hdu 5143 NPY and arithmetic progression(枚举)

最新推荐文章于 2019-08-06 10:45:46 发布

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最新推荐文章于 2019-08-06 10:45:46 发布

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原文链接：http://www.cnblogs.com/aizc/p/7576285.html

本文介绍了一道数学挑战题的解决方案，题目要求将给定数量的数字1、2、3、4分配到一个或多个长度至少为3的等差数列中，确保每个数字恰好使用一次。通过具体的样例输入输出展示了如何判断是否能成功分配。

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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, $a_1, a_2, a_3, a_4$, 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≤1000001≤T≤100000).
The next T lines,each contains 4 integers a1,a2,a3,a4(0≤a1,a2,a3,a4≤109)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

/*
解题思路：等差数列共三种情况：123 234 1234 用光所有数字，组成0到多个长度至少为3的等差数列 组不成等差数列返回No 组成返回Yes
            1. a个1 b个2 c个3 d个4 组成公差为1的等差数列后 剩余的数字大于3的话可以自身组成公差为0的等差数列（例如组成2个123 一个333 那么输入的abcd依次为2250）
            2. a个1 b个2 c个3 d个4 刚好用完，组成多个公差为1的等差数列
*/

#include<stdio.h>
#include<string.h>
#define MAX 100000
#define MAX2 1000000000
int main()
{
    int i,j,k,T,flag;
    int a,b,c,d;
    scanf("%d",&T);
    if (T<=MAX)
    {
    while(T--){
        scanf("%d%d%d%d",&a,&b,&c,&d);
        flag=0;
        if(a<=MAX2||b<=MAX2||c<=MAX2||d<=MAX2){
            for (i = 0; i <= 2; i++)    //情况1：123
                for (j = 0; j <= 2; j++)//情况2：234                
                    for (k = 0; k <= 2; k++)//情况3：1234                    
                        {if (((a-i-k==0)||(a-i-k>=3))&&((b-i-j-k==0)||(b-i-j-k>=3))&&((c-i-j-k==0)||(c-i-j-k>=3))&&((d-j-k==0)||(d-j-k>=3)))
                            {flag=1;break;}
                    }




        if (flag)
            printf("Yes\n");        
        else
            printf("No\n");
    }
    }    
    }


    return 0;
}

转载于:https://www.cnblogs.com/aizc/p/7576285.html