# 【二分+数学公式】Trailing Zeroes (III)

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’.

# Output for Sample Input

3

1

2

5

Case 1: 5

Case 2: 10

Case 3: impossible

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL sum(LL N)
{
LL ans = 0;
while (N)
{
ans += N / 5;
N /= 5;
}
return ans;
}

int main()
{
int t = 0;
cin >> t;
int cas = 0;
while (t--)
{
int q;
scanf("%d", &q);
LL left = 1, ri = 100000000000;
LL ans = 0;
while (ri >= left)
{
int mid = (left + ri) >> 1;
if (sum(mid) == q)
{
ans = mid;
ri = mid - 1;
}
else if (sum(mid) > q)
ri = mid - 1;
else
left = mid + 1;
}
printf("Case %d: ", ++cas);
if (ans)
printf("%lld\n", ans);
else
printf("impossible\n");
}
return 0;

}