HDU 5143 NPY and arithmetic progression （枚举基本类型123、234、1234的个数）

NPY and arithmetic progression

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 799    Accepted Submission(s): 254

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

31 2 2 11 0 0 03 0 0 0

 

Sample Output

YesNoYes
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}.
 

 

Source

BestCoder Round #22

 

题目大意：给出数字1,2,3,4，的数量，问能否将这些数用完组成长度不小于三的等差数列。

解题思路：由数字1,2,3,4，组成的等差数列只有123、234、1234和长度>=3的常数列。如果数列123、234和1234的数量大于等于3，那么这些数量的数列可以变为3个或4个常数列,例如：123、123、123变成111、222、333。所以可以直接枚举数列123、234、1234的个数（0~2），再判断数字1、 2、 3、 4的剩余个数，如果是0或者大于等于3，那么就可以。

代码如下：

#include <cstdio>#include <iostream>#include <cstdlib>#include <cstring>#include <cmath>#include <string>#include <algorithm>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#include <limits.h>#include <ctime>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-6)#define inf (1<<28)#define sqr(x) (x) * (x)#define mod 1000000007using namespace std;typedef long long ll;typedef unsigned long long ULL;int main(){    int i,j,k,t;    int a1,a2,a3,a4;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d%d",&a1,&a2,&a3,&a4);        int flag=0;        for(i=0; i<3; i++)        {            for(j=0; j<3; j++)            {                for(k=0; k<3; k++)                {                    if((a1-i-k)<0||(a2-i-j-k)<0||(a3-i-j-k)<0||(a4-j-k)<0)                        continue;                    if(((a1-i-k)>=3||(a1-i-k)==0)&&((a2-i-j-k)>=3||(a2-i-j-k)==0)&&((a3-i-j-k)>=3||(a3-i-j-k)==0)&&((a4-j-k)>=3||(a4-j-k)==0))                    {                        flag=1;                        break;                    }                }                if(flag)                    break;            }            if(flag)                break;        }        if(flag)            printf("Yes\n");        else            printf("No\n");    }    return 0;}