HDU2578 Dating with girls(1)（二分）

Dating with girls(1)

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4870    Accepted Submission(s): 1557

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

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

 

Sample Output

35

 

Source

HDU 2009-5 Programming Contest 

http://acm.split.hdu.edu.cn/showproblem.php?pid=2578

题意很给力 典型的二分  注意位置不同也算一种  相同数字去掉

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int aa[100010];int main(){    int t,n,k,i;    scanf("%d",&t);    while(t--)    {        scanf("%d %d",&n,&k);        for(i=0; i<n; i++)            scanf("%d",&aa[i]);        sort(aa,aa+n);        int  len=unique(aa,aa+n)-aa;        long long sum=0;        int l,r;        for(i=0; i<len; i++)        {            l=0,r=len;            int hj=k-aa[i];            while(l<r)            {                int mid=(l+r)>>1;                if(aa[mid]<hj) l=mid+1;                else if(aa[mid]>hj)r=mid-1;                else {l=mid; break;}            }            if(aa[l]==hj) sum++;        }        printf("%I64d\n",sum);    }    return 0;}