Trailing Zeroes (III) LightOJ - 1138 二分+找规律

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

 

题意：求出现一个最小的数的阶乘满足末尾有连续q个0。

思路：一个最小的数末尾要出现0一定要是能被5或10整除的，0的连续长度与数组大小正相关，故二分数字得解。

#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<limits.h>
#include<algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int INF=1e9;
typedef unsigned long long ll;
typedef long long LL;
int main()
{
    int T,cases=1;
    scanf("%d",&T);
    while(T--)
    {
        int q;
        LL ans=-1;
        scanf("%d",&q);
        int l=1,r=q*5,mid;//为什么右端点是q*5，因为q*5>=ans.想一想为什么（hint：刚才那个序列。。。）
        while(l<=r)
        {
            mid=(l+r)/2;
            int m=mid,wu=5,count=0;
            while(m>=wu)
            {
                int p=m;
                p/=wu;
                count+=p;
                wu*=5;
            }
            if(count==q)
            {
                ans=mid;
                break;
            }
            else if(count>q)
                r=mid-1;
            else
                l=mid+1;
        }
        if(ans==-1)
            printf("Case %d: impossible\n",cases++);
        else
            printf("Case %d: %lld\n",cases++,ans/5*5);
    }
}

代码转自(https://blog.csdn.net/duan_1998/article/details/72566106)