dfs(water)

sum of factorials

Time Limit: 3000  ms | Memory Limit: 65535  KB

Difficulty: 3

describe

Give you a non-negative integer n, determine whether n is the sum of the factorials of some numbers (these numbers are not allowed to be reused and are positive numbers), such as 9=1! +2!+3!, if yes, output Yes, otherwise output No;

enter

The first line has an integer 0<m<100, indicating that there are m groups of test data;
each group of test data has a positive integer n<1000000;

output

If the condition is met, output Yes, otherwise output No;

sample input

2

9

10

Sample output

Yes

No

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;
int s1[15];

int jiecheng (int a)
{
	int sum=1;
	while(a)
	{
		sum*=a;
		a--;
	}
	return sum;
}

int flag=0;
int n;
void dfs(int cur,int sum)
{
	if(cur > 10 || sum > n)
		return ;
	if(sum==n)
	{
		flag=1;
		return ;
	}
	
	dfs(cur+1,sum+s1[cur]);//dfs  
	dfs(cur+1,sum);
}
intmain()
{
	
	int k,m;
	for(int i=1;i<=11;i++)
	{
		for(int t=1;t<=i;t++)
		{
			s1[i]=jiecheng(t);
		}
	}
	
	cin>>k;
	while(k--)
	{
		cin>>n;
		dfs(1,0);
		if(flag==1)
		cout<<"Yes"<<endl;
		else
		cout<<"No"<<endl;
		flag=0;
	}
}