Lightoj 1138 （ Trailing Zeroes (III)）

 Trailing Zeroes (III)

 

Description

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

题意：问末尾有n个0的最小阶乘

题解：hdu 1124的变形，阶乘+二分。

<span style="font-size:18px;">#include<cstdio>
#define N 0x3f3f3f3f//不要写成1e8
using namespace std;

long long find (long long x){
long long cnt=0;
while (x){
cnt+=x/5;
x/=5;
}
return cnt;
}
long long solve(long long y){
long long le=1;
long long ri=N,s;
long long mid;
long long ans=-1;
while (le<=ri){
mid=(le+ri)/2;
s=find(mid);
if (s==y){
ans=mid;
ri=mid-1;
}
else if (s>y){
ri=mid-1;
}
else if (s<y)
le=mid+1;
}
return ans;
}
int main(){
int t,cas=1;
long long q,ans;
scanf ("%d",&t);
while (t--){
scanf ("%lld",&q);
printf ("Case %d: ",cas++);
ans=solve(q);
if (ans==-1)
printf ("impossible\n");
else
printf ("%lld\n",ans);
}
return 0;
}
</span>