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：

任何质因数都可以写成素数相乘的形式。所以计算一个数的阶乘后面几个0，只需计算这个数包含多少5即可。（关于这点不清楚的点：点击打开链接）。

可以用二分法，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>