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
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?
He thinks it's easy to understand,and he found a challenging problem from his talented math teacher:
You're given four integers,
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 integersa1,a2,a3,a4(0≤a1,a2,a3,a4≤109) .
The next T lines,each contains 4 integers
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 YesHintIn 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}.
Source
/* ***********************************************
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;
}