HDU 5143 NPY and arithmetic progression

NPY and arithmetic progression


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 189    Accepted Submission(s): 61


Problem Description 
NPY is learning arithmetic progression in his math class. In mathematics, an arithmetic progression (AP) is a sequence of numbers such that the difference between the consecutive terms is constant.(from wikipedia)
He thinks it's easy to understand,and he found a challenging problem from his talented math teacher:
You're given four integers, a1,a2,a3,a4, which are the numbers of 1,2,3,4 you have.Can you divide these numbers into some Arithmetic Progressions,whose lengths are equal to or greater than 3?(i.e.The number of AP can be one)
Attention: You must use every number exactly once.
Can you solve this problem? 
  
Input 
The first line contains a integer T — the number of test cases (1≤T≤100000).
The next T lines,each contains 4 integers a1,a2,a3,a4(0≤a1,a2,a3,a4≤109). 
  
Output 
For each test case,print "Yes"(without quotes) if the numbers can be divided properly,otherwise print "No"(without quotes). 
  
Sample Input 
3
1 2 2 1
1 0 0 0
3 0 0 0


  
Sample Output 
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}.
 

题解：问你能否组成若干个大于三的等差序列，一开始想着一个个减，结果爆了。。想想不对。。应该只要是一个数的个数大于3就可以自己构成一个等差序列了，这样先把大于三的返回掉，然后在考虑有些的小于三的，用搜索做掉

#include <iostream>#include <cstdio>#include <climits>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <algorithm>#define esp 1e-6#define inf 0x0f0f0f0f#define LL long long  using namespace std;bool dfs(int a,int b,int c,int d){if(a==0&b==0&&c==0&&d==0) return true;//if((a>=3&&b==0&&c==0&&d==0)||(b>=3&&a==0&&c==0&&d==0)||(c>=3&&a==0&&b==0&&d==0)||(d>=3&&a==0&&b==0&&c==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(dfs(a-1,b-1,c-1,d-1))return true;if(a-1>=0&&b-1>=0&&c-1>=0&&d>=0)if(dfs(a-1,b-1,c-1,d))return true;if(a>=0&&b-1>=0&&c-1>=0&&d-1>=0)if(dfs(a,b-1,c-1,d-1))return true;return false;}int main(){int i,j,k,l,m,n,t,ans,a,b,c,d;scanf("%d",&t);while(t--){scanf("%d%d%d%d",&a,&b,&c,&d);if(dfs(a,b,c,d)==true)printf("Yes\n");else printf("No\n");}return 0;}