标签：inf   stack   build   namespace   map   线段   const   代码   queue   

题意：

给出区间与、或、异或\(x\)操作，还有询问区间和。

思路：

因为数比较小，我们给每一位建线段树，这样每次只要更新对应位的答案。
与\(0\)和或\(1\)相当于重置区间，异或\(1\)相当于翻转区间，那么设出两个\(lazy\)搞一下。注意父区间\(pushdown\)重置标记时，子区间的翻转标记要清空。

代码：

#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 1e6 + 5;
const int MAXM = 3e6;
const ll MOD = 1e9 + 7;
const ull seed = 131;
const int INF = 0x3f3f3f3f;

#define lson (rt << 1)
#define rson (rt << 1 | 1)
int sum[5][maxn << 2];
int lazy[5][maxn << 2], rev[5][maxn << 2];
int a[maxn];
void pushup(int rt, int bit){
    sum[bit][rt] = sum[bit][lson] + sum[bit][rson];
}
void pushdown(int rt, int bit, int l, int r){
    int m = (l + r) >> 1;
    if(lazy[bit][rt] != -1){
        sum[bit][lson] = lazy[bit][rt] * (m - l + 1);
        sum[bit][rson] = lazy[bit][rt] * (r - m);
        lazy[bit][lson] = lazy[bit][rson] = lazy[bit][rt];
        rev[bit][lson] = rev[bit][rson] = 0;    //!!!!!
        lazy[bit][rt] = -1;
    }
    if(rev[bit][rt]){
        sum[bit][lson] = m - l + 1 - sum[bit][lson];
        sum[bit][rson] = r - m - sum[bit][rson];
        rev[bit][lson] ^= 1;
        rev[bit][rson] ^= 1;
        rev[bit][rt] = 0;
    }
}
void build(int l, int r, int bit, int rt){
    lazy[bit][rt] = -1;
    rev[bit][rt] = 0;
    if(l == r){
        sum[bit][rt] = (a[l] >> bit) & 1;
        return;
    }
    int m = (l + r) >> 1;
    build(l, m, bit, lson);
    build(m + 1, r, bit, rson);
    pushup(rt, bit);
}
void update(int L, int R, int l, int r, int op, int bit, int rt){
    if(L <= l && R >= r){
        if(op == 1){    //&0
            lazy[bit][rt] = 0;
            rev[bit][rt] = 0;
            sum[bit][rt] = 0;
        }
        else if(op == 2){   //|1
            lazy[bit][rt] = 1;
            rev[bit][rt] = 0;
            sum[bit][rt] = r - l + 1;
        }
        else{   //^1
            rev[bit][rt] ^= 1;
            sum[bit][rt] = r - l + 1 - sum[bit][rt];
        }
        return;
    }
    pushdown(rt, bit, l, r);
    int m = (l + r) >> 1;
    if(L <= m)
        update(L, R, l, m, op, bit, lson);
    if(R > m)
        update(L, R, m + 1, r, op, bit, rson);
    pushup(rt, bit);
}
int query(int L, int R, int l, int r, int bit, int rt){
    if(L <= l && R >= r){
        return sum[bit][rt];
    }
    pushdown(rt, bit, l, r);
    int m = (l + r) >> 1, ret = 0;
    if(L <= m)
        ret += query(L, R, l, m, bit, lson);
    if(R > m)
        ret += query(L, R, m + 1, r, bit, rson);
    return ret;
}
int main(){
    int n, m, T;
    scanf("%d", &T);
    while(T--){
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for(int i = 0; i < 5; i++) build(1, n, i, 1);
        while(m--){
            int op, l, r, x;
            char ope[10];
            scanf("%s", &ope);
            if(ope[0] == 'A') op = 1;
            else if(ope[0] == 'O') op = 2;
            else if(ope[0] == 'X') op = 3;
            else op = 4;
            if(op == 4){
                scanf("%d%d", &l, &r);
                l++, r++;
                ll ans = 0;
                for(int i = 0; i < 5; i++){
                    x = query(l, r, 1, n, i, 1);
                    ans += 1LL * (1LL << i) * x;
                }
                printf("%lld\n", ans);
            }
            else{
                scanf("%d%d%d", &x, &l, &r);
                l++, r++;
                if(op == 2 || op == 3){
                    for(int i = 0; i < 5; i++){
                        if((x >> i) & 1)
                            update(l, r, 1, n, op, i, 1);
                    }
                }
                else{
                    for(int i = 0; i < 5; i++){
                        if(!((x >> i) & 1))
                            update(l, r, 1, n, op, i, 1);
                    }
                }
            }
        }
    }
    return 0;
}

FZU2105 Digits Count（按位建线段树）题解

标签：inf   stack   build   namespace   map   线段   const   代码   queue   

原文地址：https://www.cnblogs.com/KirinSB/p/11489274.html