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(); //} }