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!
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.
For each cases,output the numbers of solutions to the equation.
2
5 4
1 2 3 4 5
8 8
1 4 5 7 8 9 2 6
lcy | We have carefully selected several similar problems for you:
2575 2579 2573 2576 2574 要去除重复的元素和大于k的元素，因为大于k的元素是肯定组不成等于k的方程的，不能确定数据里是否有0，所以等于k的元素也保留着，每存一个数都hash一下，然后一遍for循环，看 hash[ k-a[i] ]是否为1，是的话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 1234
10
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 }
