Luogu3870 [TJOI2009]开关 （分块）

线段树做法很简单，但分块好啊

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long

//#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;
#define D_e(x)  ;
#define Pause() ;
#define FileOpen() ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;

const int N = 100007;

int block[N], blockSize;
int a[N], ans[N], tag[N];
inline void Change(int l, int r){
	int minn = Min(r, block[l] * blockSize);
	R(i,l,minn){
		ans[block[i]] -= a[i] ^ tag[block[i]];
		a[i] ^= 1;
		ans[block[i]] += a[i] ^ tag[block[i]];
	}
	if(block[l] != block[r]){
		R(i, (block[r] - 1) * blockSize + 1, r){
			ans[block[i]] -= a[i] ^ tag[block[i]];
			a[i] ^= 1;
			ans[block[i]] += a[i] ^ tag[block[i]];
		}
	}
	R(i,block[l] + 1, block[r] - 1){
		ans[i] =  blockSize - ans[i]; // md, wo ge sa bi
		tag[i] ^= 1;
	}
}
inline int Query(int l, int r){
	int sum = 0;
	int minn = Min(r, block[l] * blockSize);
	R(i,l,minn){
		sum += a[i] ^ tag[block[i]];
	}
	if(block[l] != block[r]){
		R(i, (block[r] - 1) * blockSize + 1, r){
			sum += a[i] ^ tag[block[i]];
		}
	}
	R(i,block[l] + 1, block[r] - 1){
		sum += ans[i]; // I'm so sb
	}
	return sum;	
}

int main(){
	int n, m;
    io >> n >> m;
    blockSize = sqrt(n); 
    R(i,1,n){
    	block[i] = (i - 1) / blockSize + 1;
    }
    R(i,1,m){
    	int opt, l, r;
    	io >> opt >> l >> r;
    	if(opt == 0){
    		Change(l, r);
    	}
    	else{
    		printf("%d\n", Query(l, r));
    	} 
    }
    
    return 0;
}