LightOJ 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

今天脑残了。。。。。。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
#define space " "
using namespace std;
typedef long long LL;
//typedef __int64 Int;
typedef pair<int,int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double Pi = acos(-1);
const int MOD = 1e9+5;
const int MAXN = 1e5;
bool flag ;
bool judge(int x, int y) {
    int f = 5;
    int sum = 0;
    while (f <= x) {
        sum = sum + x/f;
        f = f*5;
    }
    if (sum == y) flag = true;
    return sum >= y;
}
int main() {
    int t, n;
    int cnt = 0;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        int ans; flag = false;
        int lb = 0, ub = 1e9;
        while (ub >= lb) {
            int mid = (ub + lb) >> 1;
            if (judge(mid, n)) {
                ans = mid; ub = mid - 1;
            }
            else lb = mid + 1;
        }
        if (flag) printf("Case %d: %d
", ++cnt, ans);
        else printf("Case %d: impossible
", ++cnt);
    }
    return 0;
}