N! Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 94583 Accepted Submission(s): 28107 Problem Description Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N! Input One N in…
我们都知道如何计算一个数的阶乘,可是,如果这个数很大呢,该如何计算? 当一个数很大时,利用平常的方法是求不出来它的阶乘的,因为数据超出了范围.因此我们要用数组来求一个大数的阶乘,用数组的每位表示结果的每个位数.话不多说,直接上代码 #include<stdio.h> #include<string.h> int main() { int i,j,n,temp,d=1,carry;//temp为阶乘元素与临时结果的乘积,carry是进位 ,d是位数 int a[3000];//确保数…
求阶乘序列前N项和 #include <stdio.h> double fact(int n); int main() { int i, n; double item, sum; while (scanf("%d", &n) != EOF) { sum = 0; if (n <= 12) { for (i = 1; i <= n; i++) { item = fact(i); sum = sum + item; } } printf("%.0f…
Big Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5930 Accepted Submission(s): 4146 Problem Description As we know, Big Number is always troublesome. But it's really important in our…