The next lecture in a high school requires two topics to be discussed. The ii -th topic is interesting by aiai units for the teacher and by bibi units for the students.
The pair of topics ii and jj (i<ji<j ) is called good if ai+aj>bi+bjai+aj>bi+bj (i.e. it is more interesting for the teacher).
Your task is to find the number of good pairs of topics.
Input
The first line of the input contains one integer nn (2≤n≤2⋅1052≤n≤2⋅105 ) — the number of topics.
The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109 ), where aiai is the interestingness of the ii -th topic for the teacher.
The third line of the input contains nn integers b1,b2,…,bnb1,b2,…,bn (1≤bi≤1091≤bi≤109 ), where bibi is the interestingness of the ii -th topic for the students.
Output
Print one integer — the number of good pairs of topic.
Examples
Input
5 4 8 2 6 2 4 5 4 1 3
Output
7
Input
4 1 3 2 4 1 3 2 4
Output
0这种题套路就是移一下项转化为ai-bi，然后先对ai-bi=ci进行排序。若要ci+cj>0，这两个起码要有一个为0，所以从大到小扫一遍，当ci大于0时，用upper_bound找到第一个大于-ci的数，这样一定能保证这个区间里的数的和大于0，把长度累加到答案里就行。其实也可以双指针同时往中间移动，这么写啰嗦一点不过常数比较小....能过万岁
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
struct good
{
int a,b;
}g[200005];
int res[200005];
int n;
int main()
{
cin>>n;
int i;
for(i=1;i<=n;i++)scanf("%d",&g[i].a);
for(i=1;i<=n;i++)scanf("%d",&g[i].b);
for(i=1;i<=n;i++)res[i]=g[i].a-g[i].b;
sort(res+1,res+n+1);
long long ans=0;
for(i=n;i>=1;i--)
{
if(res[i<0])break;
int pos=upper_bound(res+1,res+i+1,-res[i])-res;
if(pos>=1&&pos<i)
{
ans+=i-pos;
}
}
cout<<ans;
}
原文地址：https://www.cnblogs.com/lipoicyclic/p/12590480.html