DFS.....

多余3个的数可以自己成等比数列

和其他数组合成等比数列的有1,2,3,  2,3,4  1,2,3,4 三种情况

NPY and arithmetic progression

3
1 2 2 1
1 0 0 0
3 0 0 0

Yes
No
Yes
Hint
In the first case,the numbers can be divided into {1,2,3} and {2,3,4}.
In the second case,the numbers can‘t be divided properly.
In the third case,the numbers can be divided into {1,1,1}.
 

/* ***********************************************
Author        :CKboss
Created Time  :2014年12月13日 星期六 23时09分55秒
File Name     :B_2.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <cmath>

using namespace std;

bool flag;

bool ck(int a,int b,int c,int d)
{
	/// check
	if(a==0&&b==0&&c==0&&d==0) return true;
	if( ((a!=0&&a>=3)||(a==0)) && ((b!=0&&b>=3)||(b==0)) && ((c!=0&&c>=3)||(c==0)) && ((d!=0&&d>=3)||(d==0)) ) return true;

	if(a-1>=0&&b-1>=0&&c-1>=0&&d-1>=0)
		if(ck(a-1,b-1,c-1,d-1)) return true;

	if(a-1>=0&&b-1>=0&&c-1>=0)
		if(ck(a-1,b-1,c-1,d)) return true;

	if(b-1>=0&&c-1>=0&&d-1>=0)
		if(ck(a,b-1,c-1,d-1)) return true;

	return false;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		int a,b,c,d;
		scanf("%d%d%d%d",&a,&b,&c,&d);
		if(ck(a,b,c,d)==true) puts("Yes");
		else puts("No");
	}
    return 0;
}

HDOJ 5143 NPY and arithmetic progression DFS

原文：http://blog.csdn.net/ck_boss/article/details/41917745