HDOJ 2578 Dating with girls(1)

Dating with girls(1)
Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4218    Accepted Submission(s): 1317
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
   
   
   
   

 
題意はnとkを与え,またn個の数を与える.n個の数のうち2個の数x,yを見つけることである.x+y=kにします.
すべての可能性を計算します.x 0+y 0=kのとき.x1+y1=k . 等しくないものが必要です.x0!=x1 or y0!=y1.
たとえば
4 4
2 2 2 2
正しい出力は1です.
だから必ず重さを

#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 100000 + 10;

long long a[maxn], b[maxn], k;
int n, cnt;
int binary_search(int kk)
{
    int fir = 0, las = cnt - 1;
    while (fir <= las){
        int middle = (fir + las) / 2;
        if (a[middle] == kk)
            return 1;
        else if (a[middle] < kk)
            fir = middle + 1;
        else
            las = middle - 1;
    }
    return 0;
}

int main()
{
    int t;
    int ans;
    scanf("%d", &t);
    while (t--){
        ans = 0;
        scanf("%d%I64d", &n, &k);
        for (int i = 0; i < n; i++){
            scanf("%I64d", &a[i]);
        }

        sort(a, a + n);
        cnt = unique(a, a + n) - a;   //unique     
        for (int i = 0; i < cnt; i++){
            b[i] = k - a[i];
           // if (binary_search(b[i]) && a[i] != b[i])
               // ans = ans + 2;
           // else if (binary_search(b[i]) && a[i] == b[i])
               // ans = ans + 1;
            if (binary_search(b[i]))
                ans = ans + 1;
        }
        printf("%d
", ans);
    }
    return 0;
}