3 solutions
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0
#include<iostream> #include<iomanip> // --->setprecision所需头文件 using namespace std; int main(){ int T, n; double ans; cin >> T; for(int i = 1; i <= T; i++){ cin >> n; if(n < 60) cout << "0.0" << endl; else{ ans = 1.0 * (n - 60) / 10 + 1; cout << setprecision(1) << fixed<< ans << endl; //保留一位小数 } } return 0; }
- 1
Information
- ID
- 23
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 7
- Tags
- # Submissions
- 956
- Accepted
- 236
- Uploaded By