Luogu2574 XOR的艺术 （分块）

本是要练线段树的，却手贱打了个分块

//#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("inn.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 = 200007;

int n;

int blockSize, block[N];

int ans[N], a[N], tag[N];

inline void Updata(int l, int r){

	int minn = Min(r, blockSize * block[l]);

	R(i,l,minn){

		a[i] ^= 1;

		ans[block[i]] += ((a[i] ^ tag[block[i]]) == 1) - ((a[i] ^ tag[block[i]]) == 0);

	}

	if(block[l] != block[r]){

		R(i,(block[r] - 1) * blockSize + 1, r){

			a[i] ^= 1;

			ans[block[i]] += ((a[i] ^ tag[block[i]]) == 1) - ((a[i] ^ tag[block[i]]) == 0);

		}

	}

	R(i,block[l] + 1, block[r] - 1){

		ans[i] = blockSize - ans[i];

		tag[i] ^= 1;

	}

}

inline int Query(int l, int r){

	int sum = 0;

	int minn = Min(r, blockSize * block[l]);

	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];

	}

	return sum;

}

int main(){

	int m;

	io >> n >> m;

	blockSize = sqrt(n);

	R(i,1,n){

		scanf("%1d", a + i);

		block[i] = (i - 1) / blockSize + 1;

		ans[block[i]] += (a[i] == 1);

	}

	while(m--){

		int opt, l ,r;

		io >> opt >> l >> r;

		if(opt == 0){

			Updata(l, r);

		}

		else{

			printf("%d\n", Query(l, r));

		}

	}

	return 0;

}

/*

10 6

1011101001

0 2 4

1 1 5

0 3 7

1 1 10

0 1 4

1 2 6

0 0 0 0 0 1 0 0 0 1

*/

Luogu2574 XOR的艺术 （分块）的相关教程结束。