HDU 2578 Dating with girls(1) [補7-26]

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Dating with girls(1)

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3869    Accepted Submission(s): 1196

Problem DescriptionEveryone 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!
 

 

InputThe 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. 

 

OutputFor each cases,output the numbers of solutions to the equation. 

 

Sample Input25 41 2 3 4 58 81 4 5 7 8 9 2 6 

 

Sample Output35 

 

Source HDU 2009-5 Programming Contest  

 

Recommendlcy   |   We have carefully selected several similar problems for you:  2575 2579 2573 2576 2574  要去除重複的元素和大於k的元素，因為大於k的元素是肯定組不成等於k的方程的，不能確定資料裡是否有0，所以等於k的元素也保留著，每存一個數都hash一下，然後一遍for迴圈，看 hash[ k-a[i] ]是否為1，是的話cnt++；

 1 #include<queue> 2 #include<math.h> 3 #include<stdio.h> 4 #include<string.h> 5 #include<iostream> 6 #include<algorithm> 7 using namespace std; 8 #define N 123456 9 #define M 123410 11 int n,k;12 int a[N];13 int main()14 {15     int t;cin>>t;16     while(t--)17     {18         scanf("%d%d",&n,&k);19         int ha[N]={0};20         int j=0,c;21         for(int i=0;i<n;i++)22         {23             scanf("%d",&c);24             if(c>k || ha[c])continue;25             a[j++]=c;26             ha[c]=1;27         }28         int cnt=0;29         for(int i=0;i<j;i++)30             if(ha[ k-a[i] ])31                 cnt++;32 33         cout<<cnt<<endl;34     }35     return 0;36 }

 

HDU 2578 Dating with girls(1) [補7-26]