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HDU 1012.u Calculate e【水】【8月16】

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u Calculate e

Problem Description
A simple mathematical formula for e is

技术分享

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

Output
Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.
 

Sample Output
n e - ----------- 0 1 1 2 2 2.5 3 2.666666667 4 2.708333333
 
技术分享················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;
}


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HDU 1012.u Calculate e【水】【8月16】

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

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