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;

}

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Luogu3870 [TJOI2009]开关（分块）

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巴特西

Luogu3870 [TJOI2009]开关 （分块）

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Luogu3870 [TJOI2009]开关（分块）