Light oj 1138 - Trailing Zeroes (III) 【二分查找好题】【 给出N！末尾有连续的Q个0，让你求最小的N】

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

 

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PROBLEM SETTER: JANE ALAM JAN

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。

定理：求N！中  末尾连续0的个数

求法如下

LL sum(LL N){    LL ans = 0;    while(N)    {        ans += N / 5;        N /= 5;    }    return ans;}

本来不敢写，最后发现即使Q = 10^8也不会超long long（貌似int都不超）

又犯二了，区间开小了。WA了一次。

AC代码：用二分实现的

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <vector>#include <map>#include <string>#include <algorithm>#define LL long long#define MAXN 100+10#define MAXM 20000+10#define INF 0x3f3f3f3fusing namespace std;LL sum(LL N)//求N阶乘中 末尾连续的0的个数{    LL ans = 0;    while(N)    {        ans += N / 5;        N /= 5;    }    return ans;}int k = 1;int main(){    int t;    LL Q;    scanf("%d", &t);    while(t--)    {        scanf("%lld", &Q);        LL left = 1, right = 1000000000000;//一开始开小了 醉了        LL ans = 0;        while(right >= left)        {            int mid = (left + right) >> 1;            if(sum(mid) == Q)//相等时 要赋值给ans            {                ans = mid;                right = mid - 1;            }            else if(sum(mid) > Q)                right = mid - 1;            else                left = mid + 1;        }        printf("Case %d: ", k++);        if(ans)            printf("%lld\n", ans);        else            printf("impossible\n");    }    return 0;}