hdu 5655 CA Loves Stick

CA Loves Stick

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 981    Accepted Submission(s): 332

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

 题意：给你四条边，判断是否能组成一个四边形   

题解：任意三边之和大于第四边（第四边为最大边）   a+b+c>=d;  转化为减法a+b>=d-c;

注意：1、三个数相加可能超64位    2、如果边为0 就输出no

 

#include<stdio.h>
#include<string.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<cstdio> 
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 450
#define mod 10003
#define INF 0x3f3f3f3f
using namespace std;
unsigned __int64 s[6];
int main()
{
    int t;
    unsigned __int64 n,m,a,b,c,d;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64u%I64u%I64u%I64u",&a,&b,&c,&d);
        if(a==0||b==0||c==0||d==0)
        {
            printf("No\n");
            continue;
        }
        s[0]=a;s[1]=b;s[2]=c;s[3]=d;
        sort(s,s+4);
        if(s[1]+s[0]>=s[3]-s[2])
        printf("Yes\n");
        else
        printf("No\n");
    }
    return 0;
}