P1102 A-B 数对的三种解法

1.
利用map实现速查，优点是代码简洁，缺点是速度慢，内存大

#include<bits/stdc++.h>
using namespace std;
int a[200005]={0};
int main()
{
    int n,c;
    scanf("%d%d",&n,&c);

    map<int,int> maps;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        maps[a[i]]++;
    }

    long long ans=0;
    for(int i=1;i<=n;i++)
    {
        ans+=maps[a[i]+c];
    }
    printf("%lld\n",ans);
    return 0;
}

2.
遍历数组,二分查找，总时间复杂度为O(nlogn)

#include<bits/stdc++.h>
using namespace std;
int a[200005]={0};
int main()
{
    int n,c;
    scanf("%d%d",&n,&c);

    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    long long ans=0;
    sort(a+1,a+n+1);
    for(int i=1;i<=n;i++)
    {
        int x=a[i]+c;
        int l=i,r=n+1;
        while(l<r-1)
        {
            int mid=(l+r)/2;
            if(a[mid]>=x)r=mid;
            else l=mid;
        }
        int left=r;
        l=i,r=n+1;
        while(l<r-1)
        {
            int mid=(l+r)/2;
            if(a[mid]<=x)l=mid;
            else r=mid;

        }
        int right=l;
        ans+=right-left+1;
        //printf("%d %d : %d\n",left,right,ans);
    }
    printf("%lld\n",ans);
    return 0;
}

3.
指针法，时间复杂度O(n),常数项为4，最优

#include<bits/stdc++.h>
using namespace std;
int a[200005]={0};
int main()
{
    int n,c;
    scanf("%d%d",&n,&c);

    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    long long ans=0;
    sort(a+1,a+n+1);

    long long l1=1,r1=1,l2=2,r2=2;

    while(l2<=n)
    {
        while(a[r1]==a[l1])r1++;
        while(l2<=n&&a[l2+1]<a[l1]+c)l2++;
        r2=l2+1;
        while(r2<=n&&a[r2]<=a[l1]+c)r2++;
        ans+=(r1-l1)*(r2-l2-1);
        //printf("%d:%d  ,  %d:%d  %lld\n",l1,r1,l2,r2,ans);
        l1=r1;
    }
    printf("%lld",ans);
    return 0;
}