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LightOJ 1138 二分

2024-08-11 02:06:58   217人阅读

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

#include<bits/stdc++.h>using namespace std;#define ll long long#define mod 1000000007#define pi (4*atan(1.0))const int N=1e5+10,M=4e6+10,inf=1e9+10;int a[20]={5,25,125,625,3125,15625,78125,390625,1953125,9765625,48828125,244140625,1220703125};int check(int x){    int sum=0;    for(int i=0;i<13;i++)        sum+=x/a[i];    return sum;}int main(){    int x,y,z,i,t;    int T,cas=1;    scanf("%d",&T);    while(T--)    {        scanf("%d",&x);        int st=0;        int en=500000000;        while(st<en)        {            int mid=(st+en)>>1;            if(check(mid)>=x)            en=mid;            else            st=mid+1;        }        printf("Case %d: ",cas++);        if(check(st)==x)        printf("%d\n",st);        else        printf("impossible\n");    }    return 0;}

 

LightOJ 1138 二分


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