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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#142321 | #1373. Rating Problems | Khaled_Ramadan# | AC ✓ | 1ms | 3740kb | C++20 | 760b | 2023-08-18 23:15:14 | 2023-08-18 23:15:18 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define IO ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0)
typedef long long ll;
const long long N = 3e5 + 7, Mod = 1e9 + 7, INF = 2e18;
ll inv(ll a, ll b = Mod) { return 1 < a ? b - inv(b % a, a) * b / a : 1; }
const int dx[9] = {0, 0, 1, -1, 1, 1, -1, -1, 0};
const int dy[9] = {1, -1, 0, 0, -1, 1, 1, -1, 0};
void solve()
{
int n, k;
cin >> n >> k;
ll sm = 0;
for (int i = 1; i <= k; i++)
{
int x;
cin >> x;
sm += x;
}
cout << fixed << setprecision(10) << (sm - 3 * (n - k)) * 1.0 / n << " " << (sm + 3 * (n - k)) * 1.0 / n << '\n';
}
int main()
{
IO;
int TC = 1;
// cin >> TC;
while (TC--)
solve();
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3736kb
input:
10 0
output:
-3.0000000000 3.0000000000
result:
ok 2 numbers
Test #2:
score: 0
Accepted
time: 1ms
memory: 3736kb
input:
10 10 3 2 1 0 -1 -2 -3 -2 -1 0
output:
-0.3000000000 -0.3000000000
result:
ok 2 numbers
Test #3:
score: 0
Accepted
time: 1ms
memory: 3736kb
input:
10 10 0 0 0 0 0 0 0 0 0 0
output:
0.0000000000 0.0000000000
result:
ok 2 numbers
Test #4:
score: 0
Accepted
time: 1ms
memory: 3736kb
input:
10 10 3 3 3 3 3 3 3 3 3 3
output:
3.0000000000 3.0000000000
result:
ok 2 numbers
Test #5:
score: 0
Accepted
time: 1ms
memory: 3680kb
input:
10 10 -3 -3 -3 -3 -3 -3 -3 -3 -3 -3
output:
-3.0000000000 -3.0000000000
result:
ok 2 numbers
Test #6:
score: 0
Accepted
time: 1ms
memory: 3680kb
input:
1 0
output:
-3.0000000000 3.0000000000
result:
ok 2 numbers
Test #7:
score: 0
Accepted
time: 1ms
memory: 3740kb
input:
1 1 2
output:
2.0000000000 2.0000000000
result:
ok 2 numbers
Test #8:
score: 0
Accepted
time: 1ms
memory: 3736kb
input:
9 7 1 3 2 2 1 1 2
output:
0.6666666667 2.0000000000
result:
ok 2 numbers
Test #9:
score: 0
Accepted
time: 1ms
memory: 3632kb
input:
6 4 -2 -1 -2 0
output:
-1.8333333333 0.1666666667
result:
ok 2 numbers