C - Trailing Zeroes (III)（二分）

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

AC代码

 #include<stdio.h>
 const int MAXN=0x3f3f3f3f;//这个要足够大才能找到10^8
 int getq(int x){
     int q=;
     while(x){
         q+=x/;
         x/=;
     }
     return q;
 }
 void erfen(int n){
     int l=,r=MAXN;
     while(l<=r){
         int mid=(l+r)>>;
         if(getq(mid)>=n)r=mid-;//二分这点注意
         else l=mid+;
     }
     if(getq(l)==n)printf("%d\n",l);
     else puts("impossible");
 }
 int main(){
     int T,Q,flot=;
     scanf("%d",&T);
     while(T--){
         scanf("%d",&Q);
         printf("Case %d: ",++flot);
          erfen(Q);
     }
     return ;
 }