154. Factorial （二分+數論）

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

//首先在階乘n！中隻要把每個數都拆分為質因數的積，則隻需要
//質因數5的個數即可，因為2的個數會比5多很多，是以質因數有多少5，
//尾部就有多少0........
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int num(int n)
{
	int cnt=0;
	while(n)
	{
		cnt+=n/5;//
		n/=5;
	}
	return cnt;
}
void find(int n)
{
	int left=0,right=INF,mid;
	while(left<=right)
	{
		mid=(left+right)/2;
		if(num(mid)>=n)
			right=mid-1;
		else
			left=mid+1;
	}
	if(num(left)==n)   
		printf("%d\n",left);
	else
		printf("No solution\n");
}
int main(){
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
			printf("1\n");
		else
			find(n);
	}
	return 0;
}
           

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int fun(int x)
{
	int sum=0;
	while(x)
	{
		sum+=x/5;
		x/=5;
	}
	return sum;
}
int main(){
	int m;
	while(scanf("%d",&m)!=EOF)
	{
		if(m==0)
		{
			printf("1\n");
			continue;
		}
		int n=m*4/5*5;
		while(fun(n)<m)
		{
			n+=5;
		}
		if(fun(n)==0)
			printf("%d\n",n);
		else
			printf("No solution\n");
	}
	return 0;
}