' && ch <= '9'; ch = getchar())
x = (x << 3) + (x << 1) + (ch ^ 48);
return x;
}
bool flag;
long long a[100010];
int st[100010], R[100010], p[100010], n;
long long dfs(int l, int r) {
if (l > r) return 0;
if (l == r) return !p[l];
long long s = 0;
for (int i = l + 1; i < r; i++)
if (!p[i])
s++;
else
s += dfs(i, R[i]), i = R[i];
if (s > 4294967295ll || s * a[l] > 4294967295ll) flag = 1;
return s * a[l];
}
int main() {
n = read();
for (int i = 1, top = 0; i <= n; i++) {
p[i] = get();
if (p[i] == 1) a[i] = read(), st[++top] = i;
if (p[i] == 2) R[st[top--]] = i;
}
p[0] = 1;
a[0] = 1;
long long ans = dfs(0, n + 1);
if (flag)
printf("OVERFLOW!!!");
else
printf("%lld", 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;
inline int get() {
char ch = getchar();
for (; ch < 'a' || ch > 'z'; ch = getchar())
;
getchar(), getchar();
if (ch == 'a') return 0;
if (ch == 'e') return 2;
return 1;
}
inline int read() {
int x = 0;
char ch = getchar();
for (; ch < '0' || ch > '9'; ch = getchar())
;
for (; ch >= '0' && ch <= '9'; ch = getchar())
x = (x << 3) + (x << 1) + (ch ^ 48);
return x;
}
bool flag;
long long a[100010];
int st[100010], R[100010], p[100010], n;
long long dfs(int l, int r) {
if (l > r) return 0;
if (l == r) return !p[l];
long long s = 0;
for (int i = l + 1; i < r; i++)
if (!p[i])
s++;
else
s += dfs(i, R[i]), i = R[i];
if (s > 4294967295ll || s * a[l] > 4294967295ll) flag = 1;
return s * a[l];
}
int main() {
n = read();
for (int i = 1, top = 0; i <= n; i++) {
p[i] = get();
if (p[i] == 1) a[i] = read(), st[++top] = i;
if (p[i] == 2) R[st[top--]] = i;
}
p[0] = 1;
a[0] = 1;
long long ans = dfs(0, n + 1);
if (flag)
printf("OVERFLOW!!!");
else
printf("%lld", ans);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s;
long long l;
long long result = 0;
stack<long long> store;
cin >> l;
while (l > 0) {
cin >> s;
if (s == "add") {
result += 1;
} else if (s == "end" && !store.empty()) {
if (result > 4294967295) {
cout << "OVERFLOW!!!";
return 0;
}
result *= store.top();
store.pop();
result += store.top();
store.pop();
} else if (s == "for") {
long long loop;
cin >> loop;
store.push(result);
store.push(loop);
result = 0;
}
l--;
}
if (result <= 4294967295)
cout << result;
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() {
string s;
long long l;
long long result = 0;
stack<long long> store;
cin >> l;
while (l > 0) {
cin >> s;
if (s == "add") {
result += 1;
} else if (s == "end" && !store.empty()) {
if (result > 4294967295) {
cout << "OVERFLOW!!!";
return 0;
}
result *= store.top();
store.pop();
result += store.top();
store.pop();
} else if (s == "for") {
long long loop;
cin >> loop;
store.push(result);
store.push(loop);
result = 0;
}
l--;
}
if (result <= 4294967295)
cout << result;
else
cout << "OVERFLOW!!!";
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using res_t = uint32_t;
string op;
int l, n;
bool overflow = false;
res_t calc_for() {
res_t res = 0;
for (; l--;) {
res_t prev = res;
cin >> op;
if (op[0] == 'a')
++res;
else if (op[0] == 'f') {
res_t _t;
cin >> _t;
ll llres = res + (ll)_t * calc_for();
if (llres >= (1LL << 32)) overflow = true;
res = (res_t)llres;
} else if (op[0] == 'e')
break;
if (prev > res) overflow = true;
}
return res;
}
int main() {
cin >> l;
res_t res = calc_for();
if (overflow)
cout << "OVERFLOW!!!" << endl;
else
cout << res << 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;
using ll = long long;
using res_t = uint32_t;
string op;
int l, n;
bool overflow = false;
res_t calc_for() {
res_t res = 0;
for (; l--;) {
res_t prev = res;
cin >> op;
if (op[0] == 'a')
++res;
else if (op[0] == 'f') {
res_t _t;
cin >> _t;
ll llres = res + (ll)_t * calc_for();
if (llres >= (1LL << 32)) overflow = true;
res = (res_t)llres;
} else if (op[0] == 'e')
break;
if (prev > res) overflow = true;
}
return res;
}
int main() {
cin >> l;
res_t res = calc_for();
if (overflow)
cout << "OVERFLOW!!!" << endl;
else
cout << res << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const unsigned long long inf = (1ll << 32) - 1;
char s[N][100];
unsigned int a[N];
int stk[N], top;
unsigned long long b[N];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%s", s[i]);
if (s[i][0] == 'f') {
scanf("%u", &a[i]);
}
}
unsigned long long ans = 0;
for (int i = 0; i < n; i++) {
if (s[i][0] == 'f') {
b[top] = 0;
stk[top++] = i;
} else if (s[i][0] == 'a') {
if (!top) {
ans++;
if (ans > inf) {
return 0 * printf("OVERFLOW!!!\n");
}
continue;
}
b[top - 1]++;
if (b[top - 1] > inf) {
return 0 * printf("OVERFLOW!!!\n");
}
} else {
if (b[top - 1] * a[stk[top - 1]] > inf) {
return 0 * printf("OVERFLOW!!!\n");
} else {
b[top - 1] *= a[stk[top - 1]];
if (top - 1) {
b[top - 2] += b[top - 1];
if (b[top - 2] > inf) {
return 0 * printf("OVERFLOW!!!\n");
}
} else {
ans += b[top - 1];
if (ans > inf) {
return 0 * printf("OVERFLOW!!!\n");
}
}
top--;
}
}
}
printf("%u\n", (unsigned int)ans);
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;
const int N = 1e5 + 5;
const unsigned long long inf = (1ll << 32) - 1;
char s[N][100];
unsigned int a[N];
int stk[N], top;
unsigned long long b[N];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%s", s[i]);
if (s[i][0] == 'f') {
scanf("%u", &a[i]);
}
}
unsigned long long ans = 0;
for (int i = 0; i < n; i++) {
if (s[i][0] == 'f') {
b[top] = 0;
stk[top++] = i;
} else if (s[i][0] == 'a') {
if (!top) {
ans++;
if (ans > inf) {
return 0 * printf("OVERFLOW!!!\n");
}
continue;
}
b[top - 1]++;
if (b[top - 1] > inf) {
return 0 * printf("OVERFLOW!!!\n");
}
} else {
if (b[top - 1] * a[stk[top - 1]] > inf) {
return 0 * printf("OVERFLOW!!!\n");
} else {
b[top - 1] *= a[stk[top - 1]];
if (top - 1) {
b[top - 2] += b[top - 1];
if (b[top - 2] > inf) {
return 0 * printf("OVERFLOW!!!\n");
}
} else {
ans += b[top - 1];
if (ans > inf) {
return 0 * printf("OVERFLOW!!!\n");
}
}
top--;
}
}
}
printf("%u\n", (unsigned int)ans);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MOD = (1LL << 32) - 1;
int n;
char op[20];
long long ans, now;
stack<int> sta;
int main() {
scanf("%d", &n);
now = 1;
bool flag = false;
int stat = -1;
int deep = 0;
while (n--) {
scanf("%s", op);
if (op[0] == 'a') {
if (flag) {
ans = MOD + 1;
break;
}
ans = (ans + now);
if (ans > MOD) {
break;
}
} else if (op[0] == 'f') {
++deep;
int k;
scanf("%d", &k);
if (!flag) {
now = now * k;
if (now > MOD) {
stat = deep;
flag = true;
}
sta.push(k);
}
} else {
if (flag && deep == stat) {
stat = -1;
flag = false;
}
--deep;
if (!flag) {
int top = sta.top();
sta.pop();
now /= top;
}
}
}
if (ans > MOD) {
printf("OVERFLOW!!!");
} else {
printf("%lld", ans);
}
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;
const long long MOD = (1LL << 32) - 1;
int n;
char op[20];
long long ans, now;
stack<int> sta;
int main() {
scanf("%d", &n);
now = 1;
bool flag = false;
int stat = -1;
int deep = 0;
while (n--) {
scanf("%s", op);
if (op[0] == 'a') {
if (flag) {
ans = MOD + 1;
break;
}
ans = (ans + now);
if (ans > MOD) {
break;
}
} else if (op[0] == 'f') {
++deep;
int k;
scanf("%d", &k);
if (!flag) {
now = now * k;
if (now > MOD) {
stat = deep;
flag = true;
}
sta.push(k);
}
} else {
if (flag && deep == stat) {
stat = -1;
flag = false;
}
--deep;
if (!flag) {
int top = sta.top();
sta.pop();
now /= top;
}
}
}
if (ans > MOD) {
printf("OVERFLOW!!!");
} else {
printf("%lld", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
long long n, two = 1;
for (long long i = 0; i < 32; ++i) two *= 2;
cin >> n;
stack<long long> st;
long long ans = 0, cur = 1, more = -1, more_val = 0;
bool overflow = false;
for (long long i = 0; i < n; ++i) {
string s;
cin >> s;
if (s == "add") {
if (more >= 0 || cur > two - 1 - ans) overflow = true;
ans += cur;
} else if (s == "end") {
if (more >= 0 && st.size() - 1 == more) {
more = -1;
cur = more_val;
}
cur /= st.top();
st.pop();
} else {
long long x;
cin >> x;
st.push(x);
cur *= x;
if (cur > two && more == -1) {
more = st.size() - 1;
more_val = cur;
}
}
}
if (overflow)
cout << "OVERFLOW!!!";
else
cout << ans;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
solve();
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;
void solve() {
long long n, two = 1;
for (long long i = 0; i < 32; ++i) two *= 2;
cin >> n;
stack<long long> st;
long long ans = 0, cur = 1, more = -1, more_val = 0;
bool overflow = false;
for (long long i = 0; i < n; ++i) {
string s;
cin >> s;
if (s == "add") {
if (more >= 0 || cur > two - 1 - ans) overflow = true;
ans += cur;
} else if (s == "end") {
if (more >= 0 && st.size() - 1 == more) {
more = -1;
cur = more_val;
}
cur /= st.top();
st.pop();
} else {
long long x;
cin >> x;
st.push(x);
cur *= x;
if (cur > two && more == -1) {
more = st.size() - 1;
more_val = cur;
}
}
}
if (overflow)
cout << "OVERFLOW!!!";
else
cout << ans;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1000000007;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
long long int l, k, multx = 1, num = 0;
cin >> l;
string second;
bool flag = 1;
long long int N = pow(2, 32) - 1, c;
deque<long long int> dq;
dq.push_back(1);
for (long long int i = 0; i < l; i++) {
cin >> second;
if (second == "for") {
cin >> k;
if (dq.empty())
dq.push_back(k);
else {
c = dq.back() * k;
if (c > N or c < 0)
dq.push_back(-1);
else
dq.push_back(c);
}
} else if (second == "add") {
if (dq.empty())
num++;
else {
num += dq.back();
if (dq.back() < 0 or num > N) flag = 0;
}
} else
dq.pop_back();
}
if (num < 0 or num > N) flag = 0;
flag ? cout << num << endl : cout << "OVERFLOW!!!\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;
const long long int mod = 1000000007;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
long long int l, k, multx = 1, num = 0;
cin >> l;
string second;
bool flag = 1;
long long int N = pow(2, 32) - 1, c;
deque<long long int> dq;
dq.push_back(1);
for (long long int i = 0; i < l; i++) {
cin >> second;
if (second == "for") {
cin >> k;
if (dq.empty())
dq.push_back(k);
else {
c = dq.back() * k;
if (c > N or c < 0)
dq.push_back(-1);
else
dq.push_back(c);
}
} else if (second == "add") {
if (dq.empty())
num++;
else {
num += dq.back();
if (dq.back() < 0 or num > N) flag = 0;
}
} else
dq.pop_back();
}
if (num < 0 or num > N) flag = 0;
flag ? cout << num << endl : cout << "OVERFLOW!!!\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = (1LL << 32) - 1;
int com;
long long x;
string s;
stack<long long> cur;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> com;
x = 0LL;
long long mult = 1LL;
long long y;
int pend = 0;
while (com--) {
cin >> s;
if (s[0] == 'f') {
cin >> y;
if (pend <= 0) mult *= y;
if (mult > MAXN) pend++;
cur.push(y);
} else if (s[0] == 'e') {
if (pend <= 1) mult /= cur.top();
cur.pop();
pend = max(0, pend - 1);
} else {
x += mult;
if (x > MAXN) {
cout << "OVERFLOW!!!\n";
return 0;
}
}
}
cout << x << 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 long long MAXN = (1LL << 32) - 1;
int com;
long long x;
string s;
stack<long long> cur;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> com;
x = 0LL;
long long mult = 1LL;
long long y;
int pend = 0;
while (com--) {
cin >> s;
if (s[0] == 'f') {
cin >> y;
if (pend <= 0) mult *= y;
if (mult > MAXN) pend++;
cur.push(y);
} else if (s[0] == 'e') {
if (pend <= 1) mult /= cur.top();
cur.pop();
pend = max(0, pend - 1);
} else {
x += mult;
if (x > MAXN) {
cout << "OVERFLOW!!!\n";
return 0;
}
}
}
cout << x << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
long double PI = 4 * atan(1);
long long l, n, x, lim;
stack<long long> second;
string st;
int main() {
ios::sync_with_stdio(0), cin.tie(0);
second.push(1);
lim = pow(2, 32) - 1;
cin >> l;
while (l--) {
cin >> st;
if (st == "add") {
x += second.top();
if (x > lim) {
cout << "OVERFLOW!!!";
return 0;
}
} else if (st == "for") {
cin >> n;
second.push(min(second.top() * n, lim + 1));
} else {
second.pop();
}
}
if (x > lim) {
cout << "OVERFLOW!!!";
return 0;
}
cout << x;
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 MOD = 1000000007;
long double PI = 4 * atan(1);
long long l, n, x, lim;
stack<long long> second;
string st;
int main() {
ios::sync_with_stdio(0), cin.tie(0);
second.push(1);
lim = pow(2, 32) - 1;
cin >> l;
while (l--) {
cin >> st;
if (st == "add") {
x += second.top();
if (x > lim) {
cout << "OVERFLOW!!!";
return 0;
}
} else if (st == "for") {
cin >> n;
second.push(min(second.top() * n, lim + 1));
} else {
second.pop();
}
}
if (x > lim) {
cout << "OVERFLOW!!!";
return 0;
}
cout << x;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n;
cin >> n;
string s;
int c = 0;
char f[1000000];
vector<long long int> loop;
int j = 0;
int k = 0;
loop.push_back(1);
unsigned long long int b = 0;
unsigned long long int p = 1;
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "for") {
cin >> j;
p *= j;
if (p > 4294967295) {
p = 4294967296;
}
loop.push_back(p);
} else if (s == "add") {
b += p;
} else {
loop.pop_back();
p = loop.back();
}
if (b > 4294967295) {
cout << "OVERFLOW!!!";
return 0;
}
}
cout << b;
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() {
long long n;
cin >> n;
string s;
int c = 0;
char f[1000000];
vector<long long int> loop;
int j = 0;
int k = 0;
loop.push_back(1);
unsigned long long int b = 0;
unsigned long long int p = 1;
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "for") {
cin >> j;
p *= j;
if (p > 4294967295) {
p = 4294967296;
}
loop.push_back(p);
} else if (s == "add") {
b += p;
} else {
loop.pop_back();
p = loop.back();
}
if (b > 4294967295) {
cout << "OVERFLOW!!!";
return 0;
}
}
cout << b;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long n;
char ch[10];
struct node {
long long num, cnt;
};
stack<node> st;
int main() {
cin >> n;
st.push((node){1, 0});
while (n--) {
cin >> ch;
if (ch[0] == 'f') {
long long a;
cin >> a;
st.push((node){a, 0});
} else if (ch[0] == 'e') {
long long tmp = st.top().num * st.top().cnt;
st.pop();
st.top().cnt += tmp;
if (st.top().cnt >= 1ll * 4294967296) {
cout << "OVERFLOW!!!";
return 0;
}
} else
st.top().cnt++;
}
if (st.top().cnt >= 1ll * 4294967296) {
cout << "OVERFLOW!!!";
return 0;
}
cout << st.top().cnt;
}
| ### 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 n;
char ch[10];
struct node {
long long num, cnt;
};
stack<node> st;
int main() {
cin >> n;
st.push((node){1, 0});
while (n--) {
cin >> ch;
if (ch[0] == 'f') {
long long a;
cin >> a;
st.push((node){a, 0});
} else if (ch[0] == 'e') {
long long tmp = st.top().num * st.top().cnt;
st.pop();
st.top().cnt += tmp;
if (st.top().cnt >= 1ll * 4294967296) {
cout << "OVERFLOW!!!";
return 0;
}
} else
st.top().cnt++;
}
if (st.top().cnt >= 1ll * 4294967296) {
cout << "OVERFLOW!!!";
return 0;
}
cout << st.top().cnt;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INF64 = 3e18;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
long long n, tmp, add = 0;
stack<long long> s, q;
bool is = true;
string str;
cin >> n;
for (long long i = 0; i < n; ++i) {
cin >> str;
if (str[0] == 'a') {
add++;
} else if (str[0] == 'f') {
cin >> tmp;
s.push(tmp);
q.push(add);
add = 0;
} else if (str[0] == 'e') {
add *= s.top();
s.pop();
add += q.top();
q.pop();
}
if (add > ((1ll << 32ll) - 1ll)) {
is = false;
break;
}
}
if (is) {
cout << add;
} else {
cout << "OVERFLOW!!!";
}
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 int INF = 0x3f3f3f3f;
const long long INF64 = 3e18;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
long long n, tmp, add = 0;
stack<long long> s, q;
bool is = true;
string str;
cin >> n;
for (long long i = 0; i < n; ++i) {
cin >> str;
if (str[0] == 'a') {
add++;
} else if (str[0] == 'f') {
cin >> tmp;
s.push(tmp);
q.push(add);
add = 0;
} else if (str[0] == 'e') {
add *= s.top();
s.pop();
add += q.top();
q.pop();
}
if (add > ((1ll << 32ll) - 1ll)) {
is = false;
break;
}
}
if (is) {
cout << add;
} else {
cout << "OVERFLOW!!!";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
cerr << *it << " = " << a << "\n";
err(++it, args...);
}
long long int N = 100005;
long long int inf = 4294967296;
bool visited[100005];
vector<long long int> v(N);
vector<long long int> we(N);
long long int dfs(long long int x) {
long long int ans = 0, end_cnt = 0;
for (int i = x + 1;; i++) {
if (v[i] == -1)
ans++;
else if (v[i] > 0) {
long long int temp = dfs(i);
ans += temp % inf;
i = temp / inf;
} else {
if (ans * v[x] >= inf) {
cout << "OVERFLOW!!!"
<< "\n";
exit(0);
}
return ans * v[x] + inf * i;
}
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin.tie(NULL);
long long int n, m, i, j, k, ans = 0;
cin >> n;
v[0] = 1;
v[n + 1] = -2;
string s;
for (i = 0; i < 100005; i++) visited[i] = false;
for (i = 1; i <= n; i++) {
cin >> s;
if (s == "add") {
v[i] = -1;
} else if (s == "for") {
cin >> v[i];
} else
v[i] = -2;
}
cout << dfs(0) % inf << "\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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
cerr << *it << " = " << a << "\n";
err(++it, args...);
}
long long int N = 100005;
long long int inf = 4294967296;
bool visited[100005];
vector<long long int> v(N);
vector<long long int> we(N);
long long int dfs(long long int x) {
long long int ans = 0, end_cnt = 0;
for (int i = x + 1;; i++) {
if (v[i] == -1)
ans++;
else if (v[i] > 0) {
long long int temp = dfs(i);
ans += temp % inf;
i = temp / inf;
} else {
if (ans * v[x] >= inf) {
cout << "OVERFLOW!!!"
<< "\n";
exit(0);
}
return ans * v[x] + inf * i;
}
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin.tie(NULL);
long long int n, m, i, j, k, ans = 0;
cin >> n;
v[0] = 1;
v[n + 1] = -2;
string s;
for (i = 0; i < 100005; i++) visited[i] = false;
for (i = 1; i <= n; i++) {
cin >> s;
if (s == "add") {
v[i] = -1;
} else if (s == "for") {
cin >> v[i];
} else
v[i] = -2;
}
cout << dfs(0) % inf << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long OF = 1LL << 32;
long long LOOP(void) {
char s[5];
long long x = 0, ans = 0;
while (scanf("%s", s) != EOF && ans < OF) {
if (s[0] == 'a')
++ans;
else if (s[0] == 'f') {
scanf("%lld", &x);
x *= LOOP();
ans += x;
} else if (s[0] == 'e')
return ans;
}
printf("OVERFLOW!!!");
exit(0);
}
int main(void) {
long long x = 0, ans = 0;
char s[5];
scanf("%lld", &x);
while (scanf("%s", s) != EOF && ans < OF) {
if (s[0] == 'a') {
++ans;
continue;
}
scanf("%lld", &x);
x *= LOOP();
ans += x;
}
if (ans >= OF) {
printf("OVERFLOW!!!");
return 0;
}
printf("%lld", ans);
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 OF = 1LL << 32;
long long LOOP(void) {
char s[5];
long long x = 0, ans = 0;
while (scanf("%s", s) != EOF && ans < OF) {
if (s[0] == 'a')
++ans;
else if (s[0] == 'f') {
scanf("%lld", &x);
x *= LOOP();
ans += x;
} else if (s[0] == 'e')
return ans;
}
printf("OVERFLOW!!!");
exit(0);
}
int main(void) {
long long x = 0, ans = 0;
char s[5];
scanf("%lld", &x);
while (scanf("%s", s) != EOF && ans < OF) {
if (s[0] == 'a') {
++ans;
continue;
}
scanf("%lld", &x);
x *= LOOP();
ans += x;
}
if (ans >= OF) {
printf("OVERFLOW!!!");
return 0;
}
printf("%lld", ans);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T sqr(T a) {
return a * a;
}
template <typename T>
bool uax(T &a, const T b) {
if (a < b) {
a = b;
return true;
}
return false;
}
template <typename T>
bool uin(T &a, const T b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
in >> p.first >> p.second;
return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, pair<T1, T2> &p) {
out << p.first << ' ' << p.second;
return out;
}
mt19937_64 gen_rand;
const int N = (int)4e5 + 7;
const int N_ = (int)1007;
const int MAXN = (int)5e6 + 7;
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9 + 7;
const int ITER = (int)20;
const int LOG = 30;
const int SIZE = 1100;
const int ALPH = 256;
const int DIGIT = 10;
const int BLOCK = 350;
const int LIGHT = 350;
const long long LINF = (int64_t)2e18;
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 1e-8;
int n, cnt[200];
long long x = 0;
void read() { cin >> n; }
long long get() {
long long res = 1;
for (int i = 0; i < 100; i++) {
for (int j = 0; j < cnt[i]; j++) {
res *= i;
if (res >= (1LL << 32)) return -1;
}
}
return res;
}
void upd(int x, int sign) {
for (int i = 2; i * i <= x; i++) {
while (x % i == 0) cnt[i] += sign, x /= i;
}
if (x != 1) cnt[x] += sign;
}
void solve() {
int cnt[200] = {};
vector<int> fr;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] == 'f') {
int val;
cin >> val;
upd(val, 1);
fr.push_back(val);
}
if (s[0] == 'e') {
upd(fr.back(), -1);
fr.pop_back();
}
if (s[0] == 'a') {
long long add = get();
if (add == -1) {
cout << "OVERFLOW!!!";
exit(0);
}
x += max(add, 1LL);
if (x >= (1LL << 32)) {
cout << "OVERFLOW!!!";
exit(0);
}
}
}
cout << x;
}
double max_time = 0;
void gen() {}
void stupid() {}
void write() {}
void stress() {}
void check_time() {
double start = clock();
read(), solve();
cout << "\nTime is " << (clock() - start) / 1000.0 << "s\n";
}
void check_rand_time() {
gen_rand.seed(int(clock()));
double start = clock();
gen(), solve();
double cur_time = (clock() - start) / 1000.0;
uax(max_time, cur_time);
cerr << "cur_time = " << cur_time << " || "
<< "max_time = " << max_time << "\n";
}
int main() {
cout << fixed << setprecision(20);
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
gen_rand.seed(time(0));
read();
solve();
}
| ### 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;
template <typename T>
T sqr(T a) {
return a * a;
}
template <typename T>
bool uax(T &a, const T b) {
if (a < b) {
a = b;
return true;
}
return false;
}
template <typename T>
bool uin(T &a, const T b) {
if (a > b) {
a = b;
return true;
}
return false;
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
in >> p.first >> p.second;
return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, pair<T1, T2> &p) {
out << p.first << ' ' << p.second;
return out;
}
mt19937_64 gen_rand;
const int N = (int)4e5 + 7;
const int N_ = (int)1007;
const int MAXN = (int)5e6 + 7;
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9 + 7;
const int ITER = (int)20;
const int LOG = 30;
const int SIZE = 1100;
const int ALPH = 256;
const int DIGIT = 10;
const int BLOCK = 350;
const int LIGHT = 350;
const long long LINF = (int64_t)2e18;
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 1e-8;
int n, cnt[200];
long long x = 0;
void read() { cin >> n; }
long long get() {
long long res = 1;
for (int i = 0; i < 100; i++) {
for (int j = 0; j < cnt[i]; j++) {
res *= i;
if (res >= (1LL << 32)) return -1;
}
}
return res;
}
void upd(int x, int sign) {
for (int i = 2; i * i <= x; i++) {
while (x % i == 0) cnt[i] += sign, x /= i;
}
if (x != 1) cnt[x] += sign;
}
void solve() {
int cnt[200] = {};
vector<int> fr;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] == 'f') {
int val;
cin >> val;
upd(val, 1);
fr.push_back(val);
}
if (s[0] == 'e') {
upd(fr.back(), -1);
fr.pop_back();
}
if (s[0] == 'a') {
long long add = get();
if (add == -1) {
cout << "OVERFLOW!!!";
exit(0);
}
x += max(add, 1LL);
if (x >= (1LL << 32)) {
cout << "OVERFLOW!!!";
exit(0);
}
}
}
cout << x;
}
double max_time = 0;
void gen() {}
void stupid() {}
void write() {}
void stress() {}
void check_time() {
double start = clock();
read(), solve();
cout << "\nTime is " << (clock() - start) / 1000.0 << "s\n";
}
void check_rand_time() {
gen_rand.seed(int(clock()));
double start = clock();
gen(), solve();
double cur_time = (clock() - start) / 1000.0;
uax(max_time, cur_time);
cerr << "cur_time = " << cur_time << " || "
<< "max_time = " << max_time << "\n";
}
int main() {
cout << fixed << setprecision(20);
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
gen_rand.seed(time(0));
read();
solve();
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <typename... T>
void pp(T... args) {
((cout << args << " "), ...);
cout << "\n";
}
template <typename... T>
void po(T... args) {
((cout << args << " "), ...);
cout << "";
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int test = 1;
for (int tt = 1; tt <= test; tt++) {
long long int n;
cin >> n;
long long int val = pow(2, 32) - 1;
long long int sum = 0;
long long int ans = 0;
vector<long long int> Q;
for (long long int i = 0; i < n; i++) {
string s;
long long int k;
cin >> s;
if (s == "for") {
cin >> k;
if (!Q.empty()) k *= Q[Q.size() - 1];
if (k > val) k = -1;
Q.push_back(k);
}
if (s == "add") {
if (Q.size() != 0) {
long long int tot = Q[Q.size() - 1];
if (tot < 0) {
pp("OVERFLOW!!!");
return 0;
}
ans += tot;
if (ans > val) {
pp("OVERFLOW!!!");
return 0;
}
} else {
ans++;
if (ans > val) {
pp("OVERFLOW!!!");
return 0;
}
}
}
if (s == "end") {
Q.pop_back();
}
}
pp(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;
template <typename... T>
void pp(T... args) {
((cout << args << " "), ...);
cout << "\n";
}
template <typename... T>
void po(T... args) {
((cout << args << " "), ...);
cout << "";
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int test = 1;
for (int tt = 1; tt <= test; tt++) {
long long int n;
cin >> n;
long long int val = pow(2, 32) - 1;
long long int sum = 0;
long long int ans = 0;
vector<long long int> Q;
for (long long int i = 0; i < n; i++) {
string s;
long long int k;
cin >> s;
if (s == "for") {
cin >> k;
if (!Q.empty()) k *= Q[Q.size() - 1];
if (k > val) k = -1;
Q.push_back(k);
}
if (s == "add") {
if (Q.size() != 0) {
long long int tot = Q[Q.size() - 1];
if (tot < 0) {
pp("OVERFLOW!!!");
return 0;
}
ans += tot;
if (ans > val) {
pp("OVERFLOW!!!");
return 0;
}
} else {
ans++;
if (ans > val) {
pp("OVERFLOW!!!");
return 0;
}
}
}
if (s == "end") {
Q.pop_back();
}
}
pp(ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int TESTS = 1;
while (TESTS--) {
long long int n, x = 1, ans = 0, w = 0, q;
string s;
cin >> n;
long long int p = pow(2, 32) - 1;
stack<long long int> stackstack;
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "for") cin >> q;
if (s == "add" && stackstack.empty() == 1) {
if (ans + 1 > p) {
cout << "OVERFLOW!!!";
return 0;
} else
ans += 1;
} else if (s == "add") {
if (stackstack.top() == p + 1) {
cout << "OVERFLOW!!!";
return 0;
}
if (stackstack.top() + ans > p) {
cout << "OVERFLOW!!!";
return 0;
}
ans += stackstack.top();
} else if (s == "end") {
stackstack.pop();
} else if (s == "for") {
if (stackstack.empty())
x = 1;
else
x = stackstack.top();
if (x * q > p)
stackstack.push(p + 1);
else
stackstack.push(x * q);
}
}
cout << 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;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int TESTS = 1;
while (TESTS--) {
long long int n, x = 1, ans = 0, w = 0, q;
string s;
cin >> n;
long long int p = pow(2, 32) - 1;
stack<long long int> stackstack;
for (int i = 0; i < n; i++) {
cin >> s;
if (s == "for") cin >> q;
if (s == "add" && stackstack.empty() == 1) {
if (ans + 1 > p) {
cout << "OVERFLOW!!!";
return 0;
} else
ans += 1;
} else if (s == "add") {
if (stackstack.top() == p + 1) {
cout << "OVERFLOW!!!";
return 0;
}
if (stackstack.top() + ans > p) {
cout << "OVERFLOW!!!";
return 0;
}
ans += stackstack.top();
} else if (s == "end") {
stackstack.pop();
} else if (s == "for") {
if (stackstack.empty())
x = 1;
else
x = stackstack.top();
if (x * q > p)
stackstack.push(p + 1);
else
stackstack.push(x * q);
}
}
cout << ans;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
ifstream f1;
ofstream f2;
const long long ma = (1ll << 32);
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
f1.open("INP.txt");
f2.open("OUT.txt");
long long n;
cin >> n;
stack<long long> next;
next.push(1);
long long cur = 0;
for (long long i = 0; i < n; i++) {
string task;
cin >> task;
if (task == "add") {
cur += next.top();
} else if (task == "for") {
long long val;
cin >> val;
next.push(min(ma, val * next.top()));
} else {
next.pop();
}
}
if (cur >= ma) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << cur << 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;
ifstream f1;
ofstream f2;
const long long ma = (1ll << 32);
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
f1.open("INP.txt");
f2.open("OUT.txt");
long long n;
cin >> n;
stack<long long> next;
next.push(1);
long long cur = 0;
for (long long i = 0; i < n; i++) {
string task;
cin >> task;
if (task == "add") {
cur += next.top();
} else if (task == "for") {
long long val;
cin >> val;
next.push(min(ma, val * next.top()));
} else {
next.pop();
}
}
if (cur >= ma) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << cur << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) {
if (a > b) {
return a;
} else {
return b;
}
}
long long min(long long a, long long b) {
if (a < b) {
return a;
} else {
return b;
}
}
long long rand_int(long long l, long long r) {
static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
return uniform_int_distribution<long long>(l, r)(gen);
}
long long power(long long b, long long e) {
if (e == 0) return 1;
if (e % 2)
return ((b * power((b) * (b), (e - 1) / 2)));
else
return power((b) * (b), e / 2);
}
long long modpower(long long b, long long e, long long q) {
long long MOD = q;
if (e == 0) return 1;
if (e % 2)
return ((b % MOD) * modpower((b % MOD) * (b % MOD), (e - 1) / 2, q)) % MOD;
else
return modpower((b % MOD) * (b % MOD), e / 2, q) % MOD;
}
void dpv(vector<long long> v) {
for (long long i = 0; i < v.size(); i++) {
cout << v[i] << " ";
}
cout << endl;
}
void dpv(vector<pair<long long, long long> > v) {
for (long long i = 0; i < v.size(); i++) {
cout << v[i].first << " " << v[i].second << endl;
}
}
void dpv(set<long long> v) {
for (auto i : v) {
cout << i << " ";
}
cout << endl;
}
long long ceil1(long long x, long long y) {
long long r = x / y;
if (x % y == 0) {
return r;
} else {
return r + 1;
}
}
const long long MX = (1LL << 32) - 1;
void oblivious() {
long long n;
cin >> n;
string op;
stack<long long> rs;
rs.push(1ll);
long long res = 0;
for (long long i = 0; i < n; i++) {
cin >> op;
if (op == "add") {
res += rs.top();
if (res > MX) {
cout << "OVERFLOW!!!" << endl;
return;
}
} else if (op == "end") {
rs.pop();
} else {
long long x;
cin >> x;
rs.push(min(rs.top() * x, MX + 1));
}
}
cout << res << "\n";
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t = 1;
while (t--) {
oblivious();
}
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 max(long long a, long long b) {
if (a > b) {
return a;
} else {
return b;
}
}
long long min(long long a, long long b) {
if (a < b) {
return a;
} else {
return b;
}
}
long long rand_int(long long l, long long r) {
static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
return uniform_int_distribution<long long>(l, r)(gen);
}
long long power(long long b, long long e) {
if (e == 0) return 1;
if (e % 2)
return ((b * power((b) * (b), (e - 1) / 2)));
else
return power((b) * (b), e / 2);
}
long long modpower(long long b, long long e, long long q) {
long long MOD = q;
if (e == 0) return 1;
if (e % 2)
return ((b % MOD) * modpower((b % MOD) * (b % MOD), (e - 1) / 2, q)) % MOD;
else
return modpower((b % MOD) * (b % MOD), e / 2, q) % MOD;
}
void dpv(vector<long long> v) {
for (long long i = 0; i < v.size(); i++) {
cout << v[i] << " ";
}
cout << endl;
}
void dpv(vector<pair<long long, long long> > v) {
for (long long i = 0; i < v.size(); i++) {
cout << v[i].first << " " << v[i].second << endl;
}
}
void dpv(set<long long> v) {
for (auto i : v) {
cout << i << " ";
}
cout << endl;
}
long long ceil1(long long x, long long y) {
long long r = x / y;
if (x % y == 0) {
return r;
} else {
return r + 1;
}
}
const long long MX = (1LL << 32) - 1;
void oblivious() {
long long n;
cin >> n;
string op;
stack<long long> rs;
rs.push(1ll);
long long res = 0;
for (long long i = 0; i < n; i++) {
cin >> op;
if (op == "add") {
res += rs.top();
if (res > MX) {
cout << "OVERFLOW!!!" << endl;
return;
}
} else if (op == "end") {
rs.pop();
} else {
long long x;
cin >> x;
rs.push(min(rs.top() * x, MX + 1));
}
}
cout << res << "\n";
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long t = 1;
while (t--) {
oblivious();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 5;
const long long MAXN = ((long long)1 << 32) - 1;
pair<long long, long long> a[maxn];
int stk[maxn], top, R[maxn];
long long solve(int l, int r) {
long long sum = 0;
for (int i = l; i <= r; i++) {
if (a[i].first == 1) {
long long tmp = solve(i + 1, R[i]);
if (tmp < 0) return -1;
sum = sum + a[i].second * tmp;
if (sum > MAXN) return -1;
i = R[i];
} else if (a[i].first == 3) {
sum = sum + 1;
if (sum > MAXN) return -1;
}
}
return sum;
}
int main() {
long long l;
cin >> l;
for (int i = 1; i <= l; i++) {
string s;
cin >> s;
long long n;
if (s[0] == 'f') {
cin >> n;
a[i] = {1, n};
} else if (s[0] == 'e') {
a[i] = {2, 0};
} else {
a[i] = {3, 1};
}
}
for (int i = 1; i <= l; i++) {
if (a[i].first == 1) {
stk[++top] = i;
} else if (a[i].first == 2) {
R[stk[top]] = i;
top--;
}
}
long long ans = solve(1, l);
if (ans < 0 || ans > MAXN) {
cout << "OVERFLOW!!!" << endl;
} else {
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;
const int maxn = 3e5 + 5;
const long long MAXN = ((long long)1 << 32) - 1;
pair<long long, long long> a[maxn];
int stk[maxn], top, R[maxn];
long long solve(int l, int r) {
long long sum = 0;
for (int i = l; i <= r; i++) {
if (a[i].first == 1) {
long long tmp = solve(i + 1, R[i]);
if (tmp < 0) return -1;
sum = sum + a[i].second * tmp;
if (sum > MAXN) return -1;
i = R[i];
} else if (a[i].first == 3) {
sum = sum + 1;
if (sum > MAXN) return -1;
}
}
return sum;
}
int main() {
long long l;
cin >> l;
for (int i = 1; i <= l; i++) {
string s;
cin >> s;
long long n;
if (s[0] == 'f') {
cin >> n;
a[i] = {1, n};
} else if (s[0] == 'e') {
a[i] = {2, 0};
} else {
a[i] = {3, 1};
}
}
for (int i = 1; i <= l; i++) {
if (a[i].first == 1) {
stk[++top] = i;
} else if (a[i].first == 2) {
R[stk[top]] = i;
top--;
}
}
long long ans = solve(1, l);
if (ans < 0 || ans > MAXN) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
stack<long long> fors;
vector<pair<string, long long>> lines;
int n;
long long mpl = 1;
long long res = 0;
long long overflow_num = (1LL << 32LL) - 1;
int main() {
iostream::sync_with_stdio(false);
cin >> n;
for (int i = 0; i < n; i++) {
string cmd;
cin >> cmd;
if (cmd == "for") {
long long times;
cin >> times;
lines.push_back(make_pair("for", times));
} else {
lines.push_back(make_pair(cmd, -1));
}
}
for (int i = 0; i < n; i++) {
string cmd = lines[i].first;
if (cmd == "for") {
long long times = lines[i].second;
if (mpl * times > overflow_num) {
int opened = 1;
for (int j = i + 1; j < n; j++) {
if (lines[j].first == "for") {
opened++;
} else if (lines[j].first == "end") {
opened--;
if (opened == 0) {
i = j;
break;
}
} else if (lines[j].first == "add") {
cout << "OVERFLOW!!!" << endl;
return 0;
}
}
} else {
fors.push(mpl);
mpl *= times;
}
} else if (cmd == "end") {
if (!fors.empty()) {
mpl = fors.top();
fors.pop();
}
} else if (cmd == "add") {
res += mpl;
if (res > overflow_num) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
}
}
if (res <= overflow_num)
cout << res << endl;
else
cout << "OVERFLOW!!!" << endl;
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;
stack<long long> fors;
vector<pair<string, long long>> lines;
int n;
long long mpl = 1;
long long res = 0;
long long overflow_num = (1LL << 32LL) - 1;
int main() {
iostream::sync_with_stdio(false);
cin >> n;
for (int i = 0; i < n; i++) {
string cmd;
cin >> cmd;
if (cmd == "for") {
long long times;
cin >> times;
lines.push_back(make_pair("for", times));
} else {
lines.push_back(make_pair(cmd, -1));
}
}
for (int i = 0; i < n; i++) {
string cmd = lines[i].first;
if (cmd == "for") {
long long times = lines[i].second;
if (mpl * times > overflow_num) {
int opened = 1;
for (int j = i + 1; j < n; j++) {
if (lines[j].first == "for") {
opened++;
} else if (lines[j].first == "end") {
opened--;
if (opened == 0) {
i = j;
break;
}
} else if (lines[j].first == "add") {
cout << "OVERFLOW!!!" << endl;
return 0;
}
}
} else {
fors.push(mpl);
mpl *= times;
}
} else if (cmd == "end") {
if (!fors.empty()) {
mpl = fors.top();
fors.pop();
}
} else if (cmd == "add") {
res += mpl;
if (res > overflow_num) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
}
}
if (res <= overflow_num)
cout << res << endl;
else
cout << "OVERFLOW!!!" << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = unsigned long long int;
stack<ll> st, temp, over_stack;
ll lim = 4294967295;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n, i, j;
cin >> n;
ll ans = 0;
bool over = false, flag = false;
while (n--) {
string s;
cin >> s;
if (s[0] == 'f') {
ll val;
cin >> val;
if (flag) continue;
if (!st.size()) {
st.push(val);
} else {
if (over) {
over_stack.push(val);
continue;
}
if (val > (lim / st.top())) {
over = true;
over_stack.push(val);
continue;
}
ll u = st.top();
st.pop();
st.push(val * u);
}
temp.push(val);
} else if (s[0] == 'a') {
if (flag) continue;
if (over) {
flag = true;
continue;
}
if (!st.size()) {
ans++;
if (ans > lim) flag = true;
} else {
ans += (st.top());
if (ans > lim) flag = true;
}
} else {
if (flag) continue;
if (over) {
over_stack.pop();
if (!over_stack.size()) over = false;
continue;
}
ll u = st.top();
st.pop();
ll v = temp.top();
temp.pop();
u /= v;
st.push(u);
}
}
if (flag) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
cout << ans << endl;
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;
using ll = unsigned long long int;
stack<ll> st, temp, over_stack;
ll lim = 4294967295;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n, i, j;
cin >> n;
ll ans = 0;
bool over = false, flag = false;
while (n--) {
string s;
cin >> s;
if (s[0] == 'f') {
ll val;
cin >> val;
if (flag) continue;
if (!st.size()) {
st.push(val);
} else {
if (over) {
over_stack.push(val);
continue;
}
if (val > (lim / st.top())) {
over = true;
over_stack.push(val);
continue;
}
ll u = st.top();
st.pop();
st.push(val * u);
}
temp.push(val);
} else if (s[0] == 'a') {
if (flag) continue;
if (over) {
flag = true;
continue;
}
if (!st.size()) {
ans++;
if (ans > lim) flag = true;
} else {
ans += (st.top());
if (ans > lim) flag = true;
}
} else {
if (flag) continue;
if (over) {
over_stack.pop();
if (!over_stack.size()) over = false;
continue;
}
ll u = st.top();
st.pop();
ll v = temp.top();
temp.pop();
u /= v;
st.push(u);
}
}
if (flag) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
cout << ans << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct cmp {
bool operator()(long long a, long long b) { return (a > b); }
};
long long mod = 1000000007;
const long long N = 1ll << 32;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n;
cin >> n;
stack<long long> s;
s.push(1);
long long ans = 0;
for (long long i = 0; i < n; i++) {
string str;
cin >> str;
if (str == "for") {
long long x;
cin >> x;
s.push(min(s.top() * x, N));
} else if (str == "end") {
s.pop();
} else {
ans += s.top();
}
}
if (ans >= N)
cout << "OVERFLOW!!!";
else
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;
struct cmp {
bool operator()(long long a, long long b) { return (a > b); }
};
long long mod = 1000000007;
const long long N = 1ll << 32;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n;
cin >> n;
stack<long long> s;
s.push(1);
long long ans = 0;
for (long long i = 0; i < n; i++) {
string str;
cin >> str;
if (str == "for") {
long long x;
cin >> x;
s.push(min(s.top() * x, N));
} else if (str == "end") {
s.pop();
} else {
ans += s.top();
}
}
if (ans >= N)
cout << "OVERFLOW!!!";
else
cout << ans;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using db = double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
template <class T>
using V = vector<T>;
template <class T, size_t SZ>
using AR = array<T, SZ>;
const int MOD = 1e9 + 7;
const int MX = 2e5 + 5;
const ll INF = 1e18;
const ld PI = acos((ld)-1);
const int xd[4] = {1, 0, -1, 0}, yd[4] = {0, 1, 0, -1};
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
constexpr int pct(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return 31 - __builtin_clz(x); }
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
template <class T>
bool ckmin(T& a, const T& b) {
return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
return a < b ? a = b, 1 : 0;
}
template <class T, class U>
T fstTrue(T lo, T hi, U first) {
hi++;
assert(lo <= hi);
while (lo < hi) {
T mid = lo + (hi - lo) / 2;
first(mid) ? hi = mid : lo = mid + 1;
}
return lo;
}
template <class T, class U>
T lstTrue(T lo, T hi, U first) {
lo--;
assert(lo <= hi);
while (lo < hi) {
T mid = lo + (hi - lo + 1) / 2;
first(mid) ? lo = mid : hi = mid - 1;
}
return lo;
}
template <class T>
void remDup(vector<T>& v) {
sort(begin(v), end(v));
v.erase(unique(begin(v), end(v)), end(v));
}
template <class T, class U>
void erase(T& t, const U& u) {
auto it = t.find(u);
assert(it != end(t));
t.erase(u);
}
template <class T>
void re(complex<T>& c);
template <class T, class U>
void re(pair<T, U>& p);
template <class T>
void re(vector<T>& v);
template <class T, size_t SZ>
void re(AR<T, SZ>& a);
template <class T>
void re(T& x) {
cin >> x;
}
void re(db& d) {
str t;
re(t);
d = stod(t);
}
void re(ld& d) {
str t;
re(t);
d = stold(t);
}
template <class T, class... U>
void re(T& t, U&... u) {
re(t);
re(u...);
}
template <class T>
void re(complex<T>& c) {
T a, b;
re(a, b);
c = {a, b};
}
template <class T, class U>
void re(pair<T, U>& p) {
re(p.first, p.second);
}
template <class T>
void re(vector<T>& x) {
for (auto& a : x) re(a);
}
template <class T, size_t SZ>
void re(AR<T, SZ>& x) {
for (auto& a : x) re(a);
}
str to_string(char c) { return str(1, c); }
str to_string(const char* second) { return (str)second; }
str to_string(str second) { return second; }
str to_string(bool b) { return to_string((int)b); }
template <class T>
str to_string(complex<T> c) {
stringstream ss;
ss << c;
return ss.str();
}
str to_string(vector<bool> v) {
str res = "{";
for (int i = (0); i < ((int)(v).size()); ++i) res += char('0' + v[i]);
res += "}";
return res;
}
template <size_t SZ>
str to_string(bitset<SZ> b) {
str res = "";
for (int i = (0); i < (SZ); ++i) res += char('0' + b[i]);
return res;
}
template <class T, class U>
str to_string(pair<T, U> p);
template <class T>
str to_string(T v) {
bool fst = 1;
str res = "";
for (const auto& x : v) {
if (!fst) res += " ";
fst = 0;
res += to_string(x);
}
return res;
}
template <class T, class U>
str to_string(pair<T, U> p) {
return to_string(p.first) + " " + to_string(p.second);
}
template <class T>
void pr(T x) {
cout << to_string(x);
}
template <class T, class... U>
void pr(const T& t, const U&... u) {
pr(t);
pr(u...);
}
void ps() { pr("\n"); }
template <class T, class... U>
void ps(const T& t, const U&... u) {
pr(t);
if (sizeof...(u)) pr(" ");
ps(u...);
}
void DBG() { cerr << "]" << endl; }
template <class T, class... U>
void DBG(const T& t, const U&... u) {
cerr << to_string(t);
if (sizeof...(u)) cerr << ", ";
DBG(u...);
}
void setIn(str second) { freopen(second.c_str(), "r", stdin); }
void setOut(str second) { freopen(second.c_str(), "w", stdout); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
void setIO(str second = "") {
unsyncIO();
if ((int)(second).size()) {
setIn(second + ".in"), setOut(second + ".out");
}
}
int main() {
setIO();
int l;
cin >> l;
ll num = 0;
vl v;
v.push_back(1);
for (int a = (0); a < (l); ++a) {
string second;
cin >> second;
if (second == "add") {
num += v.back();
if (num > 4294967295) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
} else if (second == "for") {
int k;
cin >> k;
v.push_back(min(4294967296LL, k * v.back()));
} else
v.pop_back();
}
cout << num << endl;
}
| ### 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;
using ll = long long;
using ld = long double;
using db = double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
template <class T>
using V = vector<T>;
template <class T, size_t SZ>
using AR = array<T, SZ>;
const int MOD = 1e9 + 7;
const int MX = 2e5 + 5;
const ll INF = 1e18;
const ld PI = acos((ld)-1);
const int xd[4] = {1, 0, -1, 0}, yd[4] = {0, 1, 0, -1};
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
constexpr int pct(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return 31 - __builtin_clz(x); }
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
template <class T>
bool ckmin(T& a, const T& b) {
return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
return a < b ? a = b, 1 : 0;
}
template <class T, class U>
T fstTrue(T lo, T hi, U first) {
hi++;
assert(lo <= hi);
while (lo < hi) {
T mid = lo + (hi - lo) / 2;
first(mid) ? hi = mid : lo = mid + 1;
}
return lo;
}
template <class T, class U>
T lstTrue(T lo, T hi, U first) {
lo--;
assert(lo <= hi);
while (lo < hi) {
T mid = lo + (hi - lo + 1) / 2;
first(mid) ? lo = mid : hi = mid - 1;
}
return lo;
}
template <class T>
void remDup(vector<T>& v) {
sort(begin(v), end(v));
v.erase(unique(begin(v), end(v)), end(v));
}
template <class T, class U>
void erase(T& t, const U& u) {
auto it = t.find(u);
assert(it != end(t));
t.erase(u);
}
template <class T>
void re(complex<T>& c);
template <class T, class U>
void re(pair<T, U>& p);
template <class T>
void re(vector<T>& v);
template <class T, size_t SZ>
void re(AR<T, SZ>& a);
template <class T>
void re(T& x) {
cin >> x;
}
void re(db& d) {
str t;
re(t);
d = stod(t);
}
void re(ld& d) {
str t;
re(t);
d = stold(t);
}
template <class T, class... U>
void re(T& t, U&... u) {
re(t);
re(u...);
}
template <class T>
void re(complex<T>& c) {
T a, b;
re(a, b);
c = {a, b};
}
template <class T, class U>
void re(pair<T, U>& p) {
re(p.first, p.second);
}
template <class T>
void re(vector<T>& x) {
for (auto& a : x) re(a);
}
template <class T, size_t SZ>
void re(AR<T, SZ>& x) {
for (auto& a : x) re(a);
}
str to_string(char c) { return str(1, c); }
str to_string(const char* second) { return (str)second; }
str to_string(str second) { return second; }
str to_string(bool b) { return to_string((int)b); }
template <class T>
str to_string(complex<T> c) {
stringstream ss;
ss << c;
return ss.str();
}
str to_string(vector<bool> v) {
str res = "{";
for (int i = (0); i < ((int)(v).size()); ++i) res += char('0' + v[i]);
res += "}";
return res;
}
template <size_t SZ>
str to_string(bitset<SZ> b) {
str res = "";
for (int i = (0); i < (SZ); ++i) res += char('0' + b[i]);
return res;
}
template <class T, class U>
str to_string(pair<T, U> p);
template <class T>
str to_string(T v) {
bool fst = 1;
str res = "";
for (const auto& x : v) {
if (!fst) res += " ";
fst = 0;
res += to_string(x);
}
return res;
}
template <class T, class U>
str to_string(pair<T, U> p) {
return to_string(p.first) + " " + to_string(p.second);
}
template <class T>
void pr(T x) {
cout << to_string(x);
}
template <class T, class... U>
void pr(const T& t, const U&... u) {
pr(t);
pr(u...);
}
void ps() { pr("\n"); }
template <class T, class... U>
void ps(const T& t, const U&... u) {
pr(t);
if (sizeof...(u)) pr(" ");
ps(u...);
}
void DBG() { cerr << "]" << endl; }
template <class T, class... U>
void DBG(const T& t, const U&... u) {
cerr << to_string(t);
if (sizeof...(u)) cerr << ", ";
DBG(u...);
}
void setIn(str second) { freopen(second.c_str(), "r", stdin); }
void setOut(str second) { freopen(second.c_str(), "w", stdout); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
void setIO(str second = "") {
unsyncIO();
if ((int)(second).size()) {
setIn(second + ".in"), setOut(second + ".out");
}
}
int main() {
setIO();
int l;
cin >> l;
ll num = 0;
vl v;
v.push_back(1);
for (int a = (0); a < (l); ++a) {
string second;
cin >> second;
if (second == "add") {
num += v.back();
if (num > 4294967295) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
} else if (second == "for") {
int k;
cin >> k;
v.push_back(min(4294967296LL, k * v.back()));
} else
v.pop_back();
}
cout << num << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = unsigned long long int;
stack<ll> st, temp, over_stack;
ll lim = 4294967295;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n, i, j;
cin >> n;
ll ans = 0;
bool over = false, flag = false;
while (n--) {
string s;
cin >> s;
if (s[0] == 'f') {
ll val;
cin >> val;
if (flag) continue;
if (!st.size()) {
st.push(val);
} else {
if (over) {
over_stack.push(val);
continue;
}
if (val > (lim / st.top())) {
over = true;
over_stack.push(val);
continue;
}
ll u = st.top();
st.pop();
st.push(val * u);
}
temp.push(val);
} else if (s[0] == 'a') {
if (flag) continue;
if (over) {
flag = true;
continue;
}
if (!st.size()) {
ans++;
if (ans > lim) flag = true;
} else {
ans += (st.top());
if (ans > lim) flag = true;
}
} else {
if (flag) continue;
if (over) {
over_stack.pop();
if (!over_stack.size()) over = false;
continue;
}
ll u = st.top();
st.pop();
ll v = temp.top();
temp.pop();
u /= v;
st.push(u);
}
}
if (flag) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
cout << ans << 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;
using ll = unsigned long long int;
stack<ll> st, temp, over_stack;
ll lim = 4294967295;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n, i, j;
cin >> n;
ll ans = 0;
bool over = false, flag = false;
while (n--) {
string s;
cin >> s;
if (s[0] == 'f') {
ll val;
cin >> val;
if (flag) continue;
if (!st.size()) {
st.push(val);
} else {
if (over) {
over_stack.push(val);
continue;
}
if (val > (lim / st.top())) {
over = true;
over_stack.push(val);
continue;
}
ll u = st.top();
st.pop();
st.push(val * u);
}
temp.push(val);
} else if (s[0] == 'a') {
if (flag) continue;
if (over) {
flag = true;
continue;
}
if (!st.size()) {
ans++;
if (ans > lim) flag = true;
} else {
ans += (st.top());
if (ans > lim) flag = true;
}
} else {
if (flag) continue;
if (over) {
over_stack.pop();
if (!over_stack.size()) over = false;
continue;
}
ll u = st.top();
st.pop();
ll v = temp.top();
temp.pop();
u /= v;
st.push(u);
}
}
if (flag) {
cout << "OVERFLOW!!!" << endl;
return 0;
}
cout << ans << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long x = 0;
long long maxm = (1ll << 32);
stack<long long> stk;
long long n;
long long total = 1;
cin >> n;
string s;
stk.push(1);
for (int i = 0; i < n; i++) {
cin >> s;
if (s[0] == 'a') {
x = min(x + stk.top(), maxm);
} else if (s[0] == 'f') {
int s2;
cin >> s2;
long long top;
top = stk.top();
stk.push(min(s2 * top, maxm));
} else if (s[0] == 'e') {
stk.pop();
}
}
if (x == maxm)
cout << "OVERFLOW!!!" << endl;
else
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>
using namespace std;
int main() {
long long x = 0;
long long maxm = (1ll << 32);
stack<long long> stk;
long long n;
long long total = 1;
cin >> n;
string s;
stk.push(1);
for (int i = 0; i < n; i++) {
cin >> s;
if (s[0] == 'a') {
x = min(x + stk.top(), maxm);
} else if (s[0] == 'f') {
int s2;
cin >> s2;
long long top;
top = stk.top();
stk.push(min(s2 * top, maxm));
} else if (s[0] == 'e') {
stk.pop();
}
}
if (x == maxm)
cout << "OVERFLOW!!!" << endl;
else
cout << x << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("avx,avx2,fma")
using namespace std;
long long power(long long a, long long b) {
long long res = 1;
a = a % 998244353;
while (b > 0) {
if (b & 1) {
res = (res * a) % 998244353;
}
a = (a * a) % 998244353;
b >>= 1;
}
return res;
}
long long fermat_inv(long long y) { return power(y, 998244353 - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void solve() {
long long maxi = 1LL << 32LL;
long long n;
cin >> n;
stack<long long> st;
st.push(1);
long long ans = 0;
for (long long i = 0; i < n; i++) {
string s;
cin >> s;
if (s == "for") {
long long x;
cin >> x;
st.push(min(maxi, st.top() * x));
} else if (s == "end") {
st.pop();
} else {
ans += st.top();
}
}
if (ans >= maxi)
cout << "OVERFLOW!!!";
else
cout << ans;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long T = 1;
while (T--) {
solve();
}
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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("avx,avx2,fma")
using namespace std;
long long power(long long a, long long b) {
long long res = 1;
a = a % 998244353;
while (b > 0) {
if (b & 1) {
res = (res * a) % 998244353;
}
a = (a * a) % 998244353;
b >>= 1;
}
return res;
}
long long fermat_inv(long long y) { return power(y, 998244353 - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
void solve() {
long long maxi = 1LL << 32LL;
long long n;
cin >> n;
stack<long long> st;
st.push(1);
long long ans = 0;
for (long long i = 0; i < n; i++) {
string s;
cin >> s;
if (s == "for") {
long long x;
cin >> x;
st.push(min(maxi, st.top() * x));
} else if (s == "end") {
st.pop();
} else {
ans += st.top();
}
}
if (ans >= maxi)
cout << "OVERFLOW!!!";
else
cout << ans;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long T = 1;
while (T--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t;
cin >> t;
long long f_true = 0;
long long f_num = 1;
long long ans = 0;
stack<long long> v1;
stack<long long> v2;
for (long long z = 0; z < t; z++) {
string a;
cin >> a;
if (a[0] == 'a') {
ans++;
} else if (a[0] == 'f') {
long long b;
cin >> b;
v1.push(b);
v2.push(ans);
} else {
long long x = v1.top();
long long y = v2.top();
long long p = ans - y;
ans += (x - 1) * p;
v1.pop();
v2.pop();
}
if (ans >= 4294967296) {
cout << "OVERFLOW!!!";
return 0;
}
}
cout << ans;
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() {
long long t;
cin >> t;
long long f_true = 0;
long long f_num = 1;
long long ans = 0;
stack<long long> v1;
stack<long long> v2;
for (long long z = 0; z < t; z++) {
string a;
cin >> a;
if (a[0] == 'a') {
ans++;
} else if (a[0] == 'f') {
long long b;
cin >> b;
v1.push(b);
v2.push(ans);
} else {
long long x = v1.top();
long long y = v2.top();
long long p = ans - y;
ans += (x - 1) * p;
v1.pop();
v2.pop();
}
if (ans >= 4294967296) {
cout << "OVERFLOW!!!";
return 0;
}
}
cout << ans;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
unsigned long long x;
string s;
unsigned long long n, l;
vector<unsigned long long> v = {1};
void inp();
void solve();
void out();
int main() {
inp();
solve();
out();
return 0;
}
void inp() {
cin >> l;
while (l--) {
cin >> s;
if (s == "for") {
cin >> n;
n *= v.back();
v.push_back(n);
if (v.back() > UINT_MAX) v.back() = UINT_MAX;
} else {
if (s == "add") {
if (v.back() == UINT_MAX) {
cout << "OVERFLOW!!!";
exit(0);
}
x += v.back();
} else
v.pop_back();
}
if (x > UINT_MAX) {
cout << "OVERFLOW!!!";
exit(0);
}
}
cout << x;
}
void solve() {}
void out() {}
| ### 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;
unsigned long long x;
string s;
unsigned long long n, l;
vector<unsigned long long> v = {1};
void inp();
void solve();
void out();
int main() {
inp();
solve();
out();
return 0;
}
void inp() {
cin >> l;
while (l--) {
cin >> s;
if (s == "for") {
cin >> n;
n *= v.back();
v.push_back(n);
if (v.back() > UINT_MAX) v.back() = UINT_MAX;
} else {
if (s == "add") {
if (v.back() == UINT_MAX) {
cout << "OVERFLOW!!!";
exit(0);
}
x += v.back();
} else
v.pop_back();
}
if (x > UINT_MAX) {
cout << "OVERFLOW!!!";
exit(0);
}
}
cout << x;
}
void solve() {}
void out() {}
``` |
#include <bits/stdc++.h>
using namespace std;
double pi = 3.141592653589793238462;
string numkey = "0123456789";
string uni = "abcdefghijklmnopqrstuvwxyz";
long long int dx[4] = {0, 0, 1, -1};
long long int dy[4] = {1, -1, 0, 0};
long long int INF = (1LL << 62);
void increase_bit(long long int n, long long int cnt[]) {
long long int p = 0;
while (n > 0) {
if (n & 1) cnt[p]++;
p++;
n = n >> 1;
}
}
signed main() {
long long int n;
cin >> n;
stack<long long int> s;
s.push(1);
long long int great = 1LL << 32;
long long int ans = 0;
for (long long int i = 0; i < n; i++) {
string p;
cin >> p;
if (p == "for") {
long long int x;
cin >> x;
s.push(min(great, s.top() * x));
} else if (p == "end") {
s.pop();
} else if (p == "add") {
ans += s.top();
}
if (ans >= great) {
cout << "OVERFLOW!!!";
return 0;
}
}
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;
double pi = 3.141592653589793238462;
string numkey = "0123456789";
string uni = "abcdefghijklmnopqrstuvwxyz";
long long int dx[4] = {0, 0, 1, -1};
long long int dy[4] = {1, -1, 0, 0};
long long int INF = (1LL << 62);
void increase_bit(long long int n, long long int cnt[]) {
long long int p = 0;
while (n > 0) {
if (n & 1) cnt[p]++;
p++;
n = n >> 1;
}
}
signed main() {
long long int n;
cin >> n;
stack<long long int> s;
s.push(1);
long long int great = 1LL << 32;
long long int ans = 0;
for (long long int i = 0; i < n; i++) {
string p;
cin >> p;
if (p == "for") {
long long int x;
cin >> x;
s.push(min(great, s.top() * x));
} else if (p == "end") {
s.pop();
} else if (p == "add") {
ans += s.top();
}
if (ans >= great) {
cout << "OVERFLOW!!!";
return 0;
}
}
cout << ans;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int mod = 998244353;
string s;
long long a[N], b[N];
int main() {
int n, now = 1, tmp;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> s;
if (s == "add")
a[now]++;
else if (s == "for") {
cin >> tmp;
b[now] = tmp;
now++;
} else {
a[now - 1] += b[now - 1] * a[now];
if (a[now - 1] > (1ll << 32) - 1) {
puts("OVERFLOW!!!");
return 0;
}
a[now] = 0;
now--;
}
}
if (a[1] <= (1ll << 32) - 1)
cout << a[1] << endl;
else
puts("OVERFLOW!!!");
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;
const int N = 1e5 + 5;
const int mod = 998244353;
string s;
long long a[N], b[N];
int main() {
int n, now = 1, tmp;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> s;
if (s == "add")
a[now]++;
else if (s == "for") {
cin >> tmp;
b[now] = tmp;
now++;
} else {
a[now - 1] += b[now - 1] * a[now];
if (a[now - 1] > (1ll << 32) - 1) {
puts("OVERFLOW!!!");
return 0;
}
a[now] = 0;
now--;
}
}
if (a[1] <= (1ll << 32) - 1)
cout << a[1] << endl;
else
puts("OVERFLOW!!!");
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 70;
const int INF = 2.1e9;
const int MOD = 1e9 + 7;
const double eps = 1e-7;
const double PI = acos(-1.0);
int n, m, k;
char s[maxn];
int op[maxn], val[maxn], last[maxn];
long long cur[maxn];
bool flag[maxn];
long long fun() {
long long x = 0, MX = 1LL << 32;
int cnt = 0;
cur[0] = 1;
flag[0] = 1;
last[0] = 0;
for (int i = 1; i <= n; i++) {
if (op[i] == 0) {
cur[i] = cur[i - 1];
flag[i] = flag[i - 1];
if (!flag[i]) return -1;
x += cur[i];
if (x >= MX) return -1;
} else if (op[i] == 1) {
cur[i] = cur[last[cnt]] * val[i];
if (cur[i] >= MX || !flag[last[cnt]])
flag[i] = 0;
else
flag[i] = 1;
++cnt;
last[cnt] = i;
} else {
--cnt;
cur[i] = cur[last[cnt]];
flag[i] = flag[last[cnt]];
}
}
return x;
}
int main() {
while (cin >> n) {
for (int i = 1; i <= n; i++) {
scanf("%s", s);
if (s[0] == 'a')
op[i] = 0;
else if (s[0] == 'f') {
op[i] = 1;
scanf("%d", &val[i]);
} else
op[i] = 2;
}
long long ans = fun();
if (ans == -1)
cout << "OVERFLOW!!!" << endl;
else
cout << ans << endl;
}
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;
const int maxn = 2e5 + 70;
const int INF = 2.1e9;
const int MOD = 1e9 + 7;
const double eps = 1e-7;
const double PI = acos(-1.0);
int n, m, k;
char s[maxn];
int op[maxn], val[maxn], last[maxn];
long long cur[maxn];
bool flag[maxn];
long long fun() {
long long x = 0, MX = 1LL << 32;
int cnt = 0;
cur[0] = 1;
flag[0] = 1;
last[0] = 0;
for (int i = 1; i <= n; i++) {
if (op[i] == 0) {
cur[i] = cur[i - 1];
flag[i] = flag[i - 1];
if (!flag[i]) return -1;
x += cur[i];
if (x >= MX) return -1;
} else if (op[i] == 1) {
cur[i] = cur[last[cnt]] * val[i];
if (cur[i] >= MX || !flag[last[cnt]])
flag[i] = 0;
else
flag[i] = 1;
++cnt;
last[cnt] = i;
} else {
--cnt;
cur[i] = cur[last[cnt]];
flag[i] = flag[last[cnt]];
}
}
return x;
}
int main() {
while (cin >> n) {
for (int i = 1; i <= n; i++) {
scanf("%s", s);
if (s[0] == 'a')
op[i] = 0;
else if (s[0] == 'f') {
op[i] = 1;
scanf("%d", &val[i]);
} else
op[i] = 2;
}
long long ans = fun();
if (ans == -1)
cout << "OVERFLOW!!!" << endl;
else
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
long long int t = pow(2, 32);
using namespace std;
int32_t main() {
long long int n;
cin >> n;
stack<long long int> s;
s.push(1);
long long int flag = 0;
long long int sum = 0;
for (long long int i = 0; i < n; i++) {
string x;
cin >> x;
if (x == "add") {
sum += s.top();
} else if (x == "for") {
long long int a;
cin >> a;
s.push(min(s.top() * a, t));
} else {
s.pop();
}
if (sum >= t) {
flag = 1;
}
}
if (flag == 1) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << sum << endl;
}
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>
long long int t = pow(2, 32);
using namespace std;
int32_t main() {
long long int n;
cin >> n;
stack<long long int> s;
s.push(1);
long long int flag = 0;
long long int sum = 0;
for (long long int i = 0; i < n; i++) {
string x;
cin >> x;
if (x == "add") {
sum += s.top();
} else if (x == "for") {
long long int a;
cin >> a;
s.push(min(s.top() * a, t));
} else {
s.pop();
}
if (sum >= t) {
flag = 1;
}
}
if (flag == 1) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << sum << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int l = 0;
cin >> l;
string cm = "";
int fnum = 0;
stack<long long> toadd;
long long end = 0;
bool ovflw = false;
for (int i = 0; i < l; i++) {
cin >> cm;
if (toadd.empty() == true) {
toadd.push(1);
}
if (cm[0] == 'f') {
cin >> fnum;
if ((end + fnum * toadd.top()) <= 4294967295LL) {
toadd.push(fnum * toadd.top());
} else {
toadd.push(0);
}
} else if (cm[0] == 'e') {
toadd.pop();
} else {
if (toadd.top() == 0 || (end + toadd.top()) > 4294967295LL) {
cout << "OVERFLOW!!!";
ovflw = true;
break;
} else {
end = end + toadd.top();
}
}
}
if (ovflw == false) {
cout << end;
}
}
| ### 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() {
int l = 0;
cin >> l;
string cm = "";
int fnum = 0;
stack<long long> toadd;
long long end = 0;
bool ovflw = false;
for (int i = 0; i < l; i++) {
cin >> cm;
if (toadd.empty() == true) {
toadd.push(1);
}
if (cm[0] == 'f') {
cin >> fnum;
if ((end + fnum * toadd.top()) <= 4294967295LL) {
toadd.push(fnum * toadd.top());
} else {
toadd.push(0);
}
} else if (cm[0] == 'e') {
toadd.pop();
} else {
if (toadd.top() == 0 || (end + toadd.top()) > 4294967295LL) {
cout << "OVERFLOW!!!";
ovflw = true;
break;
} else {
end = end + toadd.top();
}
}
}
if (ovflw == false) {
cout << end;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll MAXV = (1ll << 32);
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
ll res = 0;
vector<ll> stack = {1};
for (int i = 0; i < n; ++i) {
string str;
cin >> str;
if (str == "add") {
res += stack.back();
} else if (str == "for") {
ll k;
cin >> k;
stack.push_back(min(MAXV, k * stack.back()));
} else if (str == "end") {
stack.pop_back();
}
}
if (res >= MAXV)
cout << "OVERFLOW!!!\n";
else
cout << res << '\n';
}
| ### 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;
using ll = long long;
const ll MAXV = (1ll << 32);
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
ll res = 0;
vector<ll> stack = {1};
for (int i = 0; i < n; ++i) {
string str;
cin >> str;
if (str == "add") {
res += stack.back();
} else if (str == "for") {
ll k;
cin >> k;
stack.push_back(min(MAXV, k * stack.back()));
} else if (str == "end") {
stack.pop_back();
}
}
if (res >= MAXV)
cout << "OVERFLOW!!!\n";
else
cout << res << '\n';
}
``` |
#include <bits/stdc++.h>
char a[100005][100];
long long int b[1000005];
int c[100005];
int main() {
int i, j, m, flag = 0;
long long int n;
scanf("%lld", &n);
for (i = 1; i <= n; i++) {
scanf("%s", a[i]);
if (a[i][0] == 'f') {
scanf("%d", &c[i]);
}
}
int num = 100000;
int flag1 = 0;
long long int sum = 0;
for (i = 1; i <= 1000000; i++) b[i] = 1;
for (i = 1; i <= n; i++) {
if (a[i][0] == 'a') {
if (flag1 > 0) {
flag = 1;
break;
}
sum += b[num];
}
if (a[i][0] == 'f') {
b[++num] = b[num - 1] * c[i];
if (b[num] >= pow(2, 32)) {
flag1++;
}
}
if (a[i][0] == 'e') {
flag1--;
num--;
}
if (sum >= pow(2, 32)) {
flag = 1;
break;
}
}
if (flag == 1)
printf("OVERFLOW!!!\n");
else
printf("%lld\n", sum);
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>
char a[100005][100];
long long int b[1000005];
int c[100005];
int main() {
int i, j, m, flag = 0;
long long int n;
scanf("%lld", &n);
for (i = 1; i <= n; i++) {
scanf("%s", a[i]);
if (a[i][0] == 'f') {
scanf("%d", &c[i]);
}
}
int num = 100000;
int flag1 = 0;
long long int sum = 0;
for (i = 1; i <= 1000000; i++) b[i] = 1;
for (i = 1; i <= n; i++) {
if (a[i][0] == 'a') {
if (flag1 > 0) {
flag = 1;
break;
}
sum += b[num];
}
if (a[i][0] == 'f') {
b[++num] = b[num - 1] * c[i];
if (b[num] >= pow(2, 32)) {
flag1++;
}
}
if (a[i][0] == 'e') {
flag1--;
num--;
}
if (sum >= pow(2, 32)) {
flag = 1;
break;
}
}
if (flag == 1)
printf("OVERFLOW!!!\n");
else
printf("%lld\n", sum);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
namespace shyutsuryoku {
const int STRSZ = 3e6 + 13;
char inbufer[STRSZ];
inline void cscan(string& i) {
int c = getc(stdin), cnt = 0;
while (c <= 32) c = getc(stdin);
while (c > 32) *(inbufer + cnt) = c, c = getc(stdin), ++cnt;
*(cnt + inbufer) = 0;
i = inbufer;
}
inline void cprint(string& i) {
char* second = new char[i.size() + 1];
strcpy(second, i.c_str());
while (*second) putchar(*second), second++;
}
inline void cprint(const char*& i) {
int cnt = 0;
while (*(i + cnt)) putc(*(i + cnt), stdout), cnt++;
}
inline void cscan(int& i) {
i = 0;
int c = getc(stdin), b = 0;
while (c <= 32) c = getc(stdin);
if (c == '-') b = 1, c = getc(stdin);
while (c >= '0' && c <= '9') i *= 10, i += c - '0', c = getc(stdin);
if (b) i = -i;
}
inline void cprint(int i) {
if (i < 0) putc('-', stdout), i = -i;
int sz = 0;
char ans[12];
while (i || !sz) ans[sz++] = i % 10 + '0', i /= 10;
while (sz--) putc(ans[sz], stdout);
}
inline void cscan(long long& i) {
i = 0;
int c = getc(stdin), b = 1;
while (c <= 32) c = getc(stdin);
if (c == '-') b = -1, c = getc(stdin);
while (c >= '0' && c <= '9') i *= 10, i += c - '0', c = getc(stdin);
i *= b;
}
inline void cprint(long long i) {
if (i < 0) putc('-', stdout), i = -i;
int sz = 0;
char ans[21];
while (i || !sz) ans[sz++] = i % 10 + '0', i /= 10;
while (sz--) putc(ans[sz], stdout);
}
inline void cscan(unsigned& i) {
i = 0;
int c = getc(stdin);
while (c <= 32) c = getc(stdin);
while (c >= '0' && c <= '9') i *= 10, i += c - '0', c = getc(stdin);
}
inline void cprint(unsigned i) {
int sz = 0;
char ans[12];
while (i || !sz) ans[sz++] = i % 10 + '0', i /= 10;
while (sz--) putc(ans[sz], stdout);
}
inline void cscan(unsigned long long& i) {
i = 0;
int c = getc(stdin);
while (c <= 32) c = getc(stdin);
while (c >= '0' && c <= '9') i *= 10, i += c - '0', c = getc(stdin);
}
inline void cprint(unsigned long long i) {
int sz = 0;
char ans[21];
while (i || !sz) ans[sz++] = i % 10 + '0', i /= 10;
while (sz--) putc(ans[sz], stdout);
}
inline void cscan(double& i) { scanf("%lf", &i); }
inline void cprint(double i) { printf("%.10lf", i); }
inline void cprint(char i) { putc(i, stdout); }
inline void cscan(char& i) {
int b = getc(stdin);
while (b <= 32) b = getc(stdin);
i = b;
}
template <class T, class K>
inline void cscan(pair<T, K>& i) {
cscan(i.first);
cscan(i.second);
}
template <class T, class K>
inline void cprint(pair<T, K>& i) {
cprint('(');
cprint(i.first);
cprint(' ');
cprint(i.second);
cprint(')');
}
template <class T>
inline void cscan(vector<T>& i) {
for (auto& j : i) cscan(j);
}
template <class T>
inline void cprint(vector<T>& i) {
for (auto& j : i) cprint(j), cprint(' ');
}
template <class T>
inline void cprint(vector<vector<T> >& i) {
for (auto& j : i) cprint(j), cprint('\n');
}
template <class T>
ostream& operator,(ostream& os, T v) {
cprint(v);
return os;
}
template <class T>
istream& operator,(istream& is, T& v) {
cscan(v);
return is;
}
} // namespace shyutsuryoku
using namespace shyutsuryoku;
void AnimeConservativeRevolution();
int main() {
srand(time(0));
AnimeConservativeRevolution();
return 0;
}
void tyloh() {
cout, "OVERFLOW!!!";
exit(0);
}
void AnimeConservativeRevolution() {
long long n, mx = (1ll << 32) - 1;
cin, n;
long long ans = 0, baka = 1, uf = 0;
vector<long long> ves;
while (n--) {
string second;
cin, second;
if (second == "for") {
long long x;
cin, x;
if (baka * x > mx || uf)
uf++;
else
baka *= x;
ves.push_back(x);
} else if (second == "end") {
if (!uf)
baka /= ves.back();
else
uf--;
ves.pop_back();
} else {
if (uf) tyloh();
if (mx - ans < baka) tyloh();
ans += baka;
}
}
cout, ans;
}
| ### 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;
namespace shyutsuryoku {
const int STRSZ = 3e6 + 13;
char inbufer[STRSZ];
inline void cscan(string& i) {
int c = getc(stdin), cnt = 0;
while (c <= 32) c = getc(stdin);
while (c > 32) *(inbufer + cnt) = c, c = getc(stdin), ++cnt;
*(cnt + inbufer) = 0;
i = inbufer;
}
inline void cprint(string& i) {
char* second = new char[i.size() + 1];
strcpy(second, i.c_str());
while (*second) putchar(*second), second++;
}
inline void cprint(const char*& i) {
int cnt = 0;
while (*(i + cnt)) putc(*(i + cnt), stdout), cnt++;
}
inline void cscan(int& i) {
i = 0;
int c = getc(stdin), b = 0;
while (c <= 32) c = getc(stdin);
if (c == '-') b = 1, c = getc(stdin);
while (c >= '0' && c <= '9') i *= 10, i += c - '0', c = getc(stdin);
if (b) i = -i;
}
inline void cprint(int i) {
if (i < 0) putc('-', stdout), i = -i;
int sz = 0;
char ans[12];
while (i || !sz) ans[sz++] = i % 10 + '0', i /= 10;
while (sz--) putc(ans[sz], stdout);
}
inline void cscan(long long& i) {
i = 0;
int c = getc(stdin), b = 1;
while (c <= 32) c = getc(stdin);
if (c == '-') b = -1, c = getc(stdin);
while (c >= '0' && c <= '9') i *= 10, i += c - '0', c = getc(stdin);
i *= b;
}
inline void cprint(long long i) {
if (i < 0) putc('-', stdout), i = -i;
int sz = 0;
char ans[21];
while (i || !sz) ans[sz++] = i % 10 + '0', i /= 10;
while (sz--) putc(ans[sz], stdout);
}
inline void cscan(unsigned& i) {
i = 0;
int c = getc(stdin);
while (c <= 32) c = getc(stdin);
while (c >= '0' && c <= '9') i *= 10, i += c - '0', c = getc(stdin);
}
inline void cprint(unsigned i) {
int sz = 0;
char ans[12];
while (i || !sz) ans[sz++] = i % 10 + '0', i /= 10;
while (sz--) putc(ans[sz], stdout);
}
inline void cscan(unsigned long long& i) {
i = 0;
int c = getc(stdin);
while (c <= 32) c = getc(stdin);
while (c >= '0' && c <= '9') i *= 10, i += c - '0', c = getc(stdin);
}
inline void cprint(unsigned long long i) {
int sz = 0;
char ans[21];
while (i || !sz) ans[sz++] = i % 10 + '0', i /= 10;
while (sz--) putc(ans[sz], stdout);
}
inline void cscan(double& i) { scanf("%lf", &i); }
inline void cprint(double i) { printf("%.10lf", i); }
inline void cprint(char i) { putc(i, stdout); }
inline void cscan(char& i) {
int b = getc(stdin);
while (b <= 32) b = getc(stdin);
i = b;
}
template <class T, class K>
inline void cscan(pair<T, K>& i) {
cscan(i.first);
cscan(i.second);
}
template <class T, class K>
inline void cprint(pair<T, K>& i) {
cprint('(');
cprint(i.first);
cprint(' ');
cprint(i.second);
cprint(')');
}
template <class T>
inline void cscan(vector<T>& i) {
for (auto& j : i) cscan(j);
}
template <class T>
inline void cprint(vector<T>& i) {
for (auto& j : i) cprint(j), cprint(' ');
}
template <class T>
inline void cprint(vector<vector<T> >& i) {
for (auto& j : i) cprint(j), cprint('\n');
}
template <class T>
ostream& operator,(ostream& os, T v) {
cprint(v);
return os;
}
template <class T>
istream& operator,(istream& is, T& v) {
cscan(v);
return is;
}
} // namespace shyutsuryoku
using namespace shyutsuryoku;
void AnimeConservativeRevolution();
int main() {
srand(time(0));
AnimeConservativeRevolution();
return 0;
}
void tyloh() {
cout, "OVERFLOW!!!";
exit(0);
}
void AnimeConservativeRevolution() {
long long n, mx = (1ll << 32) - 1;
cin, n;
long long ans = 0, baka = 1, uf = 0;
vector<long long> ves;
while (n--) {
string second;
cin, second;
if (second == "for") {
long long x;
cin, x;
if (baka * x > mx || uf)
uf++;
else
baka *= x;
ves.push_back(x);
} else if (second == "end") {
if (!uf)
baka /= ves.back();
else
uf--;
ves.pop_back();
} else {
if (uf) tyloh();
if (mx - ans < baka) tyloh();
ans += baka;
}
}
cout, ans;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void fast() {
std::ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
cerr.tie(NULL);
}
int main() {
fast();
unsigned long long over = pow(2, 32);
string s;
int i;
unsigned long long num = 0;
stack<unsigned long long> loop;
loop.push(1);
int n;
cin >> n;
while (n--) {
cin >> s;
if (s == "add") {
num += loop.top();
if (num >= over) {
cout << "OVERFLOW!!!";
return 0;
}
} else if (s == "for") {
cin >> i;
loop.push(min(over, loop.top() * i));
} else {
loop.pop();
}
}
cout << num;
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;
void fast() {
std::ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
cerr.tie(NULL);
}
int main() {
fast();
unsigned long long over = pow(2, 32);
string s;
int i;
unsigned long long num = 0;
stack<unsigned long long> loop;
loop.push(1);
int n;
cin >> n;
while (n--) {
cin >> s;
if (s == "add") {
num += loop.top();
if (num >= over) {
cout << "OVERFLOW!!!";
return 0;
}
} else if (s == "for") {
cin >> i;
loop.push(min(over, loop.top() * i));
} else {
loop.pop();
}
}
cout << num;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long inv_euclid(long long a, long long m = 998244353) {
long long m0 = m;
long long y = 0, x = 1;
if (m == 1) return 0;
while (a > 1) {
long long q = a / m;
long long t = m;
m = a % m, a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0) x += m0;
return x;
}
long long power(long long x, long long y, long long p) {
long long res = 1;
x = x % p;
if (x == 0) return 0;
while (y > 0) {
if (y & 1) res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
long long max(long long a, long long b) {
if (a > b) return a;
return b;
}
long long min(long long a, long long b) {
if (a < b) return a;
return b;
}
long long ssum(long long n) { return (n * (n + 1)) / 2; }
long long gcd(long long a, long long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / (gcd(a, b)); }
long long spf[1000000];
int main() {
long long ans = 0, add = 1, skip = 0;
stack<long long> st;
long long n;
cin >> n;
while (n--) {
string s;
cin >> s;
if (s == "for") {
long long x;
cin >> x;
st.push(x);
if (add < 4294967295)
add *= x;
else {
skip++;
}
} else if (s == "add") {
if (add > 4294967295 || skip) {
cout << "OVERFLOW!!!";
return 0;
}
ans += add;
if (ans > 4294967295) {
cout << "OVERFLOW!!!";
return 0;
}
} else {
long long x = st.top();
st.pop();
if (skip) {
skip--;
} else
add /= x;
}
}
cout << ans << "\n";
}
| ### 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 inv_euclid(long long a, long long m = 998244353) {
long long m0 = m;
long long y = 0, x = 1;
if (m == 1) return 0;
while (a > 1) {
long long q = a / m;
long long t = m;
m = a % m, a = t;
t = y;
y = x - q * y;
x = t;
}
if (x < 0) x += m0;
return x;
}
long long power(long long x, long long y, long long p) {
long long res = 1;
x = x % p;
if (x == 0) return 0;
while (y > 0) {
if (y & 1) res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
long long max(long long a, long long b) {
if (a > b) return a;
return b;
}
long long min(long long a, long long b) {
if (a < b) return a;
return b;
}
long long ssum(long long n) { return (n * (n + 1)) / 2; }
long long gcd(long long a, long long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / (gcd(a, b)); }
long long spf[1000000];
int main() {
long long ans = 0, add = 1, skip = 0;
stack<long long> st;
long long n;
cin >> n;
while (n--) {
string s;
cin >> s;
if (s == "for") {
long long x;
cin >> x;
st.push(x);
if (add < 4294967295)
add *= x;
else {
skip++;
}
} else if (s == "add") {
if (add > 4294967295 || skip) {
cout << "OVERFLOW!!!";
return 0;
}
ans += add;
if (ans > 4294967295) {
cout << "OVERFLOW!!!";
return 0;
}
} else {
long long x = st.top();
st.pop();
if (skip) {
skip--;
} else
add /= x;
}
}
cout << ans << "\n";
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long inf = 1ll << 32;
const int INF = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const int maxn = 2e5 + 5;
long long gcd(long long a, long long b) { return b != 0 ? gcd(b, a % b) : a; }
long long pow_mod(long long a, long long b, long long mod) {
long long res = 1;
while (b != 0) {
if (b & 1) res = res * a % mod;
b >>= 1;
a = a * a % mod;
}
return res;
}
stack<long long> v;
int main() {
int l;
long long ans = 0;
cin >> l;
v.push(1);
while (l--) {
string s;
cin >> s;
if (s[0] == 'f') {
int n;
cin >> n;
v.push(min(inf, v.top() * n));
} else if (s[0] == 'a') {
ans += v.top();
} else
v.pop();
}
if (ans < inf)
cout << ans << endl;
else
cout << "OVERFLOW!!!" << endl;
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;
const long long inf = 1ll << 32;
const int INF = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const int maxn = 2e5 + 5;
long long gcd(long long a, long long b) { return b != 0 ? gcd(b, a % b) : a; }
long long pow_mod(long long a, long long b, long long mod) {
long long res = 1;
while (b != 0) {
if (b & 1) res = res * a % mod;
b >>= 1;
a = a * a % mod;
}
return res;
}
stack<long long> v;
int main() {
int l;
long long ans = 0;
cin >> l;
v.push(1);
while (l--) {
string s;
cin >> s;
if (s[0] == 'f') {
int n;
cin >> n;
v.push(min(inf, v.top() * n));
} else if (s[0] == 'a') {
ans += v.top();
} else
v.pop();
}
if (ans < inf)
cout << ans << endl;
else
cout << "OVERFLOW!!!" << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int md = 1e9 + 7;
const int MX = 2e5 + 5;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
long long power(long long a, long long n) {
long long res = 1;
while (n) {
if (n % 2) res *= a;
a *= a;
n /= 2;
}
return res;
}
long long abst(long long a) { return ((a < 0) ? (-1 * a) : (a)); }
long long q;
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
q = power((long long)2, (long long)32);
long long res = 0;
int l;
cin >> l;
stack<long long> st;
st.push((long long)1);
while (l--) {
string s;
cin >> s;
if (s == "add") {
res += st.top();
} else if (s == "end") {
st.pop();
} else {
long long x;
cin >> x;
st.push(min(x * st.top(), q));
}
}
if (res >= q)
cout << "OVERFLOW!!!\n";
else
cout << res << "\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 int md = 1e9 + 7;
const int MX = 2e5 + 5;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
long long power(long long a, long long n) {
long long res = 1;
while (n) {
if (n % 2) res *= a;
a *= a;
n /= 2;
}
return res;
}
long long abst(long long a) { return ((a < 0) ? (-1 * a) : (a)); }
long long q;
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
q = power((long long)2, (long long)32);
long long res = 0;
int l;
cin >> l;
stack<long long> st;
st.push((long long)1);
while (l--) {
string s;
cin >> s;
if (s == "add") {
res += st.top();
} else if (s == "end") {
st.pop();
} else {
long long x;
cin >> x;
st.push(min(x * st.top(), q));
}
}
if (res >= q)
cout << "OVERFLOW!!!\n";
else
cout << res << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int i, j, k, l, m, n, ans = 0, mx = 1;
string s;
stack<long long int> st;
st.push(1);
for (i = 1; i <= 32; i++) mx = mx * 2;
long long int var;
scanf("%lld", &var);
while (var--) {
cin >> s;
k = 0;
if (s == "for") {
cin >> l;
st.push(min(mx, l * st.top()));
} else if (s == "end") {
st.pop();
} else
ans += st.top();
}
if (ans >= mx || k > 0)
cout << "OVERFLOW!!!" << endl;
else
cout << ans;
}
| ### 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 int i, j, k, l, m, n, ans = 0, mx = 1;
string s;
stack<long long int> st;
st.push(1);
for (i = 1; i <= 32; i++) mx = mx * 2;
long long int var;
scanf("%lld", &var);
while (var--) {
cin >> s;
k = 0;
if (s == "for") {
cin >> l;
st.push(min(mx, l * st.top()));
} else if (s == "end") {
st.pop();
} else
ans += st.top();
}
if (ans >= mx || k > 0)
cout << "OVERFLOW!!!" << endl;
else
cout << ans;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5, M = 1e9 + 7;
const unsigned long long base = 13331;
const double Pi = acos(-1.0);
const long long C = 299792458;
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
long long h[N];
int main() {
ios::sync_with_stdio(false);
int n;
cin >> n;
int now = 0;
h[0] = 1;
long long ans = 0;
long long maxx = 1;
for (int i = 1; i <= 32; i++) {
maxx *= 2;
}
int flag = 0;
maxx--;
for (int i = 1; i <= n; i++) {
string s;
cin >> s;
if (flag) continue;
if (s[0] == 'f') {
long long x;
cin >> x;
now++;
h[now] = h[now - 1] * x;
if (h[now] > maxx || h[now] < 0) {
h[now] = -1;
}
}
if (s[0] == 'a') {
ans += h[now];
if (ans > maxx || h[now] < 0) {
flag = 1;
continue;
}
}
if (s[0] == 'e') now--;
}
if (flag)
cout << "OVERFLOW!!!" << endl;
else
cout << ans << 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 int N = 2e5 + 5, M = 1e9 + 7;
const unsigned long long base = 13331;
const double Pi = acos(-1.0);
const long long C = 299792458;
inline int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
long long h[N];
int main() {
ios::sync_with_stdio(false);
int n;
cin >> n;
int now = 0;
h[0] = 1;
long long ans = 0;
long long maxx = 1;
for (int i = 1; i <= 32; i++) {
maxx *= 2;
}
int flag = 0;
maxx--;
for (int i = 1; i <= n; i++) {
string s;
cin >> s;
if (flag) continue;
if (s[0] == 'f') {
long long x;
cin >> x;
now++;
h[now] = h[now - 1] * x;
if (h[now] > maxx || h[now] < 0) {
h[now] = -1;
}
}
if (s[0] == 'a') {
ans += h[now];
if (ans > maxx || h[now] < 0) {
flag = 1;
continue;
}
}
if (s[0] == 'e') now--;
}
if (flag)
cout << "OVERFLOW!!!" << endl;
else
cout << ans << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
stack<long long> s;
long long n, x = 0;
bool f = 0;
int main() {
cin >> n;
s.push(1);
while (n--) {
string S;
cin >> S;
if (S == "for") {
long long t;
cin >> t;
if (s.top() > 4294967295) t = 1;
s.push(t * s.top());
} else if (S == "end")
s.pop();
else
x += s.top();
if (x > 4294967295) f = 1;
}
if (f == 1)
cout << "OVERFLOW!!!" << endl;
else
cout << x << 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;
stack<long long> s;
long long n, x = 0;
bool f = 0;
int main() {
cin >> n;
s.push(1);
while (n--) {
string S;
cin >> S;
if (S == "for") {
long long t;
cin >> t;
if (s.top() > 4294967295) t = 1;
s.push(t * s.top());
} else if (S == "end")
s.pop();
else
x += s.top();
if (x > 4294967295) f = 1;
}
if (f == 1)
cout << "OVERFLOW!!!" << endl;
else
cout << x << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
#pragma warning(disable : 4996)
long long sms[100002];
int main() {
int l;
string cmd;
while (cin >> l) {
int n;
int sc = 1;
long long acum = 0;
sms[0] = 1;
for (int i = int(0); i < int(l); i++) {
cin >> cmd;
if (cmd[0] == 'a') {
acum = min(acum + sms[sc - 1], (1LL << 32LL));
} else if (cmd[0] == 'e') {
sc--;
} else {
cin >> n;
sms[sc] = min(sms[sc - 1] * n, (1LL << 32LL));
sc++;
}
}
if (acum >= (1LL << 32LL))
cout << "OVERFLOW!!!" << endl;
else
cout << acum << 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;
#pragma warning(disable : 4996)
long long sms[100002];
int main() {
int l;
string cmd;
while (cin >> l) {
int n;
int sc = 1;
long long acum = 0;
sms[0] = 1;
for (int i = int(0); i < int(l); i++) {
cin >> cmd;
if (cmd[0] == 'a') {
acum = min(acum + sms[sc - 1], (1LL << 32LL));
} else if (cmd[0] == 'e') {
sc--;
} else {
cin >> n;
sms[sc] = min(sms[sc - 1] * n, (1LL << 32LL));
sc++;
}
}
if (acum >= (1LL << 32LL))
cout << "OVERFLOW!!!" << endl;
else
cout << acum << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, unsigned long long int y) {
long long int res = 1;
while (y > 0) {
if (y & 1) {
res = res * x;
}
y = y >> 1;
x = x * x;
}
return res;
}
long long int gcd(long long int a, long long int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long int powermod(long long int x, unsigned long long int y,
long long int p) {
long long int res = 1;
x = x % p;
while (y > 0) {
if (y & 1) {
res = (res * x) % p;
}
y = y >> 1;
x = (x * x) % p;
}
return res;
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int n;
cin >> n;
stack<long long int> v;
long long int ans = 0;
long long int po = 1LL << 32;
long long int idx = -1;
for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
string s;
cin >> s;
if (s == "for") {
long long int l;
cin >> l;
if (!v.empty()) l *= v.top();
v.push(min(po, l));
} else if (s == "end") {
v.pop();
} else {
long long int c = 1;
if (!v.empty()) c *= v.top();
ans += c;
}
}
if (ans >= po) ans = -1;
if (ans == -1)
cout << "OVERFLOW!!!\n";
else
cout << ans << "\n";
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;
long long int power(long long int x, unsigned long long int y) {
long long int res = 1;
while (y > 0) {
if (y & 1) {
res = res * x;
}
y = y >> 1;
x = x * x;
}
return res;
}
long long int gcd(long long int a, long long int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
long long int powermod(long long int x, unsigned long long int y,
long long int p) {
long long int res = 1;
x = x % p;
while (y > 0) {
if (y & 1) {
res = (res * x) % p;
}
y = y >> 1;
x = (x * x) % p;
}
return res;
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int n;
cin >> n;
stack<long long int> v;
long long int ans = 0;
long long int po = 1LL << 32;
long long int idx = -1;
for (long long int i = (long long int)(0); i < (long long int)(n); i++) {
string s;
cin >> s;
if (s == "for") {
long long int l;
cin >> l;
if (!v.empty()) l *= v.top();
v.push(min(po, l));
} else if (s == "end") {
v.pop();
} else {
long long int c = 1;
if (!v.empty()) c *= v.top();
ans += c;
}
}
if (ans >= po) ans = -1;
if (ans == -1)
cout << "OVERFLOW!!!\n";
else
cout << ans << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long n;
cin >> n;
long long ans = 0, j = 1;
bool poss = 1;
stack<long long> s;
for (long long i = 0; i < n; i++) {
string command;
cin >> command;
if (command == "add") {
ans += j;
if (ans > 4294967295) {
poss = 0;
}
} else if (command == "end") {
j = s.top();
s.pop();
} else {
long long k;
cin >> k;
s.push(j);
j *= k;
if (j > 4294967295) j = 4294967295 + 1;
}
}
if (!poss)
cout << "OVERFLOW!!!" << endl;
else
cout << ans << 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;
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long n;
cin >> n;
long long ans = 0, j = 1;
bool poss = 1;
stack<long long> s;
for (long long i = 0; i < n; i++) {
string command;
cin >> command;
if (command == "add") {
ans += j;
if (ans > 4294967295) {
poss = 0;
}
} else if (command == "end") {
j = s.top();
s.pop();
} else {
long long k;
cin >> k;
s.push(j);
j *= k;
if (j > 4294967295) j = 4294967295 + 1;
}
}
if (!poss)
cout << "OVERFLOW!!!" << endl;
else
cout << ans << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int input;
cin >> input;
long long int x = 0;
vector<long long int> pusing;
int pos = -1;
for (int i = 0; i < input; i++) {
string inp;
cin >> inp;
if (inp == "for") {
long long int num;
cin >> num;
if (pos >= 0) {
if (num * pusing.at(pos) > 4294967295) {
num = 4294967295 + 1;
} else {
num *= pusing.at(pos);
}
}
pos++;
pusing.push_back(num);
} else if (inp == "end") {
pusing.pop_back();
pos--;
} else if (inp == "add") {
if (pos == -1) {
x++;
} else {
x += pusing.at(pos);
}
}
if (x > 4294967295) {
break;
}
}
if (x > 4294967295) {
cout << "OVERFLOW!!!" << endl;
} else
cout << x << 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 main() {
int input;
cin >> input;
long long int x = 0;
vector<long long int> pusing;
int pos = -1;
for (int i = 0; i < input; i++) {
string inp;
cin >> inp;
if (inp == "for") {
long long int num;
cin >> num;
if (pos >= 0) {
if (num * pusing.at(pos) > 4294967295) {
num = 4294967295 + 1;
} else {
num *= pusing.at(pos);
}
}
pos++;
pusing.push_back(num);
} else if (inp == "end") {
pusing.pop_back();
pos--;
} else if (inp == "add") {
if (pos == -1) {
x++;
} else {
x += pusing.at(pos);
}
}
if (x > 4294967295) {
break;
}
}
if (x > 4294967295) {
cout << "OVERFLOW!!!" << endl;
} else
cout << x << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long M = 1000000;
long long n, ok[100005], res = 0, p[100005], t = 0;
string s;
void solve() {
cin >> s;
if (s == "for") {
cin >> n;
t++;
if (ok[t - 1]) {
ok[t] = 1;
return;
} else
ok[t] = 0;
p[t] = p[t - 1] * n;
if (p[t] >= (long long)((long long)1 << (long long)32))
ok[t] = 1;
else
ok[t] = 0;
return;
}
if (s == "end") {
t--;
} else {
res += p[t];
if (ok[t] || res >= (long long)((long long)1 << (long long)32)) {
cout << "OVERFLOW!!!";
exit(0);
}
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long T = 1;
p[0] = 1;
cin >> T;
for (int i = 1; i <= T; i++) {
solve();
}
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;
const long long M = 1000000;
long long n, ok[100005], res = 0, p[100005], t = 0;
string s;
void solve() {
cin >> s;
if (s == "for") {
cin >> n;
t++;
if (ok[t - 1]) {
ok[t] = 1;
return;
} else
ok[t] = 0;
p[t] = p[t - 1] * n;
if (p[t] >= (long long)((long long)1 << (long long)32))
ok[t] = 1;
else
ok[t] = 0;
return;
}
if (s == "end") {
t--;
} else {
res += p[t];
if (ok[t] || res >= (long long)((long long)1 << (long long)32)) {
cout << "OVERFLOW!!!";
exit(0);
}
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long T = 1;
p[0] = 1;
cin >> T;
for (int i = 1; i <= T; i++) {
solve();
}
cout << res;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int t, ans = 0, x;
long long int num = pow(2, 32) - 1;
string s;
stack<long long int> st;
cin >> t;
while (t--) {
if (ans > num) {
cout << "OVERFLOW!!!\n";
return 0;
}
cin >> s;
if (s == "end" && st.empty() == false)
st.pop();
else if (s == "add") {
if (st.empty())
ans += 1;
else {
if (st.top() == -1) {
cout << "OVERFLOW!!!\n";
return 0;
}
ans += st.top();
}
} else if (s == "for") {
cin >> x;
if (!st.empty()) {
long long int y = st.top();
if (y == -1)
x = -1;
else if (y * x > num)
x = -1;
else
x *= y;
}
st.push(x);
}
}
if (ans > num) {
cout << "OVERFLOW!!!\n";
return 0;
}
cout << ans << endl;
}
| ### 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() {
long long int t, ans = 0, x;
long long int num = pow(2, 32) - 1;
string s;
stack<long long int> st;
cin >> t;
while (t--) {
if (ans > num) {
cout << "OVERFLOW!!!\n";
return 0;
}
cin >> s;
if (s == "end" && st.empty() == false)
st.pop();
else if (s == "add") {
if (st.empty())
ans += 1;
else {
if (st.top() == -1) {
cout << "OVERFLOW!!!\n";
return 0;
}
ans += st.top();
}
} else if (s == "for") {
cin >> x;
if (!st.empty()) {
long long int y = st.top();
if (y == -1)
x = -1;
else if (y * x > num)
x = -1;
else
x *= y;
}
st.push(x);
}
}
if (ans > num) {
cout << "OVERFLOW!!!\n";
return 0;
}
cout << ans << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long INF = (1LL) << (32LL);
long long stck[100010];
int main() {
int n;
scanf("%d", &n);
stck[0] = 1;
int cur = 0;
long long ans = 0;
for (int i = 1; i <= n; i++) {
char str[10];
scanf("%s", str);
if (str[0] == 'e')
cur--;
else if (str[0] == 'a')
ans += stck[cur];
else {
cur++;
int k;
scanf("%d", &k);
stck[cur] = (long long)stck[cur - 1] * k;
stck[cur] = min(stck[cur], INF);
}
}
if (ans >= INF)
puts("OVERFLOW!!!");
else
printf("%lld\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;
const long long INF = (1LL) << (32LL);
long long stck[100010];
int main() {
int n;
scanf("%d", &n);
stck[0] = 1;
int cur = 0;
long long ans = 0;
for (int i = 1; i <= n; i++) {
char str[10];
scanf("%s", str);
if (str[0] == 'e')
cur--;
else if (str[0] == 'a')
ans += stck[cur];
else {
cur++;
int k;
scanf("%d", &k);
stck[cur] = (long long)stck[cur - 1] * k;
stck[cur] = min(stck[cur], INF);
}
}
if (ans >= INF)
puts("OVERFLOW!!!");
else
printf("%lld\n", ans);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 4294967295;
stack<long long> q;
int main() {
int l;
long long ans = 0, cnt = 1, flag = 0, tot = 0;
scanf("%I64d", &l);
getchar();
while (l--) {
char s[30];
long long x;
scanf("%s", s);
if (s[0] == 'a') {
ans += cnt;
if (ans > maxn || tot) {
flag = 1;
break;
}
} else if (s[0] == 'f') {
scanf("%I64d", &x);
if (cnt * x < maxn && !tot)
cnt *= x;
else
tot++;
q.push(x);
} else if (s[0] == 'e') {
x = q.top();
q.pop();
if (!tot)
cnt /= x;
else
tot--;
}
}
if (flag) {
printf("OVERFLOW!!!\n");
} else {
printf("%I64d\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;
const long long maxn = 4294967295;
stack<long long> q;
int main() {
int l;
long long ans = 0, cnt = 1, flag = 0, tot = 0;
scanf("%I64d", &l);
getchar();
while (l--) {
char s[30];
long long x;
scanf("%s", s);
if (s[0] == 'a') {
ans += cnt;
if (ans > maxn || tot) {
flag = 1;
break;
}
} else if (s[0] == 'f') {
scanf("%I64d", &x);
if (cnt * x < maxn && !tot)
cnt *= x;
else
tot++;
q.push(x);
} else if (s[0] == 'e') {
x = q.top();
q.pop();
if (!tot)
cnt /= x;
else
tot--;
}
}
if (flag) {
printf("OVERFLOW!!!\n");
} else {
printf("%I64d\n", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int M = 1000000;
int stk[M];
int tp = 1;
long long f[M];
int main() {
int l;
cin >> l;
long long x = 0;
long long mx = 1ll << 50;
long long up = 1ll << 32;
--up;
f[0] = 1;
bool over = 0;
for (int i = (0); i < (l); ++i) {
string s;
cin >> s;
int t;
if (s == "for") {
cin >> t;
stk[tp] = t;
f[tp] = f[tp - 1];
if (f[tp] < mx) f[tp] = f[tp - 1] * t;
++tp;
}
if (s == "end") {
--tp;
}
if (s == "add") {
x += f[tp - 1];
if (x > up) {
over = 1;
break;
}
}
}
if (over)
cout << "OVERFLOW!!!" << endl;
else
cout << x << 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 int M = 1000000;
int stk[M];
int tp = 1;
long long f[M];
int main() {
int l;
cin >> l;
long long x = 0;
long long mx = 1ll << 50;
long long up = 1ll << 32;
--up;
f[0] = 1;
bool over = 0;
for (int i = (0); i < (l); ++i) {
string s;
cin >> s;
int t;
if (s == "for") {
cin >> t;
stk[tp] = t;
f[tp] = f[tp - 1];
if (f[tp] < mx) f[tp] = f[tp - 1] * t;
++tp;
}
if (s == "end") {
--tp;
}
if (s == "add") {
x += f[tp - 1];
if (x > up) {
over = 1;
break;
}
}
}
if (over)
cout << "OVERFLOW!!!" << endl;
else
cout << x << endl;
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 T;
cin >> T;
long long ans = 0;
stack<long long> store;
long long curr = 1;
long long ind = 0;
while (T--) {
string s;
cin >> s;
if (s.at(0) == 'f') {
long long m;
cin >> m;
if (curr < ((long long)pow(2, 32) - (long long)1)) {
curr *= m;
store.push(m);
} else {
ind++;
}
} else if (s.at(0) == 'e') {
if (ind == 0) {
curr /= store.top();
store.pop();
} else {
ind--;
}
} else if (s.at(0) == 'a') {
ans += curr;
}
}
if (ans > ((long long)pow(2, 32) - (long long)1)) {
cout << "OVERFLOW!!!\n";
} 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;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long T;
cin >> T;
long long ans = 0;
stack<long long> store;
long long curr = 1;
long long ind = 0;
while (T--) {
string s;
cin >> s;
if (s.at(0) == 'f') {
long long m;
cin >> m;
if (curr < ((long long)pow(2, 32) - (long long)1)) {
curr *= m;
store.push(m);
} else {
ind++;
}
} else if (s.at(0) == 'e') {
if (ind == 0) {
curr /= store.top();
store.pop();
} else {
ind--;
}
} else if (s.at(0) == 'a') {
ans += curr;
}
}
if (ans > ((long long)pow(2, 32) - (long long)1)) {
cout << "OVERFLOW!!!\n";
} else
cout << ans << endl;
}
``` |
#include <bits/stdc++.h>
using namespace std;
template <class my>
my aabs(const my& a) {
return (a > 0) ? a : -a;
}
long long int gcd(long long int a, long long int b) {
return (b == 0) ? a : gcd(b, a % b);
}
long long int LSOne(long long int k) { return (k & (-k)); }
long long int excess = 4294967295;
void solve() {
long long int n, k, u = 0, v = 1, ans = 0, val = 1, c = 0;
cin >> n;
stack<long long int> st;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] == 'f') {
cin >> u;
}
if (val) {
if (s[0] == 'f') {
if (v <= excess) {
v *= u;
st.push(u);
} else
c++;
} else if (s[0] == 'a') {
if (v > excess)
val = 0;
else
ans += v;
} else {
if (c > 0)
c--;
else {
v /= st.top();
st.pop();
}
}
if (ans > excess) val = 0;
}
}
if (val)
cout << ans << endl;
else
cout << "OVERFLOW!!!" << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int t = 1;
while (t--) solve();
}
| ### 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;
template <class my>
my aabs(const my& a) {
return (a > 0) ? a : -a;
}
long long int gcd(long long int a, long long int b) {
return (b == 0) ? a : gcd(b, a % b);
}
long long int LSOne(long long int k) { return (k & (-k)); }
long long int excess = 4294967295;
void solve() {
long long int n, k, u = 0, v = 1, ans = 0, val = 1, c = 0;
cin >> n;
stack<long long int> st;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
if (s[0] == 'f') {
cin >> u;
}
if (val) {
if (s[0] == 'f') {
if (v <= excess) {
v *= u;
st.push(u);
} else
c++;
} else if (s[0] == 'a') {
if (v > excess)
val = 0;
else
ans += v;
} else {
if (c > 0)
c--;
else {
v /= st.top();
st.pop();
}
}
if (ans > excess) val = 0;
}
}
if (val)
cout << ans << endl;
else
cout << "OVERFLOW!!!" << endl;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int t = 1;
while (t--) solve();
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long MAXX = (1LL << (32)) - 1LL;
int n, l, idx;
string cmd;
vector<pair<char, long long> > cps;
long long forr() {
long long fn = cps[idx].second;
idx++;
long long cur = 0;
while (1) {
char c = cps[idx].first;
if (c == 'e') {
idx++;
break;
} else if (c == 'a') {
cur++;
idx++;
} else if (c == 'f') {
long long nc = forr();
if (nc < 0) return -1;
cur += nc;
}
}
if (cur > MAXX || (fn * cur) > MAXX) return -1;
return fn * cur;
}
string solve() {
idx = 0;
long long res = forr();
if (res < 0) return "OVERFLOW!!!";
stringstream ss;
ss << res;
string ress;
ss >> ress;
return ress;
}
int main() {
cin >> n;
cps.push_back(pair<char, long long>('f', 1));
for (int i = 0; i < n; i++) {
pair<char, long long> cp;
cin >> cmd;
if (cmd == "for") {
cin >> l;
cp.first = 'f';
cp.second = l;
} else {
cp.first = cmd[0];
}
cps.push_back(cp);
}
cps.push_back(pair<char, long long>('e', 0));
n += 2;
cout << solve() << endl;
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;
long long MAXX = (1LL << (32)) - 1LL;
int n, l, idx;
string cmd;
vector<pair<char, long long> > cps;
long long forr() {
long long fn = cps[idx].second;
idx++;
long long cur = 0;
while (1) {
char c = cps[idx].first;
if (c == 'e') {
idx++;
break;
} else if (c == 'a') {
cur++;
idx++;
} else if (c == 'f') {
long long nc = forr();
if (nc < 0) return -1;
cur += nc;
}
}
if (cur > MAXX || (fn * cur) > MAXX) return -1;
return fn * cur;
}
string solve() {
idx = 0;
long long res = forr();
if (res < 0) return "OVERFLOW!!!";
stringstream ss;
ss << res;
string ress;
ss >> ress;
return ress;
}
int main() {
cin >> n;
cps.push_back(pair<char, long long>('f', 1));
for (int i = 0; i < n; i++) {
pair<char, long long> cp;
cin >> cmd;
if (cmd == "for") {
cin >> l;
cp.first = 'f';
cp.second = l;
} else {
cp.first = cmd[0];
}
cps.push_back(cp);
}
cps.push_back(pair<char, long long>('e', 0));
n += 2;
cout << solve() << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long po(long long x, long long y) {
if (y == 0)
return 1;
else
return x * po(x, y - 1);
}
long long n;
string s[100010];
long long i = 0;
long long solve() {
long long ans = 0;
while (i < n) {
cin >> s[i];
if (s[i][0] == 'a') {
i++;
ans++;
} else if (s[i][0] == 'f') {
long long x;
cin >> x;
i++;
ans += (x * solve());
} else {
i++;
if (ans >= po(2, 32)) {
cout << "OVERFLOW!!!";
exit(0);
} else
return ans;
}
}
if (ans >= po(2, 32)) {
cout << "OVERFLOW!!!";
exit(0);
} else
return ans;
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n;
cout << solve();
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;
long long po(long long x, long long y) {
if (y == 0)
return 1;
else
return x * po(x, y - 1);
}
long long n;
string s[100010];
long long i = 0;
long long solve() {
long long ans = 0;
while (i < n) {
cin >> s[i];
if (s[i][0] == 'a') {
i++;
ans++;
} else if (s[i][0] == 'f') {
long long x;
cin >> x;
i++;
ans += (x * solve());
} else {
i++;
if (ans >= po(2, 32)) {
cout << "OVERFLOW!!!";
exit(0);
} else
return ans;
}
}
if (ans >= po(2, 32)) {
cout << "OVERFLOW!!!";
exit(0);
} else
return ans;
}
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n;
cout << solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int l;
long long ans = 0;
int last = 0;
long long ml = 1;
cin >> l;
stack<long long> a;
a.push(1);
while (l--) {
string s;
cin >> s;
long long t;
if (s == "for") {
cin >> t;
a.push(min(t * a.top(), 1ll << 32));
} else if (s == "end") {
if (a.size() > 0) a.pop();
} else if (s == "add") {
long long mul = 1;
if (ans == -1) {
continue;
}
mul = a.top();
if (mul == -1) {
ans = -1;
} else
ans += mul;
if (ans >= pow(2, 32)) ans = -1;
}
}
if (ans != -1) {
cout << ans;
} else {
cout << "OVERFLOW!!!";
}
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;
long long ans = 0;
int last = 0;
long long ml = 1;
cin >> l;
stack<long long> a;
a.push(1);
while (l--) {
string s;
cin >> s;
long long t;
if (s == "for") {
cin >> t;
a.push(min(t * a.top(), 1ll << 32));
} else if (s == "end") {
if (a.size() > 0) a.pop();
} else if (s == "add") {
long long mul = 1;
if (ans == -1) {
continue;
}
mul = a.top();
if (mul == -1) {
ans = -1;
} else
ans += mul;
if (ans >= pow(2, 32)) ans = -1;
}
}
if (ans != -1) {
cout << ans;
} else {
cout << "OVERFLOW!!!";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-8;
const long long mod = 1e9 + 7;
const long long INF = 4294967295;
const int N = 1e5 + 2;
const int K = 17;
void solve() {
int l;
cin >> l;
long long x = 0;
stack<long long> sfor;
long long rep = 1;
bool overfl = 0;
stack<long long> overfl_cnt;
bool is_overfl_now = false;
while (l--) {
string str;
cin >> str;
if (str == "for") {
long long cnt;
cin >> cnt;
if (overfl) continue;
if (is_overfl_now) {
overfl_cnt.push(cnt);
continue;
}
if (rep * cnt > INF) {
is_overfl_now = true;
overfl_cnt.push(cnt);
continue;
}
if (!is_overfl_now) {
sfor.push(cnt);
rep *= cnt;
}
} else if (str == "end") {
if (overfl) continue;
if (!is_overfl_now) {
long long cnt = sfor.top();
rep /= cnt;
sfor.pop();
} else {
overfl_cnt.pop();
if (overfl_cnt.empty()) is_overfl_now = false;
}
} else if (str == "add") {
if (is_overfl_now) overfl = true;
if (overfl) continue;
x += rep;
if (x > INF) overfl = true;
}
}
if (overfl)
cout << "OVERFLOW!!!\n";
else
cout << x << endl;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
while (t--) {
solve();
}
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 double eps = 1e-8;
const long long mod = 1e9 + 7;
const long long INF = 4294967295;
const int N = 1e5 + 2;
const int K = 17;
void solve() {
int l;
cin >> l;
long long x = 0;
stack<long long> sfor;
long long rep = 1;
bool overfl = 0;
stack<long long> overfl_cnt;
bool is_overfl_now = false;
while (l--) {
string str;
cin >> str;
if (str == "for") {
long long cnt;
cin >> cnt;
if (overfl) continue;
if (is_overfl_now) {
overfl_cnt.push(cnt);
continue;
}
if (rep * cnt > INF) {
is_overfl_now = true;
overfl_cnt.push(cnt);
continue;
}
if (!is_overfl_now) {
sfor.push(cnt);
rep *= cnt;
}
} else if (str == "end") {
if (overfl) continue;
if (!is_overfl_now) {
long long cnt = sfor.top();
rep /= cnt;
sfor.pop();
} else {
overfl_cnt.pop();
if (overfl_cnt.empty()) is_overfl_now = false;
}
} else if (str == "add") {
if (is_overfl_now) overfl = true;
if (overfl) continue;
x += rep;
if (x > INF) overfl = true;
}
}
if (overfl)
cout << "OVERFLOW!!!\n";
else
cout << x << endl;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
while (t--) {
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
long long mod = 1e9 + 7;
long long fmod(long long b, long long exp) {
long long res = 1;
while (exp) {
if (exp & 1ll) res = (res * b) % mod;
b = (b * b) % mod;
exp /= 2ll;
}
return res;
}
stack<long long> st;
string second;
int main() {
int n, m, i, j, k, x, q;
int t = 1;
long long cur = 1;
long long limit = (1ll << 32) - 1;
long long tot = 0;
int fl = 0;
cin >> t;
while (t--) {
cin >> second;
if (second == "for") {
long long num;
cin >> num;
if (cur <= limit) cur *= num;
st.push(cur);
} else if (second == "add") {
if (st.empty())
tot++;
else {
if (st.top() > limit)
fl = 1;
else
tot += st.top();
}
} else {
st.pop();
if (!st.empty())
cur = st.top();
else
cur = 1;
}
}
if (fl == 1 || tot > limit)
cout << "OVERFLOW!!!\n";
else
cout << tot << "\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;
const int N = 100010;
long long mod = 1e9 + 7;
long long fmod(long long b, long long exp) {
long long res = 1;
while (exp) {
if (exp & 1ll) res = (res * b) % mod;
b = (b * b) % mod;
exp /= 2ll;
}
return res;
}
stack<long long> st;
string second;
int main() {
int n, m, i, j, k, x, q;
int t = 1;
long long cur = 1;
long long limit = (1ll << 32) - 1;
long long tot = 0;
int fl = 0;
cin >> t;
while (t--) {
cin >> second;
if (second == "for") {
long long num;
cin >> num;
if (cur <= limit) cur *= num;
st.push(cur);
} else if (second == "add") {
if (st.empty())
tot++;
else {
if (st.top() > limit)
fl = 1;
else
tot += st.top();
}
} else {
st.pop();
if (!st.empty())
cur = st.top();
else
cur = 1;
}
}
if (fl == 1 || tot > limit)
cout << "OVERFLOW!!!\n";
else
cout << tot << "\n";
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long t, x;
string str;
stack<long long> st, stf;
st.push(0);
cin >> t;
while (t--) {
cin >> str;
if (str == "add") {
st.top()++;
} else if (str == "for") {
cin >> x;
stf.push(x);
st.push(0);
} else {
x = st.top();
st.pop();
x *= stf.top();
stf.pop();
st.top() += x;
if (st.top() > (1LL << 32) - 1) {
cout << "OVERFLOW!!!";
return 0;
}
}
}
if (st.top() > (1LL << 32) - 1) {
cout << "OVERFLOW!!!";
return 0;
}
cout << st.top();
}
| ### 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() {
long long t, x;
string str;
stack<long long> st, stf;
st.push(0);
cin >> t;
while (t--) {
cin >> str;
if (str == "add") {
st.top()++;
} else if (str == "for") {
cin >> x;
stf.push(x);
st.push(0);
} else {
x = st.top();
st.pop();
x *= stf.top();
stf.pop();
st.top() += x;
if (st.top() > (1LL << 32) - 1) {
cout << "OVERFLOW!!!";
return 0;
}
}
}
if (st.top() > (1LL << 32) - 1) {
cout << "OVERFLOW!!!";
return 0;
}
cout << st.top();
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
const long long MAX = (1ll << 32);
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
long long ans = 0;
stack<long long> st;
long long n, pro;
cin >> n;
st.push(1LL);
while (n--) {
string s;
cin >> s;
if (s == "add") {
ans += st.top();
} else if (s == "for") {
cin >> t;
pro = st.top() * t;
st.push(min(MAX, pro));
} else if (s == "end") {
st.pop();
}
}
if (ans >= MAX)
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 mod = 1e9 + 7;
const long long MAX = (1ll << 32);
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long t;
long long ans = 0;
stack<long long> st;
long long n, pro;
cin >> n;
st.push(1LL);
while (n--) {
string s;
cin >> s;
if (s == "add") {
ans += st.top();
} else if (s == "for") {
cin >> t;
pro = st.top() * t;
st.push(min(MAX, pro));
} else if (s == "end") {
st.pop();
}
}
if (ans >= MAX)
cout << "OVERFLOW!!!";
else
cout << ans;
}
``` |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using i2 = pair<int, int>;
using vi = vector<int>;
int main() {
ll _2_32 = 1;
for (int i = 1; i <= 32; i++) _2_32 <<= 1;
int l;
cin >> l;
stack<pair<ll, ll> > loop;
loop.push({1, 0});
bool overflow = false;
for (int i = 1; i <= l; i++) {
string s;
cin >> s;
if (s == "for") {
ll n;
cin >> n;
loop.push({n, 0});
} else if (s == "add") {
ll n = loop.top().first;
ll val = loop.top().second;
loop.pop();
loop.push({n, val + 1});
} else if (!overflow) {
ll n1 = loop.top().first;
ll val1 = loop.top().second;
loop.pop();
ll n2 = loop.top().first;
ll val2 = loop.top().second;
loop.pop();
loop.push({n2, val2 + n1 * val1});
if (n2 * (val2 + n1 * val1) >= _2_32) overflow = true;
}
}
if (overflow || loop.top().second >= _2_32)
cout << "OVERFLOW!!!";
else
cout << loop.top().second;
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;
using ll = long long;
using i2 = pair<int, int>;
using vi = vector<int>;
int main() {
ll _2_32 = 1;
for (int i = 1; i <= 32; i++) _2_32 <<= 1;
int l;
cin >> l;
stack<pair<ll, ll> > loop;
loop.push({1, 0});
bool overflow = false;
for (int i = 1; i <= l; i++) {
string s;
cin >> s;
if (s == "for") {
ll n;
cin >> n;
loop.push({n, 0});
} else if (s == "add") {
ll n = loop.top().first;
ll val = loop.top().second;
loop.pop();
loop.push({n, val + 1});
} else if (!overflow) {
ll n1 = loop.top().first;
ll val1 = loop.top().second;
loop.pop();
ll n2 = loop.top().first;
ll val2 = loop.top().second;
loop.pop();
loop.push({n2, val2 + n1 * val1});
if (n2 * (val2 + n1 * val1) >= _2_32) overflow = true;
}
}
if (overflow || loop.top().second >= _2_32)
cout << "OVERFLOW!!!";
else
cout << loop.top().second;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = (1ll << 32) - 1;
int main() {
int l;
while (cin >> l) {
string s;
stack<long long> st;
st.push(1);
long long cur = 0, n;
bool flg = true;
for (int i = 0; i < l; i++) {
cin >> s;
if (s == "for") {
cin >> n;
st.push(min(MAXN + 1, n * st.top()));
} else if (s == "add") {
cur += st.top();
if (cur > MAXN) flg = false;
} else
st.pop();
}
if (!flg)
cout << "OVERFLOW!!!" << endl;
else
cout << cur << endl;
}
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;
const long long MAXN = (1ll << 32) - 1;
int main() {
int l;
while (cin >> l) {
string s;
stack<long long> st;
st.push(1);
long long cur = 0, n;
bool flg = true;
for (int i = 0; i < l; i++) {
cin >> s;
if (s == "for") {
cin >> n;
st.push(min(MAXN + 1, n * st.top()));
} else if (s == "add") {
cur += st.top();
if (cur > MAXN) flg = false;
} else
st.pop();
}
if (!flg)
cout << "OVERFLOW!!!" << endl;
else
cout << cur << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long inf = pow(2, 32) - 1;
const int maxn = 100005;
char c[maxn][5];
int a[maxn];
int tag;
map<int, int> ma;
long long f(int l, int r) {
if (tag == -1) return 0;
long long re = 0;
for (int i = l; i <= r;) {
if (c[i][0] == 'a')
re++, i++;
else {
re += (long long)a[i] * f(i + 1, ma[i] - 1);
i = ma[i] + 1;
}
if (re > inf) return tag = -1;
}
return re;
}
void init(int l, int r) {
stack<int> ss;
for (int i = l; i <= r; i++) {
if (c[i][0] == 'f') ss.push(i);
if (c[i][0] == 'e') {
ma[ss.top()] = i;
ss.pop();
}
}
}
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%s", &c[i]);
if (c[i][0] == 'f') scanf("%d", &a[i]);
}
init(1, n);
long long ans = f(1, n);
if (tag == -1)
printf("OVERFLOW!!!\n");
else
printf("%I64d\n", 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;
const long long inf = pow(2, 32) - 1;
const int maxn = 100005;
char c[maxn][5];
int a[maxn];
int tag;
map<int, int> ma;
long long f(int l, int r) {
if (tag == -1) return 0;
long long re = 0;
for (int i = l; i <= r;) {
if (c[i][0] == 'a')
re++, i++;
else {
re += (long long)a[i] * f(i + 1, ma[i] - 1);
i = ma[i] + 1;
}
if (re > inf) return tag = -1;
}
return re;
}
void init(int l, int r) {
stack<int> ss;
for (int i = l; i <= r; i++) {
if (c[i][0] == 'f') ss.push(i);
if (c[i][0] == 'e') {
ma[ss.top()] = i;
ss.pop();
}
}
}
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%s", &c[i]);
if (c[i][0] == 'f') scanf("%d", &a[i]);
}
init(1, n);
long long ans = f(1, n);
if (tag == -1)
printf("OVERFLOW!!!\n");
else
printf("%I64d\n", ans);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<long long> tomb;
int main() {
int n;
cin >> n;
bool over = false;
long long hatar = UINT_MAX;
for (int x = 0; n > x && !over; ++x) {
string a;
cin >> a;
if (a == "for") {
int b;
cin >> b;
tomb.push_back(-b);
} else if (a == "add") {
tomb.push_back(1);
} else if (a == "end") {
long long szamolo = 0;
while (tomb.back() >= 0) {
szamolo += tomb.back();
tomb.pop_back();
}
long long ww = szamolo * tomb.back() * -1;
if (ww <= hatar) {
tomb.back() = ww;
} else {
over = true;
}
}
}
long long c = 0;
for (int x = 0; tomb.size() > x && !over; ++x) {
if (c + tomb[x] <= hatar) {
c += tomb[x];
} else {
over = true;
}
}
if (over) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << c << endl;
}
}
| ### 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;
vector<long long> tomb;
int main() {
int n;
cin >> n;
bool over = false;
long long hatar = UINT_MAX;
for (int x = 0; n > x && !over; ++x) {
string a;
cin >> a;
if (a == "for") {
int b;
cin >> b;
tomb.push_back(-b);
} else if (a == "add") {
tomb.push_back(1);
} else if (a == "end") {
long long szamolo = 0;
while (tomb.back() >= 0) {
szamolo += tomb.back();
tomb.pop_back();
}
long long ww = szamolo * tomb.back() * -1;
if (ww <= hatar) {
tomb.back() = ww;
} else {
over = true;
}
}
}
long long c = 0;
for (int x = 0; tomb.size() > x && !over; ++x) {
if (c + tomb[x] <= hatar) {
c += tomb[x];
} else {
over = true;
}
}
if (over) {
cout << "OVERFLOW!!!" << endl;
} else {
cout << c << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long n, x, ans;
string s;
stack<pair<long long, long long> > S;
bool check(long long x) {
if (x < 0 || x >= (1ll << 32)) return 0;
return 1;
}
void solve() {
cin >> n;
for (__typeof((n + 1)) i = (1); i < (n + 1); i++) {
cin >> s;
if (s == "add") {
if (!(long long)S.size())
ans++;
else {
pair<long long, long long> p = S.top();
S.pop();
p.first++;
S.push(p);
}
} else if (s == "for") {
cin >> x;
S.push({0, x});
} else if (s == "end") {
pair<long long, long long> p = S.top();
S.pop();
if (!(long long)S.size())
ans += p.first * p.second;
else {
pair<long long, long long> q = S.top();
S.pop();
q.first += p.first * p.second;
if (!check(q.first)) {
cout << "OVERFLOW!!!"
<< "\n";
return;
};
S.push(q);
}
}
if (!check(ans)) {
cout << "OVERFLOW!!!"
<< "\n";
return;
};
}
{
cout << ans << "\n";
return;
};
}
void prep() {}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long t = 1;
prep();
cout << fixed << setprecision(12);
while (t--) 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>
using namespace std;
long long n, x, ans;
string s;
stack<pair<long long, long long> > S;
bool check(long long x) {
if (x < 0 || x >= (1ll << 32)) return 0;
return 1;
}
void solve() {
cin >> n;
for (__typeof((n + 1)) i = (1); i < (n + 1); i++) {
cin >> s;
if (s == "add") {
if (!(long long)S.size())
ans++;
else {
pair<long long, long long> p = S.top();
S.pop();
p.first++;
S.push(p);
}
} else if (s == "for") {
cin >> x;
S.push({0, x});
} else if (s == "end") {
pair<long long, long long> p = S.top();
S.pop();
if (!(long long)S.size())
ans += p.first * p.second;
else {
pair<long long, long long> q = S.top();
S.pop();
q.first += p.first * p.second;
if (!check(q.first)) {
cout << "OVERFLOW!!!"
<< "\n";
return;
};
S.push(q);
}
}
if (!check(ans)) {
cout << "OVERFLOW!!!"
<< "\n";
return;
};
}
{
cout << ans << "\n";
return;
};
}
void prep() {}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long t = 1;
prep();
cout << fixed << setprecision(12);
while (t--) solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
signed main() {
long long r = 1;
for (long long i = 1; i <= 32; i++) {
r = r * 2;
}
r = r - 1;
long long q;
cin >> q;
vector<long long> bag, ex;
long long ans = 0, x = 1, g = 0, pubg = 0;
while (q--) {
string s;
cin >> s;
if (s == "add") {
if (g) {
pubg = 1;
} else {
ans += x;
if (ans > r) {
pubg = 1;
}
}
} else if (s == "for") {
long long val;
cin >> val;
if (g == 0) x = x * val;
if (x > r) {
g = 1;
x /= val;
}
if (g) {
ex.push_back(val);
} else {
bag.push_back(val);
}
} else {
if (g && ex.size()) {
long long sz = ex.size() - 1;
ex.erase(ex.begin() + sz);
} else if (!g && bag.size()) {
long long sz = bag.size() - 1;
long long a = bag[sz];
bag.erase(bag.begin() + sz);
x /= a;
}
if (ex.size() == 0) {
g = 0;
}
}
}
if (!pubg) {
cout << ans << endl;
} else {
cout << "OVERFLOW!!!" << endl;
}
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;
signed main() {
long long r = 1;
for (long long i = 1; i <= 32; i++) {
r = r * 2;
}
r = r - 1;
long long q;
cin >> q;
vector<long long> bag, ex;
long long ans = 0, x = 1, g = 0, pubg = 0;
while (q--) {
string s;
cin >> s;
if (s == "add") {
if (g) {
pubg = 1;
} else {
ans += x;
if (ans > r) {
pubg = 1;
}
}
} else if (s == "for") {
long long val;
cin >> val;
if (g == 0) x = x * val;
if (x > r) {
g = 1;
x /= val;
}
if (g) {
ex.push_back(val);
} else {
bag.push_back(val);
}
} else {
if (g && ex.size()) {
long long sz = ex.size() - 1;
ex.erase(ex.begin() + sz);
} else if (!g && bag.size()) {
long long sz = bag.size() - 1;
long long a = bag[sz];
bag.erase(bag.begin() + sz);
x /= a;
}
if (ex.size() == 0) {
g = 0;
}
}
}
if (!pubg) {
cout << ans << endl;
} else {
cout << "OVERFLOW!!!" << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
const long long maxi = (1ll << 32) - 1;
long long x = 0;
cin;
string s;
long long d = 1;
cin >> s;
vector<long long> fors = {1};
vector<long long> mul = {1};
vector<bool> is_overflowed = {false};
while (cin >> s) {
if (s[0] == 'f') {
cin >> d;
fors.push_back(d);
mul.push_back(mul.back() * d);
if (!is_overflowed.back()) {
is_overflowed.push_back(mul.back() > maxi);
} else {
is_overflowed.push_back(true);
}
} else if (s[0] == 'e') {
fors.pop_back();
mul.pop_back();
is_overflowed.pop_back();
} else {
if (is_overflowed.back()) {
cout << "OVERFLOW!!!";
return 0;
}
x += mul.back();
if (x > maxi) {
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 main() {
const long long maxi = (1ll << 32) - 1;
long long x = 0;
cin;
string s;
long long d = 1;
cin >> s;
vector<long long> fors = {1};
vector<long long> mul = {1};
vector<bool> is_overflowed = {false};
while (cin >> s) {
if (s[0] == 'f') {
cin >> d;
fors.push_back(d);
mul.push_back(mul.back() * d);
if (!is_overflowed.back()) {
is_overflowed.push_back(mul.back() > maxi);
} else {
is_overflowed.push_back(true);
}
} else if (s[0] == 'e') {
fors.pop_back();
mul.pop_back();
is_overflowed.pop_back();
} else {
if (is_overflowed.back()) {
cout << "OVERFLOW!!!";
return 0;
}
x += mul.back();
if (x > maxi) {
cout << "OVERFLOW!!!";
return 0;
}
}
}
cout << x;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long a[1000007], b[1000007];
int main() {
int n;
cin >> n;
long long x = 0LL, idx = 0LL, loop = 1LL;
int flag = 0;
string t;
for (int i = 0; i < 1000007; i++) a[i] = 1e18 + 7;
a[0] = 1;
while (n--) {
cin >> t;
loop = min(a[idx], loop);
if (t == "for") {
idx++;
cin >> a[idx];
a[idx] = a[idx - 1] * a[idx];
if (a[idx] > 4294967295) flag = 1;
b[idx] = flag;
}
if (t == "end") {
b[idx] = 0;
idx--;
}
if (t == "add") {
if (b[idx] == 1) {
cout << "OVERFLOW!!!\n";
return 0;
}
x += a[idx];
}
if (x > 4294967295) {
cout << "OVERFLOW!!!\n";
return 0;
}
if (a[idx] > 4294967295) flag = 1;
if (b[idx] == 0) flag = 0;
}
if (x > 4294967295) {
cout << "OVERFLOW!!!\n";
return 0;
}
cout << x << 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;
long long a[1000007], b[1000007];
int main() {
int n;
cin >> n;
long long x = 0LL, idx = 0LL, loop = 1LL;
int flag = 0;
string t;
for (int i = 0; i < 1000007; i++) a[i] = 1e18 + 7;
a[0] = 1;
while (n--) {
cin >> t;
loop = min(a[idx], loop);
if (t == "for") {
idx++;
cin >> a[idx];
a[idx] = a[idx - 1] * a[idx];
if (a[idx] > 4294967295) flag = 1;
b[idx] = flag;
}
if (t == "end") {
b[idx] = 0;
idx--;
}
if (t == "add") {
if (b[idx] == 1) {
cout << "OVERFLOW!!!\n";
return 0;
}
x += a[idx];
}
if (x > 4294967295) {
cout << "OVERFLOW!!!\n";
return 0;
}
if (a[idx] > 4294967295) flag = 1;
if (b[idx] == 0) flag = 0;
}
if (x > 4294967295) {
cout << "OVERFLOW!!!\n";
return 0;
}
cout << x << endl;
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;
cin >> l;
stack<long long int> S;
S.push(0);
while (l--) {
string s;
cin >> s;
if (s == "for") {
long long int n;
cin >> n;
if (multi * n > pow(2, 32) - 1 || S.top() < 0) {
long long int temp = S.top() - 1;
S.pop();
S.push(temp);
} else {
multi *= n;
S.pop();
S.push(n);
S.push(0);
}
} else if (s == "end") {
if (S.top() < 0) {
long long int temp = S.top() + 1;
S.pop();
S.push(temp);
} else {
S.pop();
multi /= S.top();
S.pop();
S.push(0);
}
} else {
if (S.size() == 1) {
ans++;
} else {
if (S.top() < 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
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() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
long long int l, ans = 0, multi = 1;
cin >> l;
stack<long long int> S;
S.push(0);
while (l--) {
string s;
cin >> s;
if (s == "for") {
long long int n;
cin >> n;
if (multi * n > pow(2, 32) - 1 || S.top() < 0) {
long long int temp = S.top() - 1;
S.pop();
S.push(temp);
} else {
multi *= n;
S.pop();
S.push(n);
S.push(0);
}
} else if (s == "end") {
if (S.top() < 0) {
long long int temp = S.top() + 1;
S.pop();
S.push(temp);
} else {
S.pop();
multi /= S.top();
S.pop();
S.push(0);
}
} else {
if (S.size() == 1) {
ans++;
} else {
if (S.top() < 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;
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
inline int GA(long long n, vector<long long>& vec) {
for (int i = 0; i < n; ++i) {
long long x;
cin >> x;
vec.push_back(x);
}
return 0;
}
inline int GAEG(long long m, vector<long long> vec[]) {
for (int i = 0; i < m; ++i) {
long long x, y;
cin >> x >> y;
x--, y--;
vec[x].push_back(y);
vec[y].push_back(x);
}
return 0;
}
const long long MD = (long long)1e9 + 7;
bool isvalid(long long xx, long long yy, long long n, long long m) {
return (xx >= 0 && xx < n && yy >= 0 && yy < m);
}
long long pw(long long x, long long y, long long dr) {
if (y == 0) return 1;
long long o = y / 2;
long long f = pw(x, o, dr) % dr;
if (y % 2) {
return (((f * f) % dr) * x) % dr;
} else
return (f * f) % dr;
}
long long p[200005];
vector<pair<long long, long long> > c;
long long deg[200005];
vector<long long> v[200005];
long long FF(long long x) { return (p[x] >= 0) ? p[x] = FF(p[x]) : x; }
int main() {
long long t;
cin >> t;
string s;
getline(cin, s);
long long x = 0;
bool bl = false;
long long p = pw(2, 32, 1000000000000) - 1;
stack<long long> st;
long long oo = 1;
long long dd = 0;
while (t--) {
getline(cin, s);
if (s[0] == 'a') {
x += oo;
if (x > p) {
bl = true;
break;
}
} else if (s[0] == 'f') {
stringstream ss;
ss << s;
ss.clear();
string zz;
ss >> zz;
long long kk;
ss >> kk;
if (oo <= p) {
st.push(kk);
oo *= kk;
} else
dd++;
} else {
if (dd) {
dd--;
continue;
}
long long kk = st.top();
st.pop();
oo /= kk;
}
}
if (bl)
cout << "OVERFLOW!!!" << endl;
else
cout << x << 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 dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};
inline int GA(long long n, vector<long long>& vec) {
for (int i = 0; i < n; ++i) {
long long x;
cin >> x;
vec.push_back(x);
}
return 0;
}
inline int GAEG(long long m, vector<long long> vec[]) {
for (int i = 0; i < m; ++i) {
long long x, y;
cin >> x >> y;
x--, y--;
vec[x].push_back(y);
vec[y].push_back(x);
}
return 0;
}
const long long MD = (long long)1e9 + 7;
bool isvalid(long long xx, long long yy, long long n, long long m) {
return (xx >= 0 && xx < n && yy >= 0 && yy < m);
}
long long pw(long long x, long long y, long long dr) {
if (y == 0) return 1;
long long o = y / 2;
long long f = pw(x, o, dr) % dr;
if (y % 2) {
return (((f * f) % dr) * x) % dr;
} else
return (f * f) % dr;
}
long long p[200005];
vector<pair<long long, long long> > c;
long long deg[200005];
vector<long long> v[200005];
long long FF(long long x) { return (p[x] >= 0) ? p[x] = FF(p[x]) : x; }
int main() {
long long t;
cin >> t;
string s;
getline(cin, s);
long long x = 0;
bool bl = false;
long long p = pw(2, 32, 1000000000000) - 1;
stack<long long> st;
long long oo = 1;
long long dd = 0;
while (t--) {
getline(cin, s);
if (s[0] == 'a') {
x += oo;
if (x > p) {
bl = true;
break;
}
} else if (s[0] == 'f') {
stringstream ss;
ss << s;
ss.clear();
string zz;
ss >> zz;
long long kk;
ss >> kk;
if (oo <= p) {
st.push(kk);
oo *= kk;
} else
dd++;
} else {
if (dd) {
dd--;
continue;
}
long long kk = st.top();
st.pop();
oo /= kk;
}
}
if (bl)
cout << "OVERFLOW!!!" << endl;
else
cout << x << endl;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int l;
stack<long long int> s;
long long x = 0;
const long long INF = 1ll << 32;
int main() {
cin >> l;
s.push(1);
while (l--) {
string c;
cin >> c;
if (c == "for") {
int a;
cin >> a;
s.push(min(INF, s.top() * a));
} else if (c == "end") {
s.pop();
} else {
x += s.top();
}
}
if (x >= INF) {
cout << "OVERFLOW!!!";
} else {
cout << x;
}
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;
int l;
stack<long long int> s;
long long x = 0;
const long long INF = 1ll << 32;
int main() {
cin >> l;
s.push(1);
while (l--) {
string c;
cin >> c;
if (c == "for") {
int a;
cin >> a;
s.push(min(INF, s.top() * a));
} else if (c == "end") {
s.pop();
} else {
x += s.top();
}
}
if (x >= INF) {
cout << "OVERFLOW!!!";
} else {
cout << x;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long INF = 0xffffffff;
const int MOD = 998244353;
const int nmax = 101234;
int l, n, pos, flg;
long long cnt[nmax], cnt2[nmax];
char s[10];
int main() {
scanf("%d", &l);
for (int i = 0; i < l; ++i) {
scanf("%s", s);
if (s[0] == 'f') {
scanf("%d", &n);
cnt[++pos] = n;
cnt2[pos] = 0;
} else if (s[0] == 'a')
++cnt2[pos];
else if (s[0] == 'e') {
cnt2[pos - 1] = cnt2[pos - 1] + cnt2[pos] * cnt[pos];
--pos;
if (cnt2[pos] > INF) {
printf("OVERFLOW!!!\n");
return 0;
}
}
}
if (cnt2[0] > INF)
printf("OVERFLOW!!!\n");
else
printf("%I64d\n", cnt2[0]);
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 = 0xffffffff;
const int MOD = 998244353;
const int nmax = 101234;
int l, n, pos, flg;
long long cnt[nmax], cnt2[nmax];
char s[10];
int main() {
scanf("%d", &l);
for (int i = 0; i < l; ++i) {
scanf("%s", s);
if (s[0] == 'f') {
scanf("%d", &n);
cnt[++pos] = n;
cnt2[pos] = 0;
} else if (s[0] == 'a')
++cnt2[pos];
else if (s[0] == 'e') {
cnt2[pos - 1] = cnt2[pos - 1] + cnt2[pos] * cnt[pos];
--pos;
if (cnt2[pos] > INF) {
printf("OVERFLOW!!!\n");
return 0;
}
}
}
if (cnt2[0] > INF)
printf("OVERFLOW!!!\n");
else
printf("%I64d\n", cnt2[0]);
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long const MAXN = 1e5 + 8;
long long const INF = 1e12 + 8;
long long const delta = 1000000007;
int main() {
std::ios::sync_with_stdio(0);
;
cin.tie(0);
cout.tie(0);
long long n;
cin >> n;
long long temp = 1;
long long x = 0;
string s;
long long t, last;
long long over = (1ll << 32) - 1;
long long bad = 0;
vector<long long> com;
com.push_back(1);
;
for (long long i = 0; i < n; i++) {
cin >> s;
if (s == "add") {
x += (temp);
} else if (s == "for") {
cin >> t;
if (temp * t > over)
temp = over + 1;
else
temp *= t;
com.push_back(temp);
;
} else {
com.pop_back();
temp = com[com.size() - 1];
}
if (x > over) bad = 1;
}
if (bad) return cout << "OVERFLOW!!!", 0;
cout << x;
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;
long long const MAXN = 1e5 + 8;
long long const INF = 1e12 + 8;
long long const delta = 1000000007;
int main() {
std::ios::sync_with_stdio(0);
;
cin.tie(0);
cout.tie(0);
long long n;
cin >> n;
long long temp = 1;
long long x = 0;
string s;
long long t, last;
long long over = (1ll << 32) - 1;
long long bad = 0;
vector<long long> com;
com.push_back(1);
;
for (long long i = 0; i < n; i++) {
cin >> s;
if (s == "add") {
x += (temp);
} else if (s == "for") {
cin >> t;
if (temp * t > over)
temp = over + 1;
else
temp *= t;
com.push_back(temp);
;
} else {
com.pop_back();
temp = com[com.size() - 1];
}
if (x > over) bad = 1;
}
if (bad) return cout << "OVERFLOW!!!", 0;
cout << x;
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int T, n, x;
cin >> T;
while (T--) {
cin >> n >> x;
cout << x * 2 << endl;
}
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int T, n, x;
cin >> T;
while (T--) {
cin >> n >> x;
cout << x * 2 << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
long long x, n;
cin >> t;
for (int i = 0; i < t; i++) {
cin >> n;
cin >> x;
cout << 2 * x << endl;
}
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
long long x, n;
cin >> t;
for (int i = 0; i < t; i++) {
cin >> n;
cin >> x;
cout << 2 * x << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t, n, x;
cin >> t;
while (t--) {
cin >> n >> x;
cout << 2 * x << endl;
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t, n, x;
cin >> t;
while (t--) {
cin >> n >> x;
cout << 2 * x << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int T;
int N[101];
int x[101];
vector<int> num;
int main(void) {
cin >> T;
for (int i = 0; i < T; i++) {
cin >> N[i];
cin >> x[i];
}
for (int i = 0; i < T; i++) {
cout << x[i] * 2 << "\n";
}
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int T;
int N[101];
int x[101];
vector<int> num;
int main(void) {
cin >> T;
for (int i = 0; i < T; i++) {
cin >> N[i];
cin >> x[i];
}
for (int i = 0; i < T; i++) {
cout << x[i] * 2 << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void fast() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
}
int main(void) {
fast();
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
cout << 2 * x << "\n";
}
}
| ### Prompt
In CPP, your task is to solve the following problem:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void fast() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
}
int main(void) {
fast();
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
cout << 2 * x << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 50001;
int r[N], c[N];
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t;
cin >> t;
while (t--) {
int x, n;
cin >> x >> n;
cout << (n << 1) << "\n";
}
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 50001;
int r[N], c[N];
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t;
cin >> t;
while (t--) {
int x, n;
cin >> x >> n;
cout << (n << 1) << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
int main() {
int T = read();
while (T--) {
int n = read(), x = read();
cout << x * 2 << endl;
}
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
int main() {
int T = read();
while (T--) {
int n = read(), x = read();
cout << x * 2 << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
scanf("%d", &n);
while (n--) {
int x, y;
scanf("%d %d", &x, &y);
printf("%d\n", y << 1);
}
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
scanf("%d", &n);
while (n--) {
int x, y;
scanf("%d %d", &x, &y);
printf("%d\n", y << 1);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t-- > 0) {
int n, x;
cin >> n >> x;
cout << 2 * x << endl;
}
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t-- > 0) {
int n, x;
cin >> n >> x;
cout << 2 * x << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
cout << 2 * x << endl;
}
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
cout << 2 * x << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d", &t);
for (int i = 0; i < t; ++i) {
int n, x;
scanf("%d %d", &n, &x);
printf("%d\n", x * 2);
}
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
scanf("%d", &t);
for (int i = 0; i < t; ++i) {
int n, x;
scanf("%d %d", &n, &x);
printf("%d\n", x * 2);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
cout << 2 * x << endl;
}
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, x;
cin >> n >> x;
cout << 2 * x << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int t, a, b, i;
cin >> t;
while (t--) {
cin >> a >> b;
cout << b * 2 << endl;
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int t, a, b, i;
cin >> t;
while (t--) {
cin >> a >> b;
cout << b * 2 << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
const long long infL = 1e18;
int setbit(int n, int pos) { return n = n | (1 << pos); }
int resetbit(int n, int pos) { return n = n & ~(1 << pos); }
bool checkbit(int n, int pos) { return (bool)(n & (1 << pos)); }
template <typename T>
T mul(T x, T y) {
x %= mod;
y %= mod;
return (x * y) % mod;
}
template <typename T>
T add(T x, T y) {
x %= mod;
y %= mod;
return (x + y) % mod;
}
template <typename T>
T modPow(T x, T k, T mod) {
if (k == 0) return 1;
if (k & 1) {
return ((x % mod) * modPow(x, k - 1, mod) % mod) % mod;
} else {
T ret = modPow(x, k / 2, mod);
ret %= mod;
return (ret * ret) % mod;
}
}
template <typename T>
T ext_gcd(T num1, T num2, T &X, T &Y) {
T r2, r1, q2, q1, x2, x1, y2, y1, x, y, r;
x2 = 1;
y2 = 0;
x1 = 0;
y1 = 1;
for (r2 = num1, r1 = num2; r1 != 0;
y2 = y1, y1 = y, x2 = x1, x1 = x, r2 = r1, r1 = r) {
q1 = r2 / r1;
x = x2 - q1 * x1;
y = y2 - q1 * y1;
r = r2 % r1;
}
X = x2;
Y = y2;
return r2;
}
template <typename T>
T modInv(T x, T m) {
return modPow(x, m - 2, m);
}
template <typename T>
void print(const T &v) {
cerr << v << ' ';
}
template <typename T1, typename... T2>
void print(const T1 &first, const T2 &...rest) {
print(first);
print(rest...);
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
return os << "[ " << p.first << ", " << p.second << " ]";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << "[ ";
for (int i = 0; i < v.size(); i++) os << v[i] << ' ';
return os << " ]";
}
int modAdd(int &x, int y) {
x += y;
if (x >= mod) x -= mod;
return x;
}
const int M = (int)3e5 + 7;
const int mx = (int)5000 + 123;
const int nx = 500 + 2;
int a[nx + 2];
void solve() {
long long t, x, n;
cin >> t;
while (t--) {
cin >> n >> x;
cout << x * 2 << "\n";
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t = 1, c = 0;
while (t--) {
solve();
}
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
const long long infL = 1e18;
int setbit(int n, int pos) { return n = n | (1 << pos); }
int resetbit(int n, int pos) { return n = n & ~(1 << pos); }
bool checkbit(int n, int pos) { return (bool)(n & (1 << pos)); }
template <typename T>
T mul(T x, T y) {
x %= mod;
y %= mod;
return (x * y) % mod;
}
template <typename T>
T add(T x, T y) {
x %= mod;
y %= mod;
return (x + y) % mod;
}
template <typename T>
T modPow(T x, T k, T mod) {
if (k == 0) return 1;
if (k & 1) {
return ((x % mod) * modPow(x, k - 1, mod) % mod) % mod;
} else {
T ret = modPow(x, k / 2, mod);
ret %= mod;
return (ret * ret) % mod;
}
}
template <typename T>
T ext_gcd(T num1, T num2, T &X, T &Y) {
T r2, r1, q2, q1, x2, x1, y2, y1, x, y, r;
x2 = 1;
y2 = 0;
x1 = 0;
y1 = 1;
for (r2 = num1, r1 = num2; r1 != 0;
y2 = y1, y1 = y, x2 = x1, x1 = x, r2 = r1, r1 = r) {
q1 = r2 / r1;
x = x2 - q1 * x1;
y = y2 - q1 * y1;
r = r2 % r1;
}
X = x2;
Y = y2;
return r2;
}
template <typename T>
T modInv(T x, T m) {
return modPow(x, m - 2, m);
}
template <typename T>
void print(const T &v) {
cerr << v << ' ';
}
template <typename T1, typename... T2>
void print(const T1 &first, const T2 &...rest) {
print(first);
print(rest...);
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
return os << "[ " << p.first << ", " << p.second << " ]";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
os << "[ ";
for (int i = 0; i < v.size(); i++) os << v[i] << ' ';
return os << " ]";
}
int modAdd(int &x, int y) {
x += y;
if (x >= mod) x -= mod;
return x;
}
const int M = (int)3e5 + 7;
const int mx = (int)5000 + 123;
const int nx = 500 + 2;
int a[nx + 2];
void solve() {
long long t, x, n;
cin >> t;
while (t--) {
cin >> n >> x;
cout << x * 2 << "\n";
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int t = 1, c = 0;
while (t--) {
solve();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int ans[n];
for (int i = 0; i < (int)n; i++) {
int a, b;
cin >> a >> b;
ans[i] = 2 * b;
}
for (int i = 0; i < (int)n; i++) cout << ans[i] << endl;
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 ≤ T ≤ 100) — the number of queries. The next T lines contain queries — one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 ≤ x < n ≤ 10^{9}) — the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) — the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
int ans[n];
for (int i = 0; i < (int)n; i++) {
int a, b;
cin >> a >> b;
ans[i] = 2 * b;
}
for (int i = 0; i < (int)n; i++) cout << ans[i] << endl;
return 0;
}
``` |
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