lightoj Trailing Zeroes (III)

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

忘记了二分
#include<stdio.h>

#include<string.h>

#define LL __int64

int sum(int n)

{
int ans=0;
while(n)
{
ans+=n/5;
n/=5;
}
return ans;

}

int main()

{
int i;
int t,q;
int l=1;
scanf("%d",&t);
while(t--)
{
scanf("%d",&q);
printf("Case %d: ",l++);
int l=1;
int r=1000000000;
int mid,ans=0;
while(l<=r)
{
mid=(l+r)/2;
if(sum(mid)>q)
r=mid-1;
else if(sum(mid)==q)
{
ans=mid;
r=mid-1;
}
else
l=mid+1;
}
if(ans==0)
printf("impossible\n");
else
printf("%d\n",ans);
}
return 0;

}