间接法判断一组数据乘积结果零的个数

分别对每个数进行2和5取余，若为零则加以1，最后取2和5的min，就是0的个数
```#include <iostream>
#include<cmath>
using namespace std;
int main()
{
 int a[100]={5650,4542,3554,473,946,4114,3871,9073,90,4329,2758,7949,6113,5659,5245,7432,3051,4434,6704,3594,9937,1173,6866,3397,4759,7557,3070,2287,1453,9899,1486,5722,3135,1170,4014,5510,5120,729,2880,9019,2049,698,4582,4346,4427,646,9742,7340,1230,7683,5693,7015,6887,7381,4172,4341,2909,2027,7355,5649,6701,6645,1671,5978,2704,9926,295,3125,3878,6785,2066,4247,4800,1578,6652,4616,1113,6205,3264,2915,3966,5291,2904,1285,2193,1428,2265,8730,9436,7074,689,5510,8243,6114,337,4096,8199,7313,3685,211};
 int s[2]={0};
 for(int i=0;i<100;i++)
 {
   int k=a[i];
   while(k%2==0)
   {
     k/=2;
     s[0]++;
   }
   while(k%5==0)
   {
     k/=5;
     s[1]++;
   }
 }
cout<<min(s[0],s[1]);
  return 0;
}