uva 10014
2013-08-06 16:15
337 查看
题意:单纯的数学推论题:自己动手吧
结果 : (n + 1)a1 = na0 + an+1 - 2(nc1 + (n-1)c2 + (n-2)c3 + …… + cn)
结果 : (n + 1)a1 = na0 + an+1 - 2(nc1 + (n-1)c2 + (n-2)c3 + …… + cn)
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int main() { int t,n; double af,al,c,sum; cin>>t; while (t--) { cin>>n>>af>>al; sum = 0 ; for (int i = 0 ; i < n ; i++) { cin>>c; sum += (n-i) * c; } printf("%0.2lf\n",(n*af + al - 2 * sum)/(n+1)); if (t) printf("\n"); } return 0; }
相关文章推荐
- UVA 10014 - Simple calculations
- UVA - 10014 Simple calculations
- uva 10014 - Simple calculations
- UVA - 10014 Simple calculations
- Uva 10014 - Simple calculations
- uva 10014 - Simple calculations
- Uva 10014 - Simple calculations
- uva 10014 Simple calculations
- UVa 10014 Simple calculations (数学)
- uva 10014 Simple calculations
- UVA - 10014 - Simple calculations (经典的数学推导题!!)
- UVA - 10014 Simple calculations
- UVa 10014 简单的计算
- uva 10014 - Simple calculations
- uva 10014 Simple calculations
- UVA - 10014 - Simple calculations (经典的数学推导题!!)
- UVA10014 - Simple calculations
- UVA 10014 Simple calculations
- UVa 10014 - Simple calculations
- uva 10014 Simple calculations(公式推导)