Trailing Zeroes (III) -；lightoj 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：如果某个数的阶乘有n个0，问这个数最小可以是多少。

题目描述：

　　假设有一个数n，它的阶乘末尾有Q个零，现在给出Q，问n最小为多少？

解题思路:

　　由于数字末尾的零等于min（因子2的个数，因子5的个数），又因为2<5,那么假设有一无限大的数n，n=2^x=5^y，可知x<<y。

所以我们可以直接根据因子5的个数，算阶乘末尾的零的个数。1<=Q<=10^8,所以可以用二分快速求解。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

long long slove(long long n)
{
    long long ans = 0;
    while(n)
    {
        ans += n/5;
        n /= 5;
    }
    return ans;
}

int main()
{
    int t, l = 1;

    scanf("%d", &t);
    while(t--)
    {
        long long n;
        scanf("%lld", &n);
        long long mid, low = 4, high = 500000000;

        while(low <= high)    // 二分查找快~
        {
            mid = (low+high)/2;
            long long num = slove(mid);
            if(num >= n)
                high = mid-1;
            else
                low = mid+1;
        }
        if(slove(low) == n)
            printf("Case %d: %lld
", l++, low);
        else
            printf("Case %d: impossible
", l++);
    }
    return 0;
}