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[];
 int t;
 int main()
 {
      scanf("%d",&t);
      for(int i=;i<=t;i++)
 {
     scanf("%I64d %I64d %I64d %I64d",&a[],&a[],&a[],&a[]);
     sort(a,a+);
     //cout<<a[0]<<a[1]<<a[2]<<a[3]<<endl;
     if(a[]==||a[]==||a[]==||a[]==)
     cout<<"No"<<endl;
     else
     {
     if(a[]>a[]-a[]-a[])
      cout<<"Yes"<<endl;
      else
      cout<<"No"<<endl;
     }
 }
     return ;
 }