LightOJ 1138

Description:

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:

3

1

2

5

Sample Output:

Case 1: 5

Case 2: 10

Case 3: impossible

题意：找阶乘尾0的个数是q的数，输出最小的那个。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

int Solve(int n) ///计算n的阶乘的尾0个数
{
    int sum = 0;

    while (n)
    {
        sum += n/5;
        n /= 5;
    }

    return sum;
}

int main ()
{
    int T, q, k = 0, i, mid, j, x, y;

    scanf("%d", &T);

    while (T--)
    {
        scanf("%d", &q);

        k++;
        x = -1;

        i = 1; j = 1e9;

        while (i <= j) ///用二分搜索枚举1e9以内的数的阶乘尾数0的个数，保存在y中
        {
            mid = (i+j)/2;

            y = Solve(mid);

            if (y >= q) ///如果找到的尾数0的个数比q大，继续查找左边区间
            {
                j = mid-1;
                if (y == q) ///如果找到的尾数0的个数等于q，保存这个值终止循环
                {
                    x = mid;
                    break;
                }
            }
            else i = mid+1; ///如果找到的尾数0的个数比q小，继续查找右边区间
        }

        if (x % 5 != 0) x = x-x%5; ///因为题目要求找到最小的x  所以要对x向下取整
                                  ///(eg: q==3 时  二分查找到的x可能是15，16，17,18,19)
        if (x == 0) x = -1; ///可能找不到某个数的阶乘有q个0
        if (x == -1) printf("Case %d: impossible\n", k);
        else printf("Case %d: %d\n", k, x);
    }

    return 0;
}