[CF#365 (Div. 2) Mishka and Interesting sum] 线段树离线处理区间不同数

题目链接：[CF#365 (Div. 2) Mishka and Interesting sum]

题意描述：给定N个数a1, a2, ..., an ， M次查询。每次查询一个区间[l, r] 中出现偶数次的数字的异或和。

解题思路： 如果求出现奇数次数字的异或和， 那么利用异或的性质，直接一个前缀异或和就分分钟搞定。出现偶数次数字的异或和可以这么考虑：

首先呢，还是求出前缀异或和。

然后呢， 求出区间不同数的异或和，即区间中每个数字不管出现多少次， 都做一次处理，求出区间异或和。嘿嘿嘿。给[hdu3333 Turing Tree] 思路一样。详情请转到我的上一篇博客[hdu3333 Turing Tree] 线段树离线 处理区间不同数。

然后呢，求区间[L, R] 的答案就是区间中所有数字的异或和(可以用之前的前缀异或和求出) 再异或 区间不同数的异或和。 

#include <map>#include <cmath>#include <queue>#include <vector>#include <cstdio>#include <string>#include <cstring>#include <iomanip>#include <iostream>#include <algorithm>using namespace std;//#pragma comment(linker, "/STACK:1024000000,1024000000")#define FIN             freopen("input.txt","r",stdin)#define FOUT            freopen("output.txt","w",stdout)#define fst             first#define snd             second#define lson            l, mid, rt << 1#define rson            mid + 1, r, rt << 1 | 1typedef __int64  LL;//typedef long long LL;const int MAXN = 1000000 + 5;const int MAXM = 1000000 + 5;struct Node {    int L, R, ID;    Node () {}    Node (int L, int R, int ID) : L (L), R (R), ID (ID) {}    bool operator < (const Node& e) const {        if (R == e.R) return L < e.L;        return R < e.R;    }} inp[MAXM];int N, M;int A[MAXN], pre[MAXN], seg[MAXN << 2];int res[MAXM];map<int, int> pos;inline void pushUp (const int& rt) {    seg[rt] = seg[rt << 1] ^ seg[rt << 1 | 1];}void build (int l, int r, int rt) {    memset (seg, 0x00, sizeof (seg) );}LL query (int L, int R, int l, int r, int rt) {    if (L <= l && r <= R) {        return seg[rt];    }    int mid = (l + r) >> 1;    LL ret = 0LL;    if (L <= mid) ret ^= query (L, R, lson);    if (R > mid) ret ^= query (L, R, rson);    return ret;}void update (int p, int v, int l, int r, int rt) {    if (l == r) {        seg[rt] = v;        return;    }    int mid = (l + r) >> 1;    if (p <= mid) update (p, v, lson);    else update (p, v, rson);    pushUp (rt);}int main() {#ifndef ONLINE_JUDGE    FIN;#endif // ONLINE_JUDGE    int L, R, ID;    while (~scanf ("%d", &N) ) {        memset (pre, 0, sizeof (pre) );        for (int i = 1; i <= N; i++) {            scanf ("%d", &A[i]);            pre[i] = A[i] ^ pre[i - 1];        }        scanf ("%d", &M);        for (int i = 1; i <= M; i++) {            scanf ("%d %d", &inp[i].L, &inp[i].R);            inp[i].ID = i;        }        sort (inp + 1, inp + M + 1);        build (1, N, 1);        pos.clear();        for (int i = 1, lst = 1, p; i <= M; i++) {            L = inp[i].L, R = inp[i].R, ID = inp[i].ID;            while (lst <= R) {                p = pos[A[lst]];                if (p) {                    update (p, 0, 1, N, 1);                }                update (lst, A[lst], 1, N, 1);                pos[A[lst]] = lst;                lst ++;            }            res[ID] = pre[R] ^ pre[L - 1] ^ query (L, R, 1, N, 1);        }        for (int i = 1; i <= M; i++) {            printf ("%d\n", res[i]);        }    }    return 0;}