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…

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…

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++) {…

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…

1012: A MST Problem 时间限制: 1 Sec  内存限制: 32 MB 提交: 7  解决: 4 题目描述 It is just a mining spanning tree ( 最小生成树 ) problem, what makes you a little difficult is that you are in a 3D space. 输入 The first line of the input contains the number of test cases in t…

母牛的故事 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 33371    Accepted Submission(s): 16459 Problem Description 有一头母牛,它每年年初生一头小母牛.每头小母牛从第四个年头开始,每年年初也生一头小母牛.请编程实现在第n年的时候,共有多少头母牛?   Input 输入数据由多…

十进制与八进制的转换(栈和队列) Description 对于输入的任意一个非负十进制整数,利用栈打印输出与其等值的八进制数. Input 111 Output 157 Sample Input 148 Sample Output 224 思想很简单,n除8取余,利用栈先进后出的特点成功实现10进制到8进制的转换 代码: #include <iostream> #include<stack> using namespace std; int main() { stack<in…

Problem A: WERTYU Description A common typing error is to place yourhands on the keyboard one row to the right of the correct position. Then ``Q'' is typed as ``W'' and ``J'' is typed as ``K'' and so on. Your task is to decode a message typed inthis…