ACM-简单题之Dating with girls1——hdu2578

Dating with girls(1)
题目：http://acm.hdu.edu.cn/showproblem.php?pid=2578
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2792 Accepted Submission(s): 886

Problem Description
Everyone in the HDU knows that the number of boys is larger than the number of girls. But now, every boy wants to date with pretty girls. The girls like to date with the boys with higher IQ. In order to test the boys ' IQ, The girls make a problem, and the boys who can solve the problem
correctly and cost less time can date with them.
The problem is that : give you n positive integers and an integer k. You need to calculate how many different solutions the equation x + y = k has . x and y must be among the given n integers. Two solutions are different if x0 != x1 or y0 != y1.
Now smart Acmers, solving the problem as soon as possible. So you can dating with pretty girls. How wonderful!


Input
The first line contain an integer T. Then T cases followed. Each case begins with two integers n(2 <= n <= 100000) , k(0 <= k < 2^31). And then the next line contain n integers.

Output
For each cases,output the numbers of solutions to the equation.

Sample Input
2
5 4
1 2 3 4 5
8 8
1 4 5 7 8 9 2 6

Sample Output
3

5

一道水题（我WA好多次。。。）

题意就是给你T个数，查看 两个数相加等于k的个数 。

明显的查找题目，之前做省赛有一道感觉类似，于是直觉般的想用排序+二分查找来搞定这道题。

以第一个sample为例，k为4，所以1+3,和2+2，还有3+1。

a+b和b+a是可以算作两种的，但是有一点要注意，人家可能给你多个1和一个3，这样也是只能算两种。

为什么我WA好多次呢？

第一次是多个1，一个3没想到，WA一次，自己看了遍题意，改回来了。

后面实在不知道自己哪错。。

于是，看人家代码。

差别在于，我在输入的时候做了一个判断，输入的T个数中，有大于等于K的，一律舍弃，不存入数组。

我把这个改回来，无论啥样全都输入进数组，结果就A了。。

我现在也不大懂为什么不能舍弃掉那些。。望各位给予解答。。Thank U~

/*
Author：Tree
From： http://blog.csdn.net/lttree
Dating with girls1 
hdu 2578
排序+二分查找
*/

#include <iostream>
#include <algorithm>

using namespace std;
int arr[100001];

// 二分查找，返回是否查找到。
bool binary_search( int v,int len )
{
	int l,h,m;
	l=0,h=len-1;
	while( l<=h )
	{
		m=(l+h)/2;
		if(arr[m]==v)	return 1;
		else if( arr[m]>v )	h=m-1;
		else	l=m+1;
	}
	return 0;
}

int main()
{
	int i,t,k,test,temp,sum;
	cin>>test;
	while(test--)
	{
		cin>>t>>k;
		for(i=0;i<t;++i)
			cin>>arr[i];	
		sort(arr,arr+t);

		sum=0;
		for(i=0;i<t;++i)
		{
			if( binary_search(k-arr[i],t) )
			{
				++sum;
				// 如果查找的是两个相同的数，那么只能算一次
				if( i!=0 && arr[i-1]==arr[i])
					--sum;
			}
		}
		cout<<sum<<endl;
	}
	return 0;
}