HDU 2578 Dating with girls(1)

Dating with girls(1)

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

25 41 2 3 4 58 81 4 5 7 8 9 2 6 

 

Sample Output

35 

 

题意
给 n 个数，找出其中两个数x，y，使其和为 k 。当 xi != xi+1 && yi != yi+1 即为一组。也就是k = 4时，1,3和3,1 是两组。

分析
sort+二分，分分钟A掉。本题有坑，例如
n = 5，k = 4
数据为 2 2 2 2 2时
只有一组满足条件 (2,2)，其他都是重复的。
注意特殊处理，就能跳坑。

本人当时被这种数据卡了两发WA，简直要哭了Orz...

AC代码如下

#include <cstdio>#include <algorithm>#define maxn 100000+2using namespace std;int a[maxn];int main(){    int t,m,n,i,j;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(i = 0 ; i < n; i++)            scanf("%d",&a[i]);        sort(a,a+n);        int ans = 0;        for(i = 0 ; i < n ; i++)        {            while(a[i] == a[i+1]) i++;            int pos = lower_bound(a,a+n,m-a[i]) - a;<span style="white-space:pre"></span>//lower_bound() 比手写二分简单很多，至少我这种懒人很喜欢，不会的朋友请自行baidu            if(a[i] + a[pos] == m) ans++;        }        printf("%d\n",ans);    }    return 0;}