Codeforces Round #340 (Div. 2) E. XOR and Favorite Number（分块 （java））

题意：

n个长度的序列， m个询问，一个询问[L, R]， 求[L, R]又多少个子区间的xor和为k

 n, m and k (1 ≤ n, m ≤ 100 000, 0 ≤ k ≤ 1 000 000)

思路：

对于原序列的前缀和序列，即询问[L, R] 中有多少对(l, r) Al ^ Ar = k

所以分块容易解决

代码：

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



public class Main {  
	  
    public static void main(String[] args) {  
        InputStream inputStream = System.in;  
        OutputStream outputStream = System.out;  
        Scanner in = new Scanner(inputStream);  
        PrintWriter out = new PrintWriter(outputStream);  
        TaskE solver = new TaskE();  
        solver.solve(1, in, out);  
        out.close();  
    }  
    
    static class MyComprator implements Comparator {
    	public int compare(Object arg0, Object arg1) {
    		node a = (node)arg0;
    		node b = (node)arg1;
    		if(a.id != b.id) return a.id - b.id;
    		return a.Y - b.Y;
    	}
    }
    
    static node []op;
    static int []cnt = new int[3000000];
    static int []a;
    static int block = (int) Math.sqrt(100000);
    static long []res;
    static class TaskE {  
        public void solve(int testNumber, Scanner in, PrintWriter out) {  
        	int n = in.nextInt();
        	int m = in.nextInt();
        	int K = in.nextInt();
        	a = new int[n + 1];
        	op = new node[m];
        	res = new long[m];
        	for(int i = 0; i < m; i ++) op[i] = new node(0, 0, 0);
        	for(int i = 1; i <= n; i ++) a[i] = in.nextInt();
        	for(int i = 0; i < m; i ++) {
        		op[i].X = in.nextInt();
        		op[i].X --;
        		op[i].Y = in.nextInt();
        		op[i].getID();
        		op[i].ID = i;
        	}
        	Arrays.sort(op, new MyComprator());
        	int curl = 0, curr = 0;
        	int L = 0, R = -1;
        	long ans = 0;

        	for(int i = 0; i < m; i ++) {
        		while(L < op[i].X) {
        			curl ^= a[L];
        			cnt[curl] --;
        			ans -= cnt[curl ^ K];
        			L ++;
        		}
        		while(L > op[i].X) {
        			L --;
        			ans += cnt[curl ^ K];
        			cnt[curl] ++;
        			curl ^= a[L];
        		}
        		while(R > op[i].Y) {
        			cnt[curr] --;
        			ans -= cnt[curr ^ K];
        			curr ^= a[R];
        			R --;
        		}
        		while(R < op[i].Y) {
        			R ++;
        			curr ^= a[R];         
        			ans += cnt[curr ^ K];
        			cnt[curr] ++;
        		}
        		res[op[i].ID] = ans;
        	}
        	
        	for(int i = 0; i < m; i ++) out.println(res[i]);
        }  
    }  
    static class node {
    	int X, Y;
    	int id, ID;
    	node(int X, int Y, int Z) {
    		this.X = X;
    		this.Y = Y;
    		this.id = Z;
    	}
    	public void getID() {
    		this.id = X / block;
    	}
    }
    static class pii implements Comparable<pii> {  
        int X, Y;  
        pii(int X, int Y) {  
            this.X = X;  
            this.Y = Y;  
        }  
        public int compareTo(pii a) {  
            if(this.X - a.X != 0) return this.X - a.X;  
            else return this.Y - a.Y;  
        }  
    }  
    static class Scanner {    
        BufferedReader br;    
        StringTokenizer st;    
                
        public Scanner(InputStream in) {  
            br = new BufferedReader(new InputStreamReader(in));  
            eat("");  
        }   
          
        private void eat(String s) {    
            st = new StringTokenizer(s);  
        }    
        
        public String nextLine() {    
            try {    
                return br.readLine();    
            } catch (IOException e) {    
                return null;    
            }    
        }    
        
        public boolean hasNext() {    
            while (!st.hasMoreTokens()) {    
                String s = nextLine();    
                if (s == null)    
                return false;    
                eat(s);    
            }    
            return true;    
        }    
        
        public String next() {    
            hasNext();    
            return st.nextToken();    
        }    
        
        public int nextInt() {    
            return Integer.parseInt(next());    
        }    
            
        public long nextLong() {    
            return Long.parseLong(next());    
        }    
                
        public double nextDouble() {    
            return Double.parseDouble(next());    
        }    
            
        public BigInteger nextBigInteger() {    
            return new BigInteger(next());    
        }    
            
    }    
}