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
3
1
2
5
Sample Output
Case 1: 5
Case 2: 10
Case 3: impossible
核心思想就是,求后缀0的个数;
ll num(ll x)
{
ll temp = 0;
while(x)
{
temp += x/5;
x/=5;
}
return temp;
}
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll num(ll x)
{
ll temp = 0;
while(x)
{
temp += x/5;
x/=5;
}
return temp;
}
int main()
{
int T,cnt = 1;
scanf("%d",&T);
while(T--)
{
ll n;
scanf("%lld",&n);
ll l = 1,r = 1e9,mid,ans = 0;
while(l <= r)
{
mid = (l+r)/2;
if(num(mid) > n)
r = mid-1;
else if(num(mid) < n)
l = mid+1;
else
{
ans = mid;
r = mid-1;
}
}
if(ans)
printf("Case %d: %lld\n",cnt++,ans);
else
printf("Case %d: impossible\n",cnt++);
}
return 0;
}