HDU 1012.u Calculate e【水】【8月16】

标签：水题   hdu   1012   c++   acm   

u Calculate e

A simple mathematical formula for e is



where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333

················计算出前9项，按照格式输出就可以。代码：

#include<cstdio>
int main(){
    int f[11];
    f[0]=1;
    for(int i=1;i<11;i++)
        f[i]=f[i-1]*i;
    printf("n e\n- -----------\n0 1\n1 2\n2 2.5\n");//前2项直接复制粘贴······
    double x=0;
    for(int i=0;i<3;i++)
        x+=(double)1/f[i];
    for(int i=3;i<10;i++){
        x+=(double)1/f[i];
        printf("%d %.9f\n",i,x);
    }
    return 0;
}

HDU 1012.u Calculate e【水】【8月16】

标签：水题   hdu   1012   c++   acm   

原文地址：http://blog.csdn.net/a995549572/article/details/47703385