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

 

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 0x3f3f3f3f
using 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;
}