ZOJ Problem Set - 2022Factorial
2012-11-08 00:00
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CSDN的:要看1到n连续相乘尾数有多少个0,就看这n个数中,有多少个是5的倍数。因为1-n个数中肯定可以提供足够的偶数,所以,只要有5的倍数就一定能够把它乘成10的倍数,就是说n以下有多少个数是5的倍数,那么它的阶乘就有多少个0。
#include <iostream> using namespace std; int main() { int T,n,sum; cin>>T; while(T--){ cin>>n; sum = 0; while(n/5){ sum += n/5; n /= 5; } cout<<sum<<endl; } return 0; }
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