LightOJ1138 Trailing Zeroes (III)

题目链接:点我

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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

3125

Sample Output

Case 1: 5Case 2: 10Case 3: impossible

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题意:

求是否存在一个数,它的阶乘末尾有q个0.

思路:

我们知道,在n!中,每一个末尾0都是由2 和 5 贡献的,那么我们只需要知道前n个数中有多少对2 和5即可,然后我们发现,5的数量比2 会少很多,所以,我们只需要知道有多少个5就可以了,然后我们直接在所有可能的n中进行二分即可

代码:

#include<cstring>#include<cstdio>#include<algorithm>#include<cmath>#include<iostream>using namespace std;typedef long long LL;LL sum(LL n){    LL ans = 0;    while(n){        ans += n / 5;        n /= 5;    }    return ans;}int main(){    int t;    scanf("%d", &t);    int kase = 0;    while( t--){        int n;        scanf("%d", &n);        LL l = 0;        LL r = 10000000000000;        while( l <= r){            LL mid = (l + r) >> 1;            if(sum(mid) >= n) r = mid -1;            else l = mid + 1;        }        printf("Case %d: ",++kase );        if(sum(l) == n)            printf("%lld\n",l);        else printf("impossible\n");    }    return 0;}