HDU u Calculate e

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Problem Description

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.

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 11 22 2.53 2.6666666674 2.708333333

Source

Greater New York 2000

看着有点吓人但是很水按照题目公式来就好了

 1 /* 2     前三个数 直接打印就行  3 */ 4 #include<cstdio> 5 #include<iostream> 6  7 using namespace std; 8  9 int main() {10     printf("n e\n");11     printf("- ");12     for(int i=1;i<=11;i++) printf("-");13     printf("\n");14     printf("0 1\n");15     printf("1 2\n");16     printf("2 2.5\n");17     for(int i=3;i<=9;i++) {18         double ans=.0,k=1;19         int t=0;20         printf("%d ",i);21         for(int j=0;j<=i;j++) {22             ans+=(1/k);23             t++;24             k*=t;25         }26         printf("%.9lf\n",ans);27     }28     return 0;29 }

代码