poj 1401 变形课 数论+二分

 变形课

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

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Description

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

3

1

2

5

Sample Output

Case 1: 5

Case 2: 10

Case 3: impossible

代码：

 

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;long a,b,i,n,m,j,k,l,left,right,mid,ans;int q(long n)//求出该数后面0的个数 {long ans=0;while(n){ans+=n/5;n=n/5;}return ans;}int main(){int flag=1;scanf("%ld",&k);while(k--){scanf("%ld",&n);left=0;right=5*n;while(left<=right)//二分判断，注意是<=:(避免死循环) {mid=(left+right)/2;long num=q(mid);if(num>n)right=mid-1;else if(num<n)left=mid+1;if(num==n){ans=mid;right=mid-1;}}if(q(left)==n)printf("Case %d: %ld\n",flag++,ans);elseprintf("Case %d: impossible\n",flag++);}return 0;}