Light OJ 1138 Trailing Zeroes (III)

题目链接

http://lightoj.com/volume_showproblem.php?problem=1138

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

 

 题意：

求第一次出现q个后缀零的n！，输出的是n

没有的话是输出“impossible"

思路：

要想出来0，一定时有5，所有就是判断n！里面5的倍数有多少个

 This is the code 

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<set>
#include<vector>
using namespace std;
#define PI acos(-1.0)
#define EPS 1e-8
#define MOD 1e9+7
#define LL long long
#define ULL unsigned long long     //1844674407370955161
#define INT_INF 0x7f7f7f7f      //2139062143
#define LL_INF 0x7f7f7f7f7f7f7f7f //9187201950435737471
const int dr[]= {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[]= {-1, 1, 0, 0, -1, 1, -1, 1};
// ios::sync_with_stdio(false);
// 那么cin, 就不能跟C的 scanf，sscanf, getchar, fgets之类的一起使用了。
const int maxn=5e8+1;
LL cal(int n)
{LL ret=0;while(n){ret+=n/5;n/=5;}return ret;
}
LL slove(LL q)
{LL l=1;LL r=maxn;LL mid;LL res=0;while(l<=r){mid=(l+r)>>1ll;LL tem=cal(mid);if(tem>=q){if(tem==q)res=mid;r=mid-1;}elsel=mid+1;}if(res==0)printf("impossible\n");elseprintf("%d\n",res);
}
int main()
{int T;scanf("%d",&T);for(int t=1;t<=T;++t){LL q;scanf("%lld",&q);printf("Case %d: ",t);slove(q);}return 0;
}