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
*/