Codeforces Round #340 (Div. 2) E. XOR and Favorite Number 【莫隊算法 + 異或和前綴和的巧妙】
任意門:http://codeforces.com/problemset/problem/617/E
E. XOR and Favorite Number
time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard outputBob has a favorite number k and ai of length n. Now he asks you to answer m queries. Each query is given by a pair l
Input
The first line of the input contains integers n, m and k (1?≤?n,?m?≤?100?000, 0?≤?k?≤?1?000?000) — the length of the array, the number of queries and Bob‘s favorite number respectively.
The second line contains n integers ai (0?≤?ai?≤?1?000?000) — Bob‘s array.
Then m lines follow. The i-th line contains integers li and ri (1?≤?li?≤?ri?≤?n) — the parameters of the i-th query.
Output
Print m lines, answer the queries in the order they appear in the input.
Examples
input6 2 3
1 2 1 1 0 3
1 6
3 5
7input
0
5 3 1output
1 1 1 1 1
1 5
2 4
1 3
9Note
4
4
In the first sample the suitable pairs of i and j for the first query are: (1, 2), (1, 4), (1, 5), (2, 3), (3, 6), (5, 6), (6, 6). Not a single of these pairs is suitable for the second query.
In the second sample xor equals 1 for all subarrays of an odd length.
題目大意:
有一串長度為 N 的數列,M次查詢 ( l,r )內有多少對 ( i, j )使得 ai ^ ai+1 ^ ... ^ aj = K;
解題思路:
這裏的區間查詢是離線的,可以用傳說中的莫隊算法(優雅而華麗的暴力算法)。
異或的題目有一個很巧妙的地方就是利用 a^b^a = b 這個性質。
這道題目我們也要預處理一下 sum(i) 前 i 個數的異或和, 那麽 ai ^ ai+1 ^ ... ^ aj = sum( i - 1) ^ sum( j ) 了。
接下來我們就可以按照查詢的區間用莫隊算法遍歷一遍,同時記錄當前區間某個前綴和的出現次數;
根據 sum( i - 1) ^ sum( j ) = K ,K ^ sum( i - 1 ) = sum( j ) || K ^ sum( j ) = sum( i - 1) ,由符合條件的前綴和次數來推出符合條件的( i,j )的個數啦。
Tip:
聽了大神的課,這道題有兩個坑:
1、答案的數據的數據範圍是爆 int 的;
2、雖然是1e6 的 K,但異或的結果要大於 1e6 ;
AC code:
1 #include <bits/stdc++.h> 2 #define LL long long int 3 using namespace std; 4 const int MAXN = 1<<20; 5 6 struct node 7 { 8 int l, r, id; //區間和查詢編號 9 }Q[MAXN]; //記錄查詢數據(離線) 10 11 int pos[MAXN]; //記錄分塊 12 LL ans[MAXN]; //記錄答案 13 LL flag[MAXN]; //維護前綴異或和出現的次數 14 int a[MAXN]; //原本數據 15 int L=1, R; //當前區間的左右結點 16 LL res; //儲存當前區間的值 17 int N, M, K; 18 bool cmp(node a, node b) //排序 19 { 20 if(pos[a.l]==pos[b.l]) return a.r < b.r; //只有左結點在同一分塊才排右結點 21 return pos[a.l] < pos[b.l]; //否則按照左結點分塊排 22 } 23 void add(int x) 24 { 25 res+=flag[a[x]^K]; 26 flag[a[x]]++; 27 } 28 void del(int x) 29 { 30 flag[a[x]]--; 31 res-=flag[a[x]^K]; 32 } 33 int main() 34 { 35 scanf("%d%d%d", &N, &M, &K); 36 int sz = sqrt(N); 37 for(int i = 1; i <= N; i++){ //讀入數據 38 scanf("%d", &a[i]); 39 a[i] = a[i]^a[i-1]; //計算前綴異或和 40 pos[i] = i/sz; //分塊 41 } 42 for(int i = 1; i <= M; i++){ //讀入查詢 43 scanf("%d%d", &Q[i].l, &Q[i].r); 44 Q[i].id = i; 45 } 46 sort(Q+1, Q+1+M, cmp); 47 flag[0] = 1; 48 for(int i = 1; i <= M; i++){ 49 while(L < Q[i].l){ //當前左結點比查詢結點小 50 del(L-1); 51 L++; 52 } 53 while(L > Q[i].l){ //當前左結點比查詢左結點大 54 L--; 55 add(L-1); 56 } 57 while(R < Q[i].r){ //當前右結點比查詢右結點小 58 R++; 59 add(R); 60 } 61 while(R > Q[i].r){ //當前右節點比查詢右節點大 62 del(R); 63 R--; 64 } 65 ans[Q[i].id] = res; 66 } 67 for(int i = 1; i <= M; i++){ 68 printf("%lld\n", ans[i]); 69 } 70 }View Code
Codeforces Round #340 (Div. 2) E. XOR and Favorite Number 【莫隊算法 + 異或和前綴和的巧妙】