Light oj 1138 - Trailing Zeroes (III) 【二分查找 && N！中末尾连续0的个数】

1138 - Trailing Zeroes (III)

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

 

题意：给你一个数Q，代表N！中末尾连续0的个数。让你求出最小的N。

求N！中尾连续0的个数：

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

AC代码

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
#define INF 0x3f3f3f3f
#define LL long long

LL 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 + 10;
        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;
}