HDU 5655 四边形判断

CA Loves Stick

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 563    Accepted Submission(s): 202

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

 

Source

BestCoder Round #78 (div.2)

 

题意： 输入四边长 a，b，c，d判断能否组成四边形

题解： 1.根据边长的范围要考虑0

          2.四边形判断条件：升序排列后 最大边小于另外三边之和

          3.根据边长的范围：a[0]>a[3]-a[2]-a[1] （加法处理会爆数据范围 应当移项 减法处理）

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<queue>
#include<stack>
#include<algorithm> 
#define ll __int64
#define mod 1 
#define PI acos(-1.0)
using namespace std;
ll a[5];
int t;
int main()
{ 
     scanf("%d",&t);
     for(int i=1;i<=t;i++)
{
    scanf("%I64d %I64d %I64d %I64d",&a[0],&a[1],&a[2],&a[3]);
    sort(a,a+4);
    //cout<<a[0]<<a[1]<<a[2]<<a[3]<<endl;
    if(a[0]==0||a[1]==0||a[2]==0||a[3]==0)
    cout<<"No"<<endl;
    else
    {
    if(a[0]>a[3]-a[2]-a[1])
     cout<<"Yes"<<endl;
     else
     cout<<"No"<<endl;
    }
}
    return 0;
}