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("
----------
")
#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
", Query(L, R, w));
}
}
return 0;
}