| Run ID | Author | Problem | Lang | Verdict | Time | Memory | Code Length | Submit Time |
|---|---|---|---|---|---|---|---|---|
| 30488 | yejiaxiangBMT | 计算多项式的值 | C++ | Accepted | 0 MS | 204 KB | 197 | 2023-11-10 15:19:20 |
#include <stdio.h> #include <math.h> int main () { double x,sum=1.0; int n; scanf("%lf %d",&x,&n); for(int i=1;i<=n;i++) { sum+=pow(x,i); } printf("%.2lf",sum); return 0; }