Light OJ 1138-trailing Zeroes (III) "Two-point search && N! The number of consecutive 0 in the end of "

1138-trailing Zeroes (III)

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Time Limit:2 second (s)

Memory limit:32 MB

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	Output for Sample Input
3 1 2 5	Case 1:5 Case 2:10 Case 3:impossible

Test instructions: give you a number q, which means n! The number of consecutive 0 in the end. Let you find the smallest n.

Ask N! Nakao number of consecutive 0:

ll change (ll x) {    ll ans = 0;    while (x) {        ans + = X/5;        x/= 5;    }    return ans;}

AC Code

#include <stdio.h> #include <string.h> #include <queue> #include <algorithm> #define INF 0x3f3f3f3f#define ll Long LONGLL Change (ll x) {    ll ans = 0;    while (x) {        ans + = X/5;        x/= 5;    }    return ans;} int main () {    int T, n;    int k = 1;    scanf ("%d", &t);    while (t--) {        scanf ("%d", &n);        LL L = 0, r = 100000000 * 5 +;        LL mid, ans;        while (R > L) {            mid = (L + r)/2;            if (change (mid) >= N) {                ans = mid;                R = Mid;            }            else                L = mid + 1;        }        printf ("Case%d:", k++);        if (change (ans) = = N)            printf ("%lld\n", ans);        else            printf ("impossible\n");    }    return 0;}

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Light OJ 1138-trailing zeroes (III) "Two-point search && N! The number of consecutive 0 in the end of "