学步园

John von Neumann, b. Dec. 28, 1903, d. Feb. 8, 1957, was a Hungarian-American mathematician who made important contributions to the foundations of mathematics, logic, quantum physics,meteorology, science, computers, and game theory.
He was noted for a phenomenal memory and the speed with which he absorbed ideas and solved problems. In 1925 he received a B.S. diploma in chemical engineering from Zurich Institute and in 1926 a Ph.D. in mathematics from the University of Budapest. His Ph.D.
dissertation on set theory was an important contribution to the subject. At the age of 20, von Neumann proposed a new definition of ordinal numbers that was universally adopted. While still in his twenties, he made many contributions in both pure and applied
mathematics that established him as a mathematician of unusual depth. His Mathematical Foundations of Quantum Mechanics (1932) built a solid framework for the new scientific discipline. During this time he also proved the mini-max theorem of GAME THEORY. He
gradually expanded his work in game theory, and with coauthor Oskar Morgenstern he wrote Theory of Games and Economic Behavior (1944).
There are some numbers which can be expressed by the sum of factorials. For example 9,9=1!+2!+3! Dr. von Neumann was very interested in such numbers. So, he gives you a number n, and wants you to tell him whether or not the number can be expressed by the sum
of some factorials.
Well, it's just a piece of cake. For a given n, you'll check if there are some xi, and let n equal to Σ_1<=i<=tx_i!. (t >=1 1, xi >= 0, xi = xj iff. i = j). If the answer is yes, say "YES"; otherwise, print out "NO".

You will get several non-negative integer n (n <= 1,000,000) from input file. Each one is in a line by itself.
The input is terminated by a line with a negative integer.

For each n, you should print exactly one word ("YES" or "NO") in a single line. No extra spaces are allowed.

9
-1
这道题是到简单的递归，也可以不用递归，以前不会写递归，写这道题的目的是学会写递归，并且熟悉他的一般模式
#include<iostream>
#include<stdio.h>
using namespace std;
int w[10]={1,1,2,6,24,120,720,5040,40320,362880};
int flage;
void  dfs(int sum, int x,int num)
{
	if(sum==x)
	{
		flage=1;
		return ;
	}
	if( sum>x)
		return ;
	
	if(num>=0)//这里可以分为从大到小或者从小到大，因为数越来越大所以就会出现先大后小比较合适更快的找到结构
	{//另外这道题可以不用递归一个循环搞定，while(x>w[i]){x-=w[i];i--;}因为数组每个数字的和都会大于前面总和
		dfs(sum+w[num],x,num-1);
		dfs(sum,x,num-1);
	}	
}
int main()
{
	int T;
	while(scanf("%d",&T) && T>=0)
	{
		flage=0;
		dfs(0,T,9);
		if(flage==1)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}









Sample Output
YES
Source
Asia Guangzhou 2003