luoguP2574 XOR的艺术

思路

01串的区间求和，区间翻转 lazy%2==0 则不用翻转，lazt%2==1则要翻转
模板题

代码

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define ls rt<<1
#define rs rt<<1|1
#define ll long long
#define FOR(i,a,b) for(int i=a;i<=b;++i)
using namespace std;
const int maxn = 2e5 + 7;
int read() {
    int x = 0, f = 1; char s = getchar();
    for (; s < '0' || s > '9'; s = getchar()) if (s == '-') f = -1;
    for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
    return x * f;
}
int n, m,a[maxn];
struct node{
    int l,r,size,sum,lazy;
}e[maxn<<2];
void pushup(int rt) {
    e[rt].sum=e[ls].sum+e[rs].sum;
}
void pushdown(int rt) {
    if(e[rt].lazy%2) {
        e[ls].lazy++;
        e[rs].lazy++;
        e[ls].sum=e[ls].size-e[ls].sum;
        e[rs].sum=e[rs].size-e[rs].sum;
        e[rt].lazy=0;
    }
}
void build(int l,int r,int rt) {
    e[rt].l=l,e[rt].r=r,e[rt].size=r-l+1;
    if(l==r) {
        e[rt].sum=a[l];
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,ls);
    build(mid+1,r,rs);
    pushup(rt);
}
void modfity(int L,int R,int rt) {
    if(L<=e[rt].l&&e[rt].r<=R) {
        e[rt].sum=e[rt].size-e[rt].sum;
        e[rt].lazy++;
        return;
    }
    pushdown(rt);
    int mid=(e[rt].l+e[rt].r)>>1;
    if(L<=mid) modfity(L,R,ls);
    if(R>mid) modfity(L,R,rs);
    pushup(rt);
}
int query(int L,int R,int rt) {
    if(L<=e[rt].l&&e[rt].r<=R) {
        return e[rt].sum;
    }
    pushdown(rt);
    int mid=(e[rt].l+e[rt].r)>>1,ans=0;
    if(L<=mid) ans+=query(L,R,ls);
    if(R>mid) ans+=query(L,R,rs);
    return ans;
}
int main() {
    n=read(),m=read();
    FOR(i,1,n) scanf("%1d",&a[i]);
    build(1,n,1);
    FOR(i,1,m) {
        int p=read(),x=read(),y=read();
        if(p) {
            printf("%d\n",query(x,y,1));
        } else {
            modfity(x,y,1);
        }
    }
    return 0;
}