hdu 5655 CA Loves Stick（简单题）（Bestcoder #78 1001）

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 416    Accepted Submission(s): 153

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)

 

Recommend

wange2014

题目大意：

       给四根木棍长度，问能否拼成一个四边形

解题思路：

      参见http://ziyuan.wmw.cn/WLKT/LWMWMZ/RJB/classonline/content0409/2b/2b58001/html/241hb_12.htm证明的很清楚，有中学生的风格

      所以最小的三条边加起来大于最大边就可以，反之不可以，但是小心高精度，unsigned long long 可以解决两个longlong数加的问题，所以改写为a[0]+a[2]>a[3]-a[1]即可，但是注意有长为0的木棍!!神棍啊!有一个0就输出no即可。

<pre name="code" class="cpp">#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<map>
#include<queue>
#include<stack>
#define ll unsigned long long
#define INF 0x3f3f3f3f
#define C(a) memset(a,0,sizeof a)
#define C_1(a) memset(a,-1,sizeof a)
#define C_i(a) memset(a,0x3f,sizeof a)
#define F(i,n) for(int i=0;i<n;i++)
#define F(n) for(int i=0;i<n;i++)
#define F_1(n) for(int i=n;i>0;i--)
#define S(a) scanf("%d",&a);
#define S2(a,b) scanf("%d%d",&a,&b);
#define SL(a) cin>>a; 
#define SD(a) scanf("%lf",&a);
#define P(a) printf("%d\n",a);
#define PL(a) printf("%I64d\n",a);
#define PD(a)printf("%lf\n",a);
#define rush() int t;scanf("%d",&t);while(t--)
using namespace std;
int main()
{
	ll a[4];
	rush()
	{
		SL(a[0])SL(a[1])SL(a[2])SL(a[3]);
		sort(a, a + 4);
		if (!a[0] || !a[1] || !a[2] || !a[3])printf("No\n");
		else if (a[3] - a[1] < a[2]+a[0])printf("Yes\n");
		else printf("No\n");
	}
}