LightOJ 1138 - Trailing Zeroes (III) （求末尾0为x的最小N---二分）

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 thatN! contains exactly Q zeroes on the trail in decimal notation. Asyou know N! = 1*2*...*N. For example, 5! = 120, 120 contains one zero onthe 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 nosolution is found then print 'impossible'.

Sample Input	Output for Sample Input
3 1 2 5	Case 1: 5 Case 2: 10 Case 3: impossible

 

思路：直接二分就好了，刚开始我还煞笔的在找规律，= =，不过这些个什么迷之PE搞不太懂




ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 101000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
int fun(int x)
{
    int sum=0;
    while(x)
    {
        sum+=x/5;
        x/=5;
    }
    return sum;
}
int main()
{
    int t;
    int cas=0;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        int l=5,r=500000000;
        int mid;
        while(l<=r)
        {
            mid=(l+r)/2;
            int k=fun(mid);
            if(k<n)
                l=mid+1;
            else
                r=mid-1;
        }
        int ans=fun(l);
        if(ans==n)
            printf("Case %d: %d\n",++cas,l);
        else
            printf("Case %d: impossible\n",++cas);
    }
    return 0;
}