cf242 E

题意:

$n$ 个数 $a_i$, 

两种询问

$1, l, r$ 查询 $[l, r]$ 的和

$2, l, r, x$ 将区间 $[l, r]$ 所有数异或 $x$

建立 $30$ 课线段树

第 $i$ 颗线段树维护所有 $a$ 二进制的第 $i$ 为上的数字 $0, 1$

异或操作分别以 $x$ 的二进制相应位异或相应线段树

可见只有当 $x$ 的二进制位为 $1$ 是操作有效

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>

using namespace std;

#define LL long long

#define gc getchar()
inline int read() {int x = 0; char c = gc; while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x;}
inline LL read_LL() {LL x = 0; char c = gc; while(c < '0' || c > '9') c = gc;
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc; return x;}
#undef gc

const int N = 1e5 + 10;

int Size[N << 2];
int n, m, Ans;

#define lson jd << 1
#define rson jd << 1 | 1

struct Node {
    int W[N << 2], F[N << 2];
    
    void Push_down(int jd) {
        F[lson] ^= 1, F[rson] ^= 1;
        W[lson] = Size[lson] - W[lson];
        W[rson] = Size[rson] - W[rson];
        F[jd] = 0;
    }
    
    void Push_up(int jd) {
        W[jd] = W[lson] + W[rson];
    }
    
    void Sec_G(int l, int r, int jd, int x, int y) {
        if(x <= l && r <= y) {
            F[jd] ^= 1;
            W[jd] = Size[jd] - W[jd];
            return ;
        }
        if(F[jd]) Push_down(jd);
        int mid = (l + r) >> 1;
        if(x <= mid) Sec_G(l, mid, lson, x, y);
        if(y > mid ) Sec_G(mid + 1, r, rson, x, y);
        Push_up(jd);
    }
    
    void Sec_A(int l, int r, int jd, int x, int y) {
        if(x <= l && r <= y) {
            Ans += W[jd];
            return ;
        }
        if(F[jd]) Push_down(jd);
        int mid = (l + r) >> 1;
        if(x <= mid) Sec_A(l, mid, lson, x, y);
        if(y > mid)  Sec_A(mid + 1, r, rson, x, y);
    }
} Tree[35];

void Build_tree(int l, int r, int jd) {
    Size[jd] = r - l + 1;
    if(l == r) {
        int x = read();
        for(int i = 0; (1 << i) <= x; i ++) {
            Tree[i + 1].W[jd] = (bool) ((1 << i) & x);
        }
        return ;
    }
    int mid = (l + r) >> 1;
    Build_tree(l, mid, lson), Build_tree(mid + 1, r, rson);
    for(int i = 1; i <= 30; i ++) {
        Tree[i].W[jd] = Tree[i].W[lson] + Tree[i].W[rson];
    }
}

int main() {
    n = read();
    Build_tree(1, n, 1);
    m = read();
    for(; m; m --) {
        int opt = read(), l = read(), r = read();
        if(opt == 2) {
            int x = read();
            for(int i = 0; (1 << i) <= x; i ++) {
                if(((1 << i) & x)) {
                    Tree[i + 1].Sec_G(1, n, 1, l, r);
                }
            }
        } else {
            LL Answer = 0;
            for(int i = 1; i <= 30; i ++) {
                Ans = 0;
                Tree[i].Sec_A(1, n, 1, l, r);
                Answer += (1ll * Ans * (LL) pow(2, i - 1));
            }
            cout << Answer << "\n";
        }
    }
    
    return 0;
}

猜你喜欢

转载自www.cnblogs.com/shandongs1/p/9574811.html