hdu 5655 CA Loves Stick【思维】

最新推荐文章于 2019-01-02 08:35:28 发布

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本文介绍了一种用于判断四边形是否能形成的算法，包括输入解析、排序、边界条件检查及输出结果。

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CA Loves Stick

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 328 Accepted Submission(s): 116

Problem Description

CA loves to play with sticks.
One day he receives four pieces of sticks, he wants to know these sticks can spell a quadrilateral.
(What is quadrilateral? Click here: https://en.wikipedia.org/wiki/Quadrilateral)

Input

First line contains

T denoting the number of testcases.

T testcases follow. Each testcase contains four integers

a,b,c,d in a line, denoting the length of sticks.

1≤T≤1000, 0≤a,b,c,d≤263−1

Output

For each testcase, if these sticks can spell a quadrilateral, output "Yes"; otherwise, output "No" (without the quotation marks).

Sample Input
2
1 1 1 1
1 1 9 2

Sample Output
Yes
No

四边形简单判断定理：a+b+c>d。如果有0的边当然不行。

问题关键所在于爆long long int，每一个值都是在long long int的数据边缘，所以我们需要改变式子为：

a>d-b-c,就能解决爆的问题。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
long long int a[5];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=0;i<4;i++)
        {
            scanf("%I64d",&a[i]);
        }
        sort(a,a+4);
        if(a[0]==0)printf("No\n");
        else
        {
            if(a[3]-a[0]-a[1]<a[2])
            {
                printf("Yes\n");
            }
            else 
            printf("No\n");
        }

    }
    return 0;
}