P1102 A-B 数对

P1102 A-B 数对

#include<map>
#include<iostream>
#include<vector>
using namespace std;
map<int,int> m;
vector <int> ma;
long long ans=0;
int main()
{
	int N,C;
	cin>>N>>C;
	for (int i=0;i<N;i++)
	{
		int k;
		cin>>k;
		if (m[k]==0)
		{
			ma.push_back(k);
		}
		m[k]++;
	}
	for (int i=0;i<ma.size();i++)
	{
		ans+=(long long)m[ma[i]]*m[ma[i]+C];
	}
	cout<<ans<<endl;
	return 0;
}

方法二

//P2421 A-B数对（增强版）
#include<iostream>
#include<algorithm>
using namespace std;
int a[100001];
int main()
{
	int n,c;
	cin>>n>>c;
	for (int i=0;i<n;i++)
	{
		cin>>a[i];
	}
	long long ans=0;
	for (int i=0;i<n;i++)
	{
		long long s1=lower_bound(a,a+n,a[i]+c)-a;
		long long s2=upper_bound(a,a+n,a[i]+c)-a;
		ans=ans+s2-s1;
	}
	cout<<ans<<endl;
	
}

标签：std,main,P1102,int,数对,long,include
来源： https://www.cnblogs.com/smghj/p/15962844.html