Trailing Zeroes (III) （二分法）

Link：https://cn.vjudge.net/contest/173312#problem/C

题目：

You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*…*N. For example, 5! = 120, 120 contains one zero on the trail.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer Q (1 ≤ Q ≤ 108) in a line.

Output

For each case, print the case number and N. If no solution is found then print ‘impossible’.

Sample Input

3125

Sample Output

Case 1: 5Case 2: 10Case 3: impossible

Code：

#include<cstdio>#include<cstdlib>#include<iostream>#include<cmath>#include<cstring>#include<algorithm>#include<stack>#include<queue>#define MAX 0x7fffffff;     //最大的 long int型数using namespace std;int f(int m)                //计算n的阶乘后0的个数{    int n=0;    while(m)    {        n+=m/5;        m/=5;    }return n;}int main(){    int t,q,c=1;    scanf("%d",&t);    while(t--)    {        scanf("%d",&q);        int l=0,r=MAX;        int mid;        while(l<=r)        {            mid=(l+r)>>1;          //相当于（l+r）/2;             if(f(mid)>=q)                r=mid-1;            else                l=mid+1;        }        if(f(l)==q)            printf("Case %d: %d\n",c,l);      //若最后一次逼近的时候满足了条件             //此时left==mid（无论left与right相差1还是相等）          else            printf("Case %d: impossible\n",c);        c++;    }return 0;}