标签：force 有一个 ble for arrays fine file rop 数据
任意门：http://codeforces.com/problemset/problem/617/E
Bob 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 li and ri and asks you to count the number of pairs of integers i and j, such that l?≤?i?≤?j?≤?r and the xor of the numbers ai,?ai?+?1,?...,?aj is equal to k.
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.
Print m lines, answer the queries in the order they appear in the input.
6 2 3
1 2 1 1 0 3
1 6
3 5
7
0
5 3 1
1 1 1 1 1
1 5
2 4
1 3
9
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.
有一串长度为 N 的数列，M次查询 （ l，r ）内有多少对 （ i， j ）使得 ai ^ ai+1 ^ ... ^ aj = K;
这里的区间查询是离线的，可以用传说中的莫队算法（优雅而华丽的暴力算法）。
异或的题目有一个很巧妙的地方就是利用 a^b^a = b 这个性质。
这道题目我们也要预处理一下 sum(i) 前 i 个数的异或和， 那么 ai ^ ai+1 ^ ... ^ aj = sum( i - 1) ^ sum( j ) 了。
接下来我们就可以按照查询的区间用莫队算法遍历一遍，同时记录当前区间某个前缀和的出现次数；
根据 sum( i - 1) ^ sum( j ) = K ，K ^ sum( i - 1 ) = sum( j ) || K ^ sum( j ) = sum( i - 1) ，由符合条件的前缀和次数来推出符合条件的（ i，j ）的个数啦。
听了大神的课，这道题有两个坑：
1、答案的数据的数据范围是爆 int 的；
2、虽然是1e6 的 K，但异或的结果要大于 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; //区间和查询编号 9 }Q[MAXN]; //记录查询数据（离线） 10 11 int pos[MAXN]; //记录分块 12 LL ans[MAXN]; //记录答案 13 LL flag[MAXN]; //维护前缀异或和出现的次数 14 int a[MAXN]; //原本数据 15 int L=1, R; //当前区间的左右结点 16 LL res; //储存当前区间的值 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; //只有左结点在同一分块才排右结点 21 return pos[a.l] < pos[b.l]; //否则按照左结点分块排 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++){ //读入数据 38 scanf("%d", &a[i]); 39 a[i] = a[i]^a[i-1]; //计算前缀异或和 40 pos[i] = i/sz; //分块 41 } 42 for(int i = 1; i <= M; i++){ //读入查询 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){ //当前左结点比查询结点小 50 del(L-1); 51 L++; 52 } 53 while(L > Q[i].l){ //当前左结点比查询左结点大 54 L--; 55 add(L-1); 56 } 57 while(R < Q[i].r){ //当前右结点比查询右结点小 58 R++; 59 add(R); 60 } 61 while(R > Q[i].r){ //当前右节点比查询右节点大 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 }
Codeforces Round #340 (Div. 2) E. XOR and Favorite Number 【莫队算法 + 异或和前缀和的巧妙】
标签：force 有一个 ble for arrays fine file rop 数据
原文地址：https://www.cnblogs.com/ymzjj/p/9563453.html
