(binary search number theory) (factorial of 0 numbers in factorial) light OJ--1138

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Description

You task was 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! = 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

Code:

#include <cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<queue>#include<stack>#include<algorithm>using namespacestd;#defineN 0x3f3f3f3fintNUM0 (intN) {    intsum =0;  while(n) {sum+ = n/5; N/=5; }    returnsum;}voidSearch (intW///Two-point search{    intL=0, r=N, Mid;  while(l<=R) {Mid= (l+r) >>1; if(W <=NUM0 (mid)) R= Mid-1; ElseL= Mid +1; }    if(NUM0 (L) = =W) printf ("%d\n", L); Elseprintf ("impossible\n");}intMain () {intIcase=1, N, t; scanf ("%d", &t);  while(t--) {scanf ("%d", &N); printf ("Case %d:", icase++);    Search (n); }    return 0;}

(binary search number theory) (factorial of 0 numbers in factorial) light OJ--1138