HDU 2578 Dating with girls (1)

2 have done, put 1 to fill.

Test instructions gives N and K, and the number of N. is the n number to find the two number x, Y. Make x+y=k.

Calculate all the possibilities. When X0+y0=k. X1+y1=k. Must have unequal. X0!=x1 or Y0!=y1.

For example

4 4

2 2 2 2

The correct output should be 1.

I was sort of, went heavy, and then two points. It seems that many people are using the pointers.

Forget the STL's two-point return subscript function, the mood is not good, too lazy to check, hand hit two points.

#include <cstdio> #include <cstring> #include <string> #include <queue> #include <algorithm > #include <map> #include <stack> #include <iostream> #include <list> #include <set># include<bitset> #include <vector> #include <cmath> #define INF 0x7fffffff#define EPS 1e-8#define LL Long Long#define PI 3.141592654#define CLR (A, B) memset (A,b,sizeof (a)) #define for (I,A,B) for (int i=a;i<b;i++) #define For_ (I,A,B) for (int. i=a;i>=b;i--) #define SF scanf#define PF Printf#define All (v) (v). Begin (), (v). End () #define Acfun Std::ios::sync_with_stdio (False) #define SIZE (100000 +2) #define MOD 1000000007using namespace std;int binsearch (int *a,        int L,int R,int key) {while (l<=r) {int m= (L+R) >>1;        if (A[m]==key) return m;        else if (A[m]>key) r=m-1;    else l=m+1; } return-1;}    int main () {int t;    SF ("%d", &t);        while (t--) {int n,m; SF ("%d%d", &n,&AMP;M);        int a[size];        for (I,0,n) SF ("%d", &a[i]);        Sort (a,a+n);        int cnt =unique (a,a+n)-A;        int ans=0;        for (i,0,cnt) {if (A[i]<=m&&binsearch (a,0,cnt-1,m-a[i))!=-1) ans++;    } PF ("%d\n", ans); }}

HDU 2578 Dating with girls (1)