Problem
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! = 12…*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
题意:
给x,求一个n,满足n的阶乘末尾有x个0
思路:
首先要知道怎么求n的阶乘的末尾的零的个数:这里
然后就是随着n的增大,n的阶乘末尾的0的个数满足非递减,即有序,满足二分性质
因此二分n,求出满足条件的n就行了
code:
#include<bits/stdc++.h>
using namespace std;
//#define int long long
//typedef long long ll;
int ask(int i){
int ans=0;
while(i){
ans+=i/5;
i/=5;
}
return ans;
}
signed main(){
int T;
scanf("%d",&T);
int cas=1;
while(T--){
int n;
scanf("%d",&n);
int ans=0;
int l=0,r=1e9;//1e9的阶乘含大于2e8个0,所以够了
while(l<=r){
int mid=(l+r)/2;
if(ask(mid)>=n){
ans=mid;
r=mid-1;
}else{
l=mid+1;
}
}
printf("Case %d: ",cas++);
if(ask(ans)==n){
printf("%d\n",ans);
}else{
puts("impossible");
}
}
return 0;
}