/*tips:本题只有输入,没有输出,在线测试只检测结果,所以将前面几个结果罗列出来就OK了.为了格式输出问题纠结了半天,最后答案竟然还是错的....所以啊,做题还是得灵活变通.*/ #include<iostream> using namespace std; double Ecal(int n); double Cal(int n); void main() { printf("n e\n"); printf("- -----------\n");…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1012 u Calculate e Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 52607    Accepted Submission(s): 24106 Problem Description A simple mathematical…

u Calculate e Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 46844    Accepted Submission(s): 21489 Problem Description A simple mathematical formula for e is where n is allowed to go to infini…

u Calculate e Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 28686    Accepted Submission(s): 12762 Problem Description A simple mathematical formula for e is where n is allowed to go to infini…

u Calculate e Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37350    Accepted Submission(s): 16905 Problem Description A simple mathematical formula for e is where n is allowed to go to infini…

u Calculate e 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 gener…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1012 解题思路:对阶乘递归求和 反思:前面3个的输出格式需要注意,可以自己单独打印出来,也可以在for循环里面更改输出小数的位数,另外读题还是要仔细,输出的有9位小数. #include<stdio.h> double sum(int n) { int i; double x=1,s=0; if(n==0) return 1; else { s=1; for(i=1;i<=n;i++) {…

题解:直接模拟 #include <cstdio> int main(){ puts("n e");puts("- -----------");puts("0 1"); double ans=1.0,f=1.0; for(int i=1;i<=9;i++){ if(i==1)printf("%d %.0lf\n",i,(ans+=(f/=(double)i))); else if(i==2)printf(&q…

A simple mathematical formula for e iswhere 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 v…

分析:注意格式. #include<stdio.h> int main() { int i,j,k; double sum=0; printf("n e\n- -----------\n"); printf("0 1\n1 2\n2 2.5\n"); for(i=3;i<=9;i++) { k=1; for(j=1;j<=i;j++) k*=j; sum+=1.0/k; printf("%d %.9lf\n",i,sum+…