1775 Sum of Factorials 全排列解法+规律解法

Sum of Factorials
Time Limit: 1000MS      Memory Limit: 30000K
Total Submissions: 13391        Accepted: 4430
Description
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 Σ1<=i<=txi!. (t >=1 1, xi >= 0, xi = xj iff. i = j). If the answer is yes, say "YES"; otherwise, print out "NO".
Input
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.
Output
For each n, you should print exactly one word ("YES" or "NO") in a single line. No extra spaces are allowed.
Sample Input
9
-1
Sample Output
YES
Source
Asia Guangzhou 2003
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来源： http://poj.org/problem?id=1775

//正写
#include <cstdio>
#include <iostream>
using namespace std;
int n,flag,A[10]={1};//在 1! ~ 9! 之间
void DFS(int value,int Set)
{
if(Set>9||value>n) return;
if(value==n) {flag=1;return;}
DFS(value+A[Set+1],Set+1);
DFS(value,Set+1); //不加下一个 留着给后面加//看了题解
}
int main(void)
{
for(int i=1;i<=9;i++) A[i]=i*A[i-1]; //阶乘表
while(cin>>n)
{
if(n<0) break;
flag=0;DFS(0,-1);
printf("%s\n",n!=0&&flag?"YES":"NO");
}
}
//反写
#include <cstdio>
#include <iostream>
using namespace std;
int n,flag,A[10]={1};//在 1! ~ 9! 之间
void DFS(int value,int Set)
{
if(Set<0||value>n) return;
if(value==n) {flag=1;return;}
DFS(value+A[Set-1],Set-1);
DFS(value,Set-1); //不加下一个 留着给后面加
}
int main(void)
{
for(int i=1;i<=9;i++) A[i]=i*A[i-1]; //阶乘表
while(cin>>n)
{
if(n<0) break;
flag=0;DFS(0,10);
printf("%s\n",n!=0&&flag?"YES":"NO");
}
}

/*!这个思路非常简单 如果是a!+b!+c! 那么从后往前面减
1! < 2!
1!+2！< 3!
1!+2!+3! < 4!
...............
1!+2!+3!+n! < (n+1)!=(n+1)*n!
------->那么只是从中抽取的元素阶乘和就更小于n!
------->若S=a!+b!+c!+..+j!+k! ,a<b<c..<k  于是直到出现k! 减去它 化成子问题
------->S' = a!+b!+c!+..+j!
------->最终S能变为0则满足条件
*/

#include<iostream>
using namespace std;
int n,A[10]={1};
int main()
{
for(int i=1;i<=9;i++) A[i]=i*A[i-1]; //阶乘表
while(cin>>n&&n>=0)
{
if(n==0){cout<<"NO"<<endl;continue;}
for(int i=9;i>=0&&n;i--) if(A[i]<=n) n-=A[i];
n==0?cout<<"YES"<<endl:cout<<"NO"<<endl;
}
return 0;
}