HDU 2578 二分

Dating with girls(1)

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3918    Accepted Submission(s): 1216

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
直接k-x得到一组数据 然后 2分查找 注意去重 利用hash  也可以 只有hash
#include<cstdio>#include<iostream>#include<algorithm>using namespace std;int a[100005]={0},b[100005]={0};bool hash1[1000005]={0};bool ok(int x);int n=0;int y=0;int main(){int t=0;scanf("%d",&t);while(t--){int k=0;int sum=0;y=0;memset(hash1,0,sizeof(hash1));scanf("%d%d",&n,&k);for(int i=0;i<n;i++){scanf("%d",&a[i]);   if(hash1[k-a[i]]==0)   {b[y++]=k-a[i];hash1[k-a[i]]=1;   }}sort(a,a+n);for(int i=0;i<y;i++){if(ok(b[i])) sum++;}printf("%d\n",sum);}return 0;}bool ok(int x){int u1=0;int u2=n-1;int mid=(n-1)/2;if(a[0]==x||a[n-1]==x) return 1;while(u2-u1>1){if(x>a[mid]){u1=mid;}else if(x<a[mid]){u2=mid;}else{return 1;}mid=(u1+u2)/2;}return 0;}
别人的
#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;int a[200010],n;int find(int x){    int i=1,j=n;    while(i<=j)    {        int mid=(i+j)/2;        if(a[mid]==x||a[mid+1]==x)            return 1;        else if(a[mid]<x&&a[mid+1]>x)            return 0;        else if(a[mid]>x)            j=mid-1;        else            i=mid+1;    }    return 0;}int main(){    int t;    cin>>t;    while(t--)    {        int k;        cin>>n>>k;        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        sort(a+1,a+n+1);        __int64 sum=0;        a[0]=-1;        for(int i=1;i<=n;i++)        {            if(a[i]!=a[i-1]&&a[i]<=k&&find(k-a[i]))                sum++;        }        printf("%I64d\n",sum);    }}