SGU-154 Factorial (末尾0 & 三分）

154. Factorial

time limit per test: 0.25 sec.
memory limit per test: 4096 KB

input: standard input
output: standard output

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

One number Q written in the input (0<=Q<=10^8).

Output

Write "No solution", if there is no such number N, and N otherwise.

Sample test(s)

Input

2

Output

10

分析：要求出n!的末尾0的个数也就是寻找n!中5的个数，如5！的5因子个数是1，15！的5因子的个数是3，25的5因子个数是6，可以得到这样的递推式s[n!]=n/5+s[(n/5)!]。在(0<=Q<=10^8)内寻找这样的值n，为了抢时间用三分：

#include<cstdio>using namespace std;const int maxn=4e8+15;int sum;int getnum(int n){    if(n==0)return 0;    sum=n/5+getnum(n/5);    return sum;}int ternarysearch(int low,int high,int x){   int res;   if(high<low) res= -1;   else {       int mid1=low+(high-low)/3,mid2=high-(high-low)/3;       int q1=getnum(mid1),q2=getnum(mid2);       if(x==q1) res=mid1;       else if(x<q1) res=ternarysearch(low,mid1-1,x);       else if(x==q2) res=mid2;       else if(x>q2) res=ternarysearch(mid2+1,high,x);       else res=ternarysearch(mid1+1,mid2-1,x);   }    return res;}int main(){    //freopen("cin.txt","r",stdin);    int n;    while(~scanf("%d",&n)){        if(n==0){            printf("1\n");            continue;        }        int ans=ternarysearch(0,maxn,n);        if(ans==-1)printf("No solution\n");        else {            ans=ans/5*5;            printf("%d\n",ans);        }    }    return 0;}