C - Trailing Zeroes (III) LightOJ - 1138
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
/*****
思路:对0--5*x进行二分(别问我为什莫,自己好好想想)
*****/
AC:代码
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
ll judge(ll mid,ll n){
ll s = 0;
while(mid>=5){
mid /= 5;
s += mid;
}
if(s>=n) return 0;
else return 1;
}
int main()
{
int t,ncase = 1;
cin>>t;
while(t--)
{
ll n;
scanf("%lld",&n);
ll low = 0,high = 5*n,mid;
for(int i = 0;i<300;i++){
mid = low + (high - low)/2;
if(judge(mid,n)) low = mid + 1;
else high = mid;
}
ll a = low,s = 0;
while(low>=5){
low /= 5;
s += low;
}
printf("Case %d: ",ncase++);
if(s == n){
printf("%lld\n",a);
}
else printf("impossible\n");
}
return 0;
}