Trailing Zeroes (III)(n!后有多少零+二分）

                                      Trailing Zeroes (III)
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld
& %llu
Submit Status

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

n!中零的个数等于5出现的次数

n!=1*2*3*4*5*6*7*8*9*10*..........*n

=1*2*3*4*5**6*7*8*9*(2*5)*....*(3*5)....*n;

每出现一次五结果末尾会增加一个零

所以只需找出n!中有多少5的倍数就行了

#include <stdio.h>
long long judge(long long x) //找有多少数是5的倍数
{
long long cnt=0;
while(x)
{
cnt=cnt+x/5;
x=x/5;
}
return cnt;
}
int main()
{
int t;
scanf("%d",&t);
int cnt=1;
while(t--)
{
long long q;
scanf("%lld",&q);
long long l=0,r=500000000;
long long mid;
long long ans;
while(r>=l) //二分查找
{
mid=(l+r)/2;
if(judge(mid)>=q)
{
ans=mid;
r=mid-1;
}
else
{
l=mid+1;
}
}
if(judge(ans)!=q)
{
printf("Case %d: impossible\n",cnt);
cnt++;
}
else
{
printf("Case %d: %lld\n",cnt,ans);
cnt++;
}
}
return 0;
}