题意：
有一个1e6的数组，t次操作：将[l,r]内的值增加w，或者查询[l,r]内的值大于等于add的
思路：
分块，块大小为sqrt(n)，每次只需要暴力头尾两块，中间的整块打标记，
对于查询查操作，块内排序然后二分即可
复杂度O(T(sqrt(n)+sqrt(n)logn))
代码：
可以弄两个数组对应着写，会比我这样stl乱写快一倍
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map> #define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
//#define lowbit(x) ((x)&(-x)) using namespace std; typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL; const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 1e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f; //const db pi = acos(-1.0); struct node{
int id;
ll v;
};
int n, q;
vector<node>blc[maxn];
int block;
bool cmp(node a, node b){
return a.v<b.v;
}
ll add[maxn];
int main(){
scanf("%d %d", &n, &q);
int block = (int)sqrt(n);
for(int i = ; i < n; i++){
node tmp;
tmp.id = i;
scanf("%lld", &tmp.v);
blc[i/block].pb(tmp);
}
for(int i = ; i <= (n-)/block; i++){
sort(blc[i].begin(), blc[i].end(), cmp);
}
while(q--){
char op[];
int x, y;
ll w;
scanf("%s",op+);
scanf("%d %d %lld", &x, &y, &w);
x--;y--;
int bx = x/block;
int by = y/block;
if(op[]=='M'){
if(bx==by){
for(int i = ; i < (int)blc[bx].size(); i++){
node a = blc[bx][i];
if(a.id>=x&&a.id<=y)blc[bx][i].v+=w;
}
sort(blc[bx].begin(), blc[bx].end(),cmp);
}
else{
int rx = min(n-, (bx+)*block-);
int ly = by*block;
for(int i = ; i < (int)blc[bx].size(); i++){
int id = blc[bx][i].id;
if(id>=x&&id<=y)blc[bx][i].v+=w;
}sort(blc[bx].begin(), blc[bx].end(),cmp);
for(int i = ; i < (int)blc[by].size(); i++){
int id = blc[by][i].id;
if(id>=x&&id<=y)blc[by][i].v+=w;
}sort(blc[by].begin(), blc[by].end(),cmp);
for(int i = bx+; i < by; i++){
add[i]+=w;
}
}
}
else{
int ans = ;
if(bx==by){
for(int i = ; i < (int)blc[bx].size(); i++){
int id = blc[bx][i].id;
if(id>=x&&id<=y&&blc[bx][i].v+add[bx]>=w)ans++;
}
}
else{
for(int i = ; i < (int)blc[bx].size(); i++){
int id = blc[bx][i].id;
if(id>=x&&id<=y&&blc[bx][i].v+add[bx]>=w)ans++;
}
for(int i = ; i < (int)blc[by].size(); i++){
int id = blc[by][i].id;
if(id>=x&&id<=y&&blc[by][i].v+add[by]>=w)ans++;
}
//printf("--%d\n",ans);
for(int i = bx+; i < by; i++){
ll x = w-add[i];
int l = ;
int r = block-;
int pos = block;
while(l<=r){
int mid = (l+r)>>;
if(blc[i][mid].v>=x){
pos = mid;
r = mid-;
}
else l = mid+;
}
ans+=block-pos;
}
}
printf("%d\n",ans);
}
}
return ;
}
/*
5 3
1 2 3 4 5
A 1 5 4
M 3 5 1
A 1 5 4 5 7
3 1 4 2 5
A 1 5 4
M 3 5 1
M 3 5 2
A 1 5 4
M 1 1 3
A 1 5 4
A 1 5 10 5 3
3 1 4 2 5
A 1 5 10
M 1 5 2
A 1 5 10
*/