二分查找(D. Pair of Topics)

http://codeforces.com/contest/1324/problem/D

题意:给出两组长度为n的数组ai,bi。问满足(i < j) ai + aj > bi + bj 有多少对?

解法:变形ai-bi + aj - bj > 0 ,记数组ci = ai-bi .可知该题就是统计ci+cj > 0 有多少对。

对数组ci排序,对于ci贡献 等于 在ci后面找 大于 -ci 有多少个数。因为有序,直接upper_bound()找。

时间复杂度为(nlogn)。

//#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd __gcd
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
//ll lcm(ll a , ll b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define cin(x) scanf("%lld" , &x);
using namespace std;
const int N = 1e7+9;
const int maxn = 2e5+9;
const double esp = 1e-6;
int a[maxn] , b[maxn] , c[maxn];
 
 
void solve(){
    int n ;
    cin >> n ;
    rep(i , 1 , n){
        cin >> a[i];
    }
    rep(i , 1 , n){
        cin >> b[i];
        c[i] = a[i] - b[i] ;
    }
    sort(c + 1 , c + 1 + n);
    int ans = 0 ;
    rep(i , 1 , n){
        int cnt = upper_bound(c + 1 + i , c + 1 + n , -c[i]) - c;
        ans += n - cnt + 1 ;
    }
    cout << ans << endl;
 
}
 
signed main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0); cout.tie(0);
    //int t ;
    //cin >> t ;
    //while(t--){
        solve();
    //}
}

  

posted @ 2020-03-13 01:39  无名菜鸟1  阅读(395)  评论(0编辑  收藏  举报