Search is not available for this dataset
name
stringlengths 2
88
| description
stringlengths 31
8.62k
| public_tests
dict | private_tests
dict | solution_type
stringclasses 2
values | programming_language
stringclasses 5
values | solution
stringlengths 1
983k
|
|---|---|---|---|---|---|---|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int x, p;
cin >> x >> p;
assert(p == 100);
cout << p / 2 << endl;
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main(void) {
int x, p;
cin >> x >> p;
if (p != 100) return 0;
cout << (x % 2 == 0 ? x / 2 : x / 2 + 1) << endl;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int x, p;
int main() {
cin >> x >> p;
time_t start = clock();
int total = 0;
int num = 0;
while ((double)(clock() - start) / CLOCKS_PER_SEC < 1.95) {
int a = x;
int b = 0;
int count = 0;
while ((double)(clock() - start) / CLOCKS_PER_SEC < 1.95) {
count++;
if (rand() % 100 < p) {
a--;
} else {
a++;
}
if (count >= a) break;
}
if ((double)(clock() - start) / CLOCKS_PER_SEC < 1.95) {
num++;
total += count;
}
}
cout << fixed << setprecision(6) << (double)total / num << endl;
return (0);
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double EPS = 1e-8;
const int inf = 1e8;
int main() {
int x;
double p;
cin >> x >> p;
vector<double> dp(401);
dp[x] = 1;
double out = 0;
for (int i = 0; i < 400; i++) {
vector<double> ndp(400);
for (int j = i + 1; j < 400; j++) {
if (j) ndp[j - 1] += dp[j] * p / 100;
if (j != 399) ndp[j + 1] += dp[j] * (100 - p) / 100;
}
dp = ndp;
for (int j = 0; j < i + 2; j++) out += dp[j] * (i + 1);
}
cout << fixed << setprecision(9) << out << endl;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007;
const int INF = 1 << 29;
const double EPS = 1e-8;
long long x;
double p;
void update_prop_list(list<double>& prop) {
double left = 0;
prop.push_back(0);
for (double& e : prop) {
double copy = e;
e = (1 - p) * left + p * e;
left = copy;
}
}
int main() {
cin >> x >> p;
p /= 100;
assert(x <= 10);
list<double> prop(1, 1);
long long y = (x + 1) / 2;
for (long long t = 1; t <= y; t++) {
update_prop_list(prop);
}
double ans = 0;
for (long long t = y; t < 10000; t++) {
ans += t * prop.front();
prop.pop_front();
update_prop_list(prop);
}
printf("%.8f\n", ans);
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout << fixed << setprecision(20);
long long x;
cin >> x;
double p;
cin >> p;
if (p == 100) {
x++;
cout << x / 2 << endl;
return 0;
}
p /= 100.0;
if (x % 2 == 0) {
cout << x / 2.0 / p << "\n";
} else {
x++;
cout << x / 2.0 / p + 1 << "\n";
}
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
int x, p;
int main(void) {
scanf("%d%d", &x, &p);
int ans = x / 2;
if (x % 2 == 1) {
ans++;
}
if (ans == 0) {
ans = 1;
}
printf("%d\n", ans);
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double EPS = 1e-8;
const int inf = 1e8;
int main() {
int x;
double p;
cin >> x >> p;
if (p == 100) {
cout << (x + 1) / 2 << endl;
return 0;
}
vector<double> dp(2000);
dp[x - 1] = 1;
double out = 0;
for (int i = 0; i < 2000; i++) {
vector<double> ndp(2000);
for (int j = i; j < 2000; j++) {
if (j > i)
ndp[j - 1] += dp[j] * p / 100;
else
out += dp[i] * p / 100;
if (j != 1999) ndp[j + 1] += dp[j] * (100 - p) / 100;
}
dp = ndp;
for (int j = 0; j < i + 1; j++) out += dp[j] * (i + 1);
}
cout << fixed << setprecision(9) << out << endl;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template <typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
inline T gcd(T a, T b) {
while (b) {
a %= b;
swap(a, b);
}
return a;
}
template <class T>
inline T lcm(T a, T b) {
return a * b / gcd(a, b);
}
clock_t startTime;
double getCurrentTime() {
return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}
const ll INF = (ll)1e18;
ll MOD;
ll add(ll x, ll y) {
x += y;
if (x >= MOD) return x - MOD;
return x;
}
ll sub(ll x, ll y) {
x -= y;
if (x < 0) return x + MOD;
return x;
}
ll mul(ll x, ll y) { return (x * y) % MOD; }
ll bin_pow(ll x, ll p) {
if (p == 0) return 1;
if (p & 1) return mul(x, bin_pow(x, p - 1));
return bin_pow(mul(x, x), p / 2);
}
const int N = 2000100;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
startTime = clock();
ld p, ans = 0;
ll x;
cin >> x >> p;
p /= 100;
if (x % 2 == 0) {
ans = 1.0 * x / 2 * p;
} else {
ans = 1.0 + (1 - p) * (x + 1) / 2 / p + (x - 1) / 2;
}
cout << setprecision(10) << ans << '\n';
;
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | #include <bits/stdc++.h>
int main() {
int n, m;
scanf("%d %d", &n, &m);
if (m == 100) printf("%d\n", (n + 1) / 2);
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int x, p;
cin >> x >> p;
if (p == 100) {
int tim = (x + 1) / 2;
cout << tim << endl;
}
double pp = p / 100.0;
if (x % 2) {
printf("%.14f\n", (double)(1 - pp) / pp + 1 + (double)(x / 2) / pp);
} else
printf("%.14f\n", (double)(x / 2) / pp);
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const double eps = 1e-10;
const double pi = acos(-1.0);
const double inf = (int)1e8;
int main() {
long long x, p;
std::cin >> x >> p;
x++;
if (x % 2 == 0)
std::cout << x / 2 << ".00000000" << std::endl;
else
std::cout << x / 2 << ".50000000" << std::endl;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
using ll = long long;
template <class T>
bool set_min(T &a, const T &b) {
return a > b ? a = b, true : false;
}
template <class T>
bool set_max(T &a, const T &b) {
return a < b ? a = b, true : false;
}
template <class T>
istream &operator>>(istream &is, vector<T> &v) {
for (T &a : v) is >> a;
return is;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
for (const T &t : v) os << "\t" << t;
return os << endl;
}
template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &v) {
return os << "<" << v.first << ", " << v.second << ">";
}
const int INF = 1 << 30;
const ll INFL = 1LL << 60;
class Solver {
public:
bool solve() {
int x;
cin >> x;
double p;
cin >> p;
if (abs(100 - p) > 1e-6) return 0;
cout << x / 2 + x % 2 << endl;
return 0;
}
};
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
Solver s;
s.solve();
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | $0=($1+$1%2)*50/$2 |
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | UNKNOWN | say-get+>1*100/-get |
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int x, p;
cin >> x >> p;
if (p == 100) {
int tim = (x + 1) / 2;
cout << tim << endl;
}
if (x % 2) {
printf("%.14f\n", (double)(1 - p) / p + (double)x / p);
} else
printf("%.14f\n", (double)x / p);
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
const int INF = 1 << 29;
const double EPS = 1e-9;
const ll MOD = 1000000007;
const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
ll X, P;
int main() {
cin >> X >> P;
double res = X / 2 + (X % 2);
printf("%.20lf\n", res);
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
const int INF = 1 << 29;
const double EPS = 1e-9;
const ll MOD = 1000000007;
const int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};
ll X, P;
const int MAX_TURN = 300;
const int MAX_POS = 300;
double dp[MAX_TURN][MAX_POS][MAX_POS];
bool visited[MAX_TURN][MAX_POS][MAX_POS];
double memo(int turn, int pos_ao, int pos_cho) {
if (pos_ao == pos_cho) {
return turn;
}
if (turn >= MAX_TURN or pos_ao < 0 or pos_ao >= MAX_POS or pos_cho < 0 or
pos_cho >= MAX_POS) {
return INF;
}
double pro = (double)P / 100;
if (visited[turn][pos_ao][pos_cho]) {
return dp[turn][pos_ao][pos_cho];
}
visited[turn][pos_ao][pos_cho] = true;
double result = INF;
result = min(result, memo(turn + 1, pos_ao - 1, pos_cho - 1) * pro +
memo(turn + 1, pos_ao - 1, pos_cho + 1) * (1 - pro));
result = min(result, memo(turn + 1, pos_ao, pos_cho - 1) * pro +
memo(turn + 1, pos_ao, pos_cho + 1) * (1 - pro));
result = min(result, memo(turn + 1, pos_ao + 1, pos_cho - 1) * pro +
memo(turn + 1, pos_ao + 1, pos_cho + 1) * (1 - pro));
return dp[turn][pos_ao][pos_cho] = result;
}
int main() {
cin >> X >> P;
if (X > 10) return 0;
int geta = MAX_POS / 2;
double result = memo(0, geta, X + geta);
printf("%.20lf\n", result);
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int X, P;
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> X >> P;
if (P == 100) {
cout << (X + 1) / 2 << "\n";
}
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <class S, class T>
istream& operator>>(istream& is, pair<S, T>& p) {
return is >> p.first >> p.second;
}
const double EPS = 1e-10;
const double PI = acos(-1.0);
const long long MOD = 1e9 + 7;
double C[1010][1010];
double ps[1010], qs[1010];
int main() {
cin.tie(0);
ios_base::sync_with_stdio(false);
long long x, p;
cin >> x >> p;
if (p == 100) {
cout << (x + 1) / 2 << endl;
return 0;
}
if (x > 10) return 1;
ps[0] = qs[0] = 1.;
for (int i = (0); i < (1000); ++i) {
ps[i + 1] = ps[i] * p / 100.;
qs[i + 1] = qs[i] * (100 - p) / 100.;
}
for (int i = 1; i < 1000; ++i) {
C[i][0] = ps[i];
C[i][i] = qs[i];
for (int j = 1; j < i; ++j)
C[i][j] = C[i - 1][j] * qs[1] + C[i - 1][j - 1] * ps[1];
}
const int MAX = 10020;
vector<vector<double>> xs(2, vector<double>(MAX));
int crt = 0, nxt = 1;
xs[crt][MAX / 2 + x] = 1.;
int px = MAX / 2;
double ans = 0.;
for (int t = (0); t < (MAX / 2 - 20); ++t) {
fill((xs[nxt]).begin(), (xs[nxt]).end(), 0);
for (int a = (0); a < (MAX); ++a) {
if (a - 1 >= 0) xs[nxt][a - 1] += xs[crt][a] * p / 100.;
if (a + 1 < MAX) xs[nxt][a + 1] += xs[crt][a] * (100. - p) / 100.;
}
swap(crt, nxt);
if (xs[crt][px] < EPS && xs[crt][px - 1] < EPS && xs[crt][px + 1] < EPS)
++px;
else if (xs[crt][px] > xs[crt][px - 1] && xs[crt][px] > xs[crt][px + 1])
;
else if (xs[crt][px - 1] > xs[crt][px] && xs[crt][px - 1] > xs[crt][px + 1])
--px;
else if (xs[crt][px + 1] > xs[crt][px - 1] && xs[crt][px + 1] > xs[crt][px])
++px;
ans += (t + 1) * xs[crt][px];
xs[crt][px] = 0.;
}
cout << fixed << setprecision(9) << ans << endl;
return 0;
}
|
p03901 CODE FESTIVAL 2016 Elimination Tournament Round 2 (Parallel) - Takahashi is Missing! | Aoki is in search of Takahashi, who is missing in a one-dimentional world. Initially, the coordinate of Aoki is 0, and the coordinate of Takahashi is known to be x, but his coordinate afterwards cannot be known to Aoki.
Time is divided into turns. In each turn, Aoki and Takahashi take the following actions simultaneously:
* Let the current coordinate of Aoki be a, then Aoki moves to a coordinate he selects from a-1, a and a+1.
* Let the current coordinate of Takahashi be b, then Takahashi moves to the coordinate b-1 with probability of p percent, and moves to the coordinate b+1 with probability of 100-p percent.
When the coordinates of Aoki and Takahashi coincide, Aoki can find Takahashi. When they pass by each other, Aoki cannot find Takahashi.
Aoki wants to minimize the expected number of turns taken until he finds Takahashi. Find the minimum possible expected number of turns.
Constraints
* 1 ≦ x ≦ 1,000,000,000
* 1 ≦ p ≦ 100
* x and p are integers.
Input
The input is given from Standard Input in the following format:
x
p
Output
Print the minimum possible expected number of turns. The output is considered correct if the absolute or relative error is at most 10^{-6}.
Examples
Input
3
100
Output
2.0000000
Input
6
40
Output
7.5000000
Input
101
80
Output
63.7500000 | {
"input": [
"6\n40",
"101\n80",
"3\n100"
],
"output": [
"7.5000000",
"63.7500000",
"2.0000000"
]
} | {
"input": [],
"output": []
} | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const double eps = 1e-10;
const int inf = 1000000009;
int i, j, k, m, n, l, x, p;
int ans;
int main() {
cin >> x >> p;
if (p == 100) {
cout << (x + 1) / 2 << endl;
return 0;
}
return 0;
}
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python2 | #encoding=utf-8
inp = input()
x = 100000
for i in xrange(inp):
x = x * 1.05
if int(x % 1000) != 0:
x += 1000 - int(x % 1000)
print int(x) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import math
n = int(input())
a = 100000
for _ in range(n):
a *= 1.05
a = 1000 * math.ceil(a / 1000)
print(a)
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python2 | n=int(raw_input())
g,r=100000,1.05
for i in range(n):
g*=r
if int(g)%1000>0:
g=(int(g)/1000)*1000+1000
print g |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner s=new Scanner(System.in);
int n=s.nextInt();
double ans=100000;
for(int i=0;i<n;i++){
ans=ans*1.05;
if(ans%1000!=0){
ans=ans+1000-ans%1000;
}
}
int a=(int)ans;
System.out.println(a);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <stdio.h>
#include <string.h>
int main(){
int n;
scanf("%d", &n);
int m = 100000;
for(int i=0;i<n;i++){
m *= 1.05;
if(m%1000 != 0) m = (m/1000+1)*1000;
}
printf("%d\n", m);
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python2 | a = input()
ret = 100000
while 0<a:
ret *= 1.05
if(ret%1000):
ret = int((ret//1000+1)*1000)
a -= 1
print ret |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
using namespace std;
int main(){
int i,n=100000,m;
cin >> i;
while(i--){
n+=n*0.05;
m=n%1000;
if(m){
n+=1000-m;
}
}
cout << n << endl;
return 0;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import sys
import math
n = int(input())
debt = 100000
for i in range(1,n+1):
debt = math.ceil((debt*1.05)/1000)*1000
print(debt) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
double debt = 100000;
int N = in.nextInt();
for(int i=0;i<N;i++){
debt = Math.ceil(debt*1.05/1000)*1000;
}
System.out.printf("%.0f\n",debt);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<string>
using namespace std;
int main()
{
int n;
float b=100000;
cin>>n;
for(int i=0;i<n;i++){
b*=1.05;
b=(int)((b+999)*0.001);
b*=1000;
}
cout<<(int)b<<endl;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <string>
using namespace std;
int main() {
int m = 100000, n;
cin >> n;
for (int i = 0; i < n; i++) {
m *= 1.05;
if(m%1000)
m += 1000 - m % 1000;
}
cout << m << endl;
return 0;
}
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <string>
using namespace std;
int main(){
int n;
cin >> n;
int a;
a=100000;
int b;
b=0;
while(b<n){
a=a/20*21;
if((a%1000)!=0){a=a/1000*1000+1000;}
b=b+1;}
cout << a<<endl;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.io.PrintWriter;
import java.util.*;
import static java.lang.Math.*;
public class Main{
public static void main(String args[]){
Scanner sc = new Scanner(System.in);
PrintWriter out=new PrintWriter(System.out);
double money = 100000;
final double tax = 1.05;
int n= sc.nextInt();
for(int i=0;i<n;i++){
money*=tax;
money/=1000;
money=ceil(money);
money*=1000;
}
out.println((int)money);
out.flush();
}
}
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<stdio.h>
int main()
{
int n;
long long int s=100000;
scanf("%d",&n);
for(int i=0;i<n;i++){
s*=1.05;
s+=999;
s/=1000;
s*=1000;
}
printf("%d\n",s);
return 0;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import math
debt = 100000
rate = 1.05
week = int(input())
for i in range(week):
debt = math.ceil(debt*1.05/1000)*1000
print(debt)
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
using namespace std;
int main(){
int n,i;
double m=100.0;
cin>>n;
for(i=0;i<n;i++){
m*=1.05;
if(m-(int)m>0.0){
m+=1.0;
m=(int)m;
}
}
cout<<m<<"000"<<endl;
return 0;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main()
{
ll n, res = 100000;
cin >> n;
while (n--) {
res = (res + res / 20 + 999) / 1000 * 1000;
}
cout << res << endl;
return 0;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | from math import ceil
n=int(input())
d=int(1e5)
for x in range(n):
d=(1.05*d)
d=int(int(ceil(d/1e3))*1e3)
print(d) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
double p = 100000;
int n = s.nextInt();
for(int i = 0; i < n; ++i) {
p *= 1.05;
p = Math.ceil(p / 1000) * 1000;
}
System.out.println((int)p);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.io.*;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader br =
new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
int sum = 100000;
double debt = sum * 0.05;
for (int i = 0; i < n; i++) {
sum += debt;
if (sum % 1000 != 0) {
sum -= sum % 1000;
sum += 1000;
}
debt = sum * 0.05;
}
System.out.println(sum);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.io.*;
public class Main{
public static void main(String argv[]) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
int weeks = Integer.parseInt(line);
float debt = 100000;
for(int i=0; i<weeks; i++){
debt *= 1.05f;
if((debt % 1000) > 0.0){
debt = debt+1000-debt%1000;
}
}
System.out.print(Integer.toString((int)debt)+"\n");
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | x=int(input())
i=0
a=100000
while x>i :
a=a+a*0.05
if a%1000!=0 : a=a+1000
a=int(a/1000)
a=a*1000
i=i+1
print(a)
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import math
debt = 100000
n = int(input())
for w in range(n):
debt *= 1.05
debt = math.ceil(debt / 1000) * 1000
print(debt) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import math
n = int(input())
money = 100000
for i in range(n):
money = money*1.05
money = math.ceil(money/1000)*1000
print(money)
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | x = 100000
n = int(input())
for _ in range(n):
x = x*1.05
if x%1000>0:
x = x-x%1000+1000
print(int(x))
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
using namespace std;
double money = 100000;
int i;
int main()
{
cin>>i;
for(i;i>0;--i)
{
money *= 1.05;
money = (long)(money/1000+0.99)*1000;
}
printf("%.0lf\n",money);
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | public class Main {
static java.util.Scanner scan = new java.util.Scanner(System.in);
public static void main(String[] args) {
int input = scan.nextInt();
double total = 100;
while (0 < input--) {
total = Math.ceil(total * 1.05);
}
total *= 1000;
print((int) total);
}
public static void print(Object out) {
System.out.println(out);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<stdio.h>
int main(void){
int n,i,x=100000,z;
scanf("%d",&n);
for(i=0;i<n;i++)
{
x = x * 1.05;
if(x % 1000 != 0.0)
{
z = (x / 1000)+1;
x = z * 1000;
}
}
printf("%d\n",x);
return 0;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | n = int(input())
d = 100000
for i in range(n):
d *= 1.05
d = d - (d - 1) % 1000 + 999
print(int(d)) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <stdio.h>
int main(){
int i,n,debt = 100000;
scanf("%d",&n);
for(i = 0;i < n;i++){
debt = (((int)(debt * 1.05) - 1) / 1000 + 1) * 1000;
}
printf("%d\n",debt);
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.*;
import static java.lang.System.*;
public class Main {
Scanner sc = new Scanner(in);
void run() {
int n = sc.nextInt();
int m = 100000;
for (int i = 0; i < n; i++) {
m += (int)Math.ceil(m*0.05/1000)*1000;
}
out.println(m);
}
public static void main(String[] args) {
new Main().run();
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import math
debt = 100000
for n in range(int(input())):
debt = math.ceil(debt*1.05/1000)*1000
print(str(debt))
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | n=int(input())
x=100000
for i in range(n):
x*=1.05
if x%1000>0:
x=x-(x%1000)+1000
else:
pass
print(int(x))
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python2 | def f(a,b):
if (b == 0): return a
a = a * 1.05
if (a%1000 != 0): a = a - a%1000 + 1000
return f(a,b-1)
print int(f(100000,input())) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args)
throws IOException {
BufferedReader br = new BufferedReader(
new InputStreamReader(System.in));
int v = Integer.parseInt(br.readLine());
int m = 100000;
for (int i = 0; i < v; i++) {
m *= 1.05;
if ((m % 1000) > 0) {
m += 1000 - m % 1000;
}
}
System.out.println(m);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.*;
import java.math.*;
public class Main {
public static void main(String[] args){
Scanner in = new Scanner(System.in);
int n = in.nextInt();
double money = 100.00;
for(int i=0; i<n; i++){
money = money*1.05;
BigDecimal tmp = new BigDecimal(money);
money = tmp.setScale(0,BigDecimal.ROUND_UP).doubleValue();
}
System.out.println((int)money*1000);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <stdio.h>
int main(void)
{
int n;
int ans = 100000;
scanf("%d", &n);
for (int i = 0; i < n; i++){
ans += ans * 0.05;
ans = (ans + 999) / 1000 * 1000;
}
printf("%d\n", ans);
return (0);
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python2 | import math as k
n=input()
m=105
for i in range(n-1):
m=k.ceil(m*1.05)
print int(m*1000) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int money = 100;
for (int i = 0; i < n; i++){
double x = (double)money * 1.05;
money = (int)(Math.ceil(x));
}
System.out.println(money * 1000);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | n = int(input())
amount = 100000
for i in range(n):
amount = int(-(-amount * 1.05 // 1000) * 1000)
print(amount)
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import math
debt = 100000
n = int(input())
for i in range(n):
debt *= 1.05
debt = math.ceil(debt * 0.001)*1000
print(debt)
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <cmath>
using namespace std;
int main() {
int n;
cin >> n;
int ans = 100000;
for (int i = 0; i < n; i++)
ans = ceil(ans * 1.05 / 1000) * 1000;
cout << ans << endl;
}
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | s=100000
n=int(input())
for i in range (n):
s*=1.05
if s%1000!=0:
s=s-(s%1000)+1000
print(int(s))
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
public class Main{
static int debt(int n){
int x = 100000;
while( n > 0 ){
x *= (1 + 0.05);
if( x%1000 != 0 ){
x = x - x%1000 + 1000;
}
n--;
}
return x;
}
public static void main(String[] args){
Scanner stdIn = new Scanner(System.in);
int n = stdIn.nextInt();
System.out.println(debt(n));
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | n=int(input())
s=100000
for i in range(n):
s*=1.05
p=s%1000
if p!=0:
s+=1000-p
print(int(s))
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.*;
public class Main {
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
int debt=100000;
int n=sc.nextInt();
for(int i=0;i<n;i++){
debt=(int)Math.ceil((debt*1.05)/1000)*1000;
}
System.out.println(debt);
sc.close();
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | n=int(input())
a=100000
for i in range(n):
a=a*1.05
if a%1000 != 0:
a=int(a//1000*1000+1000)
print(a)
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
using namespace std;
int main(){
int n;
cin>>n;
int lend=100000;
for(int i=1;i<=n;i++){
lend=(lend*1.05-1)/1000;
lend=(lend+1)*1000;
}
cout<<lend<<endl;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int week = scan.nextInt();
double debt = 100000;
for(int i = 0; i < week; i++){
debt = debt * 1.05 / 1000;
debt = Math.ceil(debt);
debt = debt * 1000;
}
int ans = (int)debt;
System.out.println(ans);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python2 | import math
n = int(raw_input())
debt = 100000
for i in range(n):
debt = int(math.ceil(debt * 1.05 / 1000)) * 1000
print debt |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include "bits/stdc++.h"
using namespace std;
int main() {
long long int N; cin >> N;
long long int num = 100;
while (N--) {
num =num* 1.0499999999+1;
}
cout << num * 1000 << endl;
return 0;
}
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
class Main{
public static void main(String[] args){
final Scanner sc=new Scanner(System.in);
int syakkin=100000;
int n=sc.nextInt();
for(int i=0;i<n;i++){
syakkin*=1.05;
if(syakkin%1000!=0){
syakkin=((syakkin/1000)+1)*1000;
}
}
System.out.println(syakkin);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python2 | #!/usr/bin/python
# -*- coding: utf-8 -*-
import math
result = 100000
num = int(raw_input())
for i in range(num):
result *= 1.05
result /= 1000
result = math.ceil(result)
result *= 1000
print int(result) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import math
n = eval(input())
yen = 100000
for i in range(n):
yen += yen * 0.05
yen = int(math.ceil(yen / 1000.0) * 1000)
print(yen) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import math
v = int(input())
cal = 100000
for w in range(v):
cal = math.ceil(cal * 1.05 * 0.001) * 1000
print("%d" % cal) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<stdio.h>
int main(){
int n,i,a,b,c;
std::cin>>a;
for(n=0,i=100000;a>n;++n){
i=i+i/20;
if(i%1000!=0){
b=i%1000;
i=i+1000-b;
}
}
std::cout<<i<<std::endl;
return 0;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
int a=sc.nextInt();
int money=100000;
for(int i=0;i<a;i++){
money*=1.05;
int amari=money%1000;
money-=amari;
money+=amari==0 ? 0 : 1000;
}
System.out.println(money);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import math
debt = 100
n = int(input())
for _ in range(n):
debt = math.ceil(debt * 1.05)
print(debt * 1000) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin >> n;
int y = 100000;
for(int i = 0 ; i < n ; i++){
y = y * 105 / 100;
if( y % 1000 ) y += 1000 - y % 1000;
}
cout << y << endl;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
public class Main {
private static final int origin = 100000;
private static final double interest = 1.05;
static Scanner sc = new Scanner(System.in);
public static void main(String[] args) {
int num = sc.nextInt();
double ans = origin;
for(int count = 0; count < num; count++) {
ans = ans * interest / 1000;
ans = Math.ceil(ans) * 1000;
}
System.out.println((int)ans);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
public class Main {
/**
* @param args
*/
public static void main(String[] args) {
// TODO 自動生成されたメソッド・スタブ
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int money = 100;
for (int i = 0; i < n; i++) {
money = (int) Math.ceil(money * 1.05);
}
System.out.println(money * 1000);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python2 | import math
debt = 100000
for i in range(int(raw_input())):
debt = int(math.ceil((debt * 1.05) / 1000)) * 1000
print debt |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
using namespace std;
int main()
{
int m = 100000;
int n;
cin >> n;
for(int i=0; i<n; i++)
{
m += m / 20;
if(m % 1000 != 0) m = (m / 1000 + 1) * 1000;
}
printf("%d\n", m);
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin >> n;
int s=0;
int a=100000;
while (s<n){
s++;
a=a+a*0.05;
a=((a+999)/1000)*1000;
}
printf("%d\n",a);
}
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import math
N = int(input())
A = 100000
for _ in range(N):
A = A * 1.05
A = (int(math.ceil(A / 1000))) * 1000
print(A)
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
class Main {
public static void main(String[] args) {
int weeks = new Scanner(System.in).nextInt();
double debt = 100000;
for(int i = 0; i < weeks;i++) {
debt *= 1.05;
double fraction = debt % 1000;
if (fraction != 0)debt += 1000 - fraction;
}
System.out.println((long)debt);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | a = 100000;
n = int(input())
while n > 0:
n -= 1
a = a + (a * 5) // 100
a = ((a - 1) // 1000 + 1) * 1000 # 1000未満の切り上げ
print(a)
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <stdio.h>
int main() {int kane=100000,N;scanf("%d",&N);while (N--) {kane=kane+kane/20;int rest=kane%1000;if (rest) kane+=(1000-rest);}printf("%d\n",kane);return 0;} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n, debt = 100000;
n = sc.nextInt();
for(int i = 0; i < n; i++){
debt *= 1.05;
if(debt % 1000 != 0){
debt += 1000 - debt % 1000;
}
}
System.out.println(debt);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python2 | n = int(raw_input())
debt=100000
for i in range(n):
debt*=1.05
if debt % 1000 != 0:
debt = (int(debt / 1000)+1) * 1000
else:
debt = int(debt)
print debt |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
using namespace std;
int main()
{
int n;
int m = 100000;
cin >> n;
for(int i = 0; i < n; i++){
m += m / 20;
if(m % 1000){
m += 1000 - m % 1000;
}
}
cout << m << endl;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
#include <cmath>
using namespace std;
int main()
{
int n;
double s = 100000;
cin >> n;
while(n--)
s = ceil(s * 1.05 * 0.001) * 1000;
cout << (int)s << endl;
return 0;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include<iostream>
#include<cmath>
using namespace std;
int main(){
float base = 100;
int n;
cin >> n;
for(int i = 0; i < n; i++){
base *= 1.05;
base = ceil(base);
}
cout << (int)(base*1000) << endl;
}
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | a=100000
for _ in range(int(input())):
a=((a*1.05)//1000+1)*1000 if (a*1.05)%1000 else a*1.05
print(int(a)) |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | x=100000
for i in range(int(input())):
x*=1.05
if x%1000!=0:
x+=1000-x%1000
print(int(x))
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | java | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
double debt = 100000;
Scanner stdIn = new Scanner(System.in);
int n = Integer.parseInt(stdIn.next());
for (int i = 1; i <= n; i++) {
debt += (debt * 0.05);
int tmp = (int)(debt / 1000);
if((debt/1000) != tmp) debt = ((tmp*1000) + 1000);
}
System.out.println((long)debt);
}
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
using namespace std;
int main()
{
int n, k = 100000;
cin >> n;
for (int i = 1; i <= n; i++)
{
k = static_cast<int>(k * 1.05 + 999) / 1000 * 1000;
}
cout << k << endl;
return 0;
} |
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | python3 | import math
n = int(input())
a = 100
for i in range(n):
a = math.ceil(a*1.05)
print(a*1000)
|
p00007 Debt Hell | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week.
Write a program which computes the amount of the debt in n weeks.
Input
An integer n (0 ≤ n ≤ 100) is given in a line.
Output
Print the amout of the debt in a line.
Example
Input
5
Output
130000 | {
"input": [
"5"
],
"output": [
"130000"
]
} | {
"input": [],
"output": []
} | CORRECT | cpp | #include <iostream>
using namespace std;
int main(){
int n;
int debt = 100000;
cin >> n;
for(int i=0; i<n;i++){
debt *= 1.05;
if(debt % 1000) debt += (1000-(debt % 1000));
}
cout << debt << endl;
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.