hdoj1042(N!，大数乘)

HDOJ1042(N!)(大数乘)

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 67802    Accepted Submission(s): 19420

Problem Description

Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!

 

Input

One N in one line, process to the end of file.

 

Output

For each N, output N! in one line.

 Sample Input

123

 Sample Output

126

 

My   solution:

/*2015.10.29*/

#include<stdio.h>#include<string.h>int t1[10],t2[40000],t3[40000],n,m;void mul(){int i,j,k,g;k=n;for(i=0;i<m;i++){k=i;for(j=0;j<n;j++){t3[k++]+=t1[i]*t2[j];}}for(i=0;i<k;i++){if(t3[i]>9)/*t3起转换容器的变量，暂时存放计算结果*/{g=t3[i]/10;t3[i]%=10;t3[i+1]+=g;if(i==k-1)/*最高位要进位，需要把数组长度再加1*/k=k+1; }t2[i]=t3[i];t3[i]=0;/*一定要把t3数组清空,否则下一次运算时,会再次加上，导致结果出错*/}     n=k;/*更新t2数组长度*/}void  tiaozhen(){int i,j;t1[0]+=1;/*每乘一次，待乘数要加1,t1[0]一定要初始化为1*/ for(i=0;i<m;i++){if(t1[i]>9)/*逢10进1*/{j=t1[i]/10;t1[i]%=10;t1[i+1]+=j;if(i==m-1)/*最高位要进位，需要把数组长度再加1*/m+=1;} } }void Njie(int e){int i,j;for(i=2;i<=e;i++){tiaozhen();mul(); } }void print(){int i;for(i=n-1;i>=0;i--)printf("%d",t2[i]);printf("\n");}int main(){int i,j,e;while(scanf("%d",&e)==1){if(e>1){memset(t1,0,sizeof(t1));memset(t2,0,sizeof(t2));memset(t3,0,sizeof(t3));n=m=1;t1[0]=1;t2[0]=1;Njie(e);print();}elseprintf("1\n");}return 0;}