【Codeforces 459D】Pashmak and Parmida's problem

【链接】 我是链接,点我呀:)
【题意】

定义两个函数 f和g f(i)表示a[1..i]中等于a[i]的数字的个数

g(i)表示a[i..n]中等于a[i]的数字的个数
让你求出来(i,j) 这里i<j
的二元组个数
且f(i)>g(j)

【题解】

求出来两个数组g[N]和f[N]; (用map就行) 要算出(i< j)且f[i]>g[j]的个数 我们可以先把 g[1..n]全都加入到树状数组中。 然后顺序枚举i 遇到i就把g[i]从树状数组中删掉. 这样就只包括g[i+1..n]这些数字了 则,我们求一个1..f[i]-1的前缀和就好了 这样就是左端点为i满足题意的j的个数了

【代码】

import java.io.*;
import java.util.*;

public class Main {
    
    
    static InputReader in;
    static PrintWriter out;
        
    public static void main(String[] args) throws IOException{
        //InputStream ins = new FileInputStream("E:\\rush.txt");
        InputStream ins = System.in;
        in = new InputReader(ins);
        out = new PrintWriter(System.out);
        //code start from here
        new Task().solve(in, out);
        out.close();
    }
    static int N = (int)1e6;
    
    static class BIT{
		int a[];
		BIT(){
			a = new int[N+10];
		}
		
		public int lowbit(int x) {
			return x&(-x);
		}
		
		public void Add(int x,int y) {
			while (x<=N) {
				a[x] = a[x]+y;
				x = x + lowbit(x);
			}
		}
		
		public long sum(int x) {
			long temp = 0;
			while (x>=1) {
				temp = temp + a[x];
				x = x - lowbit(x);
			}
			return temp;
		}
	}
    
    static class Task{
    	
        int n;
        int a[],b[],c[];
        HashMap<Integer,Integer> pre,aft;
        
        public void solve(InputReader in,PrintWriter out) {
        	a = new int[N+10];b = new int[N+10];c = new int[N+10];
        	n = in.nextInt();
        	pre = new HashMap<Integer,Integer>();
        	aft = new HashMap<Integer,Integer>();
        	for (int i = 1;i <= n;i++) a[i] = in.nextInt();
        	for (int i = 1;i <= n;i++) {
        		int cnt;
        		if (pre.containsKey(a[i])) {
        			cnt = pre.get(a[i]);
        			cnt++;
        		}else {
        			cnt = 1;
        		}
        		pre.put(a[i], cnt);
        		b[i] = cnt;
        	}
        	
        	for (int i = n;i >= 1;i--) {
        		int cnt;
        		if (aft.containsKey(a[i])) {
        			cnt = aft.get(a[i]);
        			cnt++;
        		}else {
        			cnt = 1;
        		}
        		aft.put(a[i], cnt);
        		c[i] = cnt;
        	}
        	
        	BIT mybit = new BIT();
        	for (int i = 1;i <= n;i++) mybit.Add(c[i], 1);
        	long ans = 0;
        	for (int i = 1;i <= n;i++) {
        		mybit.Add(c[i], -1);
        		ans = ans + mybit.sum(b[i]-1);
        	}
        	out.println(ans);
        	
        	
        	
        }
    }

    

    static class InputReader{
        public BufferedReader br;
        public StringTokenizer tokenizer;
        
        public InputReader(InputStream ins) {
            br = new BufferedReader(new InputStreamReader(ins));
            tokenizer = null;
        }
        
        public String next(){
            while (tokenizer==null || !tokenizer.hasMoreTokens()) {
                try {
                tokenizer = new StringTokenizer(br.readLine());
                }catch(IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}