阶乘最后非零位

Last non-zero Digit in N! http://acm.hdu.edu.cn/showproblem.php?pid=1066

#include<cstdio>
#include<cstring>
const int M=1024;
const int tomod[20]= {1,1,2,6,4,2,2,4,2,8,4,4,8,4,6,8,8,6,8,2};
int lastdigit(char buf[]){//阶乘最后非0位O(nlogn)返回该位,n以字符串方式传入
    int len=strlen(buf),ret=1,a[M];
    if(len==1) return tomod[buf[0]-'0'];
    for(int i=0;i<len;i++) a[i]=buf[len-1-i]-'0';
    for(;len;len-=!a[len-1]){
        ret=ret*tomod[a[1]%2*10+a[0]]%5;
        for(int c=0,i=len-1;i>=0;i--){
            c=c*10+a[i];
            a[i]=c/5;
            c%=5;
        }
    }
    return ret+ret%2*5;
}
char str[M];
int main() {
    while(~scanf("%s",str)) {
        printf("%d\n",lastdigit(str));
    }
    return 0;
}

 

 

 

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