Luogu2801 教主的魔法 （分块）

与hzw的分块2类似，放vector排序

#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;
#include <vector>

const int N = 1000007;

int n;

int block[N], blockSize;
int a[N], tag[N];
vector<int> V[1007];

inline void Reset(int x){
	V[x].clear();
	R(i,(x - 1) * blockSize + 1, block[x] * blockSize){
		V[x].push_back(a[i]);
	}
	sort(V[x].begin(), V[x].end());
}
inline void Updata(int L, int R, int w){
	int minn = Min(R, block[L] * blockSize);
	R(i,L,minn){
		a[i] += w;
	}
	Reset(block[L]);
	if(block[L] != block[R]){
		R(i, (block[R] - 1) * blockSize + 1, R){
			a[i] += w;
		}
		Reset(block[R]);
	}
	R(i,block[L] + 1, block[R] - 1){
		tag[i] += w;
	}
}
inline int Query(int L, int R, int w){
	int sum = 0;
	int minn = Min(R, block[L] * blockSize);
	R(i,L,minn){
		sum += ((a[i] + tag[block[i]]) >= w);
	}
	if(block[L] != block[R]){
		R(i, (block[R] - 1) * blockSize + 1, R){
			sum += ((a[i] + tag[block[i]]) >= w);
		}
	}
	R(i,block[L] + 1, block[R] - 1){
		sum += blockSize - (lower_bound(V[i].begin(), V[i].end(), w - tag[i]) - V[i].begin());
	}

	return sum;
}

int main(){
	int m;
	io >> n >> m;
	blockSize = sqrt(n);
	R(i,1,n){
		io >> a[i];
		block[i] = (i - 1) / blockSize + 1;
		V[block[i]].push_back(a[i]);
	}

	int cnt = n / blockSize + 1;
	R(i,1,cnt)
		sort(V[i].begin(), V[i].end());

	while(m--){
		char opt;
		int L, R, w;
		cin >> opt;
		io >> L >> R >> w;
		if(opt == 'M'){
			Updata(L, R, w);
		}
		else{
			printf("%d\n", Query(L, R, w));
		}
	}

	return 0;
}