N!
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41380 Accepted Submission(s): 11478
基礎題。(大數處理)
解題思路:
For循環、數組處理大數
注意:0!=1;
代碼:(剛開始刷題時寫的比較濫、勿噴)
#include<stdio.h>
#include<string.h>
const int maxn=40000;
int main()
{
int i,j,n;
int f[40000];
while(scanf("%d",&n)!=EOF)
{
if(n==0){
printf("1\n");
continue;
}
memset(f,0,sizeof(f));
f[0]=1;
for(i=2;i<=n;i++)
{
int c=0;
for(j=0;j<maxn;j++)
{
int s=f[j]*i+c;
f[j]=s%10;
c=s/10;
}
}
for(j=maxn-1;j>=0;j--) if(f[j]) break;
for(i=j;i>=0;i--) printf("%d",f[i]);
printf("\n");
}
return 0;
}