LightOJ Trailing Zeroes (III) 1138【二分搜索+阶乘分解】

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：给你一个数字，这个数字代表N！后面有几个0。给出这个数字，计算N的值。

解题思路：

不论什么质因数都能够写成素数相乘的形式。所以计算一个数的阶乘后面几个0。仅仅需计算这个数包括多少5就可以。（关于这点不清楚的点：点击打开链接）。

能够用二分法，找出这个点。

想到用二分这道题也就没什么难度了。

AC代码;

<span style="font-size:18px;">#include <stdio.h>
#include <math.h>
#include <vector>
#include <queue>
#include <string>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

LL solve(LL n)
{
    LL num=0;
    while(n){
        num+=n/5;
        n/=5;
    }
    return num;
}

LL er(LL n)
{
    LL x=1;
    LL y=1844674407370;
    LL mid;
    LL res=-1;
    while(y>=x){
        mid=(x+y)/2;
        LL ans=solve(mid);
        if(ans==n){
            res=mid;
            y=mid-1;
            //return mid;
        }
        else if(ans>n){
            y=mid-1;
        }
        else if(ans<n){
            x=mid+1;
        }
    }
    return res;
}
int main()
{

    int t;
    scanf("%d",&t);
    int xp=1;
    while(t--){
        LL n;
        scanf("%lld",&n);
        LL ans=er(n);
        if(ans==-1)  printf("Case %d: impossible\n",xp++);
        else printf("Case %d: %d\n",xp++,ans);
    }
    return 0;
}
</span>