大数阶乘模板 hdoj1042

Time Limit: 10000/5000 MS (Java/Others)??? Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 53465??? Accepted Submission(s): 15109

Problem Description
Given an integer N(0 ≤ N ≤ 10000),your task is to calculate N!
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Input
One N in one line,process to the end of file.
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Output
For each N,output N! in one line.
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Sample Input
1
2
3
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Sample Output
1
2
6
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Author
JGShining（极光炫影）
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#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define M (1000000)??????????????????? //注意此处的M值，乘法不同于加减法，数组竟可能大
int jc[M];????????????????????????????????????????? //开辟数组储存

int main() { ? int n; ? while(~scanf("%d",&n)) ? { ? ?memset(jc,sizeof(jc)); ? ? ??? int i,j,k,len,digit,t; ????? jc[0]=len=1;????????????????????????? ?//将结果初始化为1 ????? ????? for(i=2;i<=n;i++)???????????????? ?//开始阶乘，从2 开始 ????? { ????? ?for(j=1,digit=0;j<=len;j++) //应阶乘中的一项与当前所得临时结果的某位//相乘（加上进位） ????? ?{ ????? ??t=jc[j-1]*i+digit;??????????????? ?//digit 进位 ????? ??jc[j-1]=t%10; ????? ??digit=t/10; ????? ?} ????? ?while(digit)?????????????????????? //判断最高位是否进位 ????? ?{ ????? ??jc[++len-1]=digit%10;?? //有进位，不仅各项有变化，而且常盾也要变 ????? ??digit/=10; ????? ?} ????? } ??? ????? for(k=len;k>0;--k) ????? ?printf("%d",jc[k-1]);??? //此处应注意 ????? ?printf("n"); ??? ? }? ? return 0; }