题目链接：https://codeforces.com/contest/1439/problem/C

题目大意：

给定一个单调不升的序列
操作一：输入 \(x,y\), 将 \([1,x]\) 内的所有元素变成 \(max[a_i,y]\)
操作二：输入 \(x,y\), 从 \(a_x\) 开始一直到 \(a_n\)，如果 \(y\) 大于 \(a_i\)，则减去 \(a_i\), 问一共减了几次

题解：

注意到序列元素一直是单调不升的
第一个操作，找到左边最远的比 \(y\) 小的位置，然后区间覆盖即可
第二个操作，在线段树上二分，找到比剩余钱数少的位置，具体实现参考代码

时间复杂度 \(O(nlogn)\)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 200010;

int n, q;
int a[maxn];

struct Node{
	int mi, tag;
	ll sum;
}t[maxn << 2];

inline void pushup(int i){ 
	t[i].sum = t[i << 1].sum + t[i << 1 | 1].sum;
	t[i].mi = min(t[i << 1].mi, t[i << 1 | 1].mi); 
}

inline void build(int i, int l, int r){
	if(l == r) {
		t[i].mi = t[i].sum = a[l];
		return;
	}
	int mid = (l + r) >> 1;
	build(i << 1, l, mid);
	build(i << 1 | 1, mid + 1, r);
	pushup(i);
}

inline void pushdown(int i, int l, int r){
	if(t[i].tag){
		t[i << 1].mi = t[i << 1 | 1].mi = t[i].tag;
		int mid = (l + r) >> 1;
		t[i << 1].sum = 1ll * (mid - l + 1) * t[i].tag;
		t[i << 1 | 1].sum = 1ll * (r - mid) * t[i].tag;
		
		t[i << 1].tag = t[i << 1 | 1].tag = t[i].tag;
		t[i].tag = 0;
	}
}

inline void modify(int i, int k, int l, int r, int x, int y){
	if(x <= l && r <= y){
		t[i].mi = k;
		t[i].sum = 1ll * (r - l + 1) * k;
		t[i].tag = k;
		return;
	}
	
	pushdown(i, l, r);
	
	int mid = (l + r) >> 1;
	if(x <= mid) modify(i << 1, k, l, mid, x, y);
	if(y > mid) modify(i << 1 | 1, k, mid + 1, r, x, y);
	pushup(i);
}

inline ll query_sum(int i, int l, int r, int x, int y){
	if(x <= l && r <= y){
		return t[i].sum;
	}
	pushdown(i, l, r);
	int mid = (l + r) >> 1;
	ll res = 0;
	if(x <= mid) res += query_sum(i << 1, l, mid, x, y);
	if(y > mid) res += query_sum(i << 1 | 1, mid + 1, r, x, y);
	return res;
}

inline int find(int i, int l, int r, int k, int p){
	if(l == r){
		return l;
	}
	
	pushdown(i, l, r);
	
	int mid = (l + r) >> 1;
	if(t[i << 1].mi < k) return find(i << 1, l, mid, k, p);
	else if(p >= mid + 1) return find(i << 1 | 1, mid + 1, r, k, p); 
	else return -1;
}

int solve(int i, ll &k, int l, int r, int p){
	if(t[i].sum <= k && p <= l){
		k -= t[i].sum;
		return r - l + 1;
	}
	if(l == r) return 0;
	pushdown(i, l, r);
	int mid = (l + r) >> 1;
	int res = 0;
	if(t[i << 1].mi <= k && p <= mid) res += solve(i << 1, k, l, mid, p);
	if(t[i << 1 | 1].mi <= k) res += solve(i << 1 | 1, k, mid + 1, r, p);
	return res;
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	n = read(), q = read();
	for(int i = 1 ; i <= n ; ++i) a[i] = read();
	
	build(1, 1, n);
	
	int t, x; ll y;
	for(int i = 1 ; i <= q ; ++i){
		t = read(), x = read(), y = read();
		if(t == 1){
			int pos = find(1, 1, n, y, x);
			if(pos == -1) continue;
			modify(1, y, 1, n, pos, x);
		} else{
			printf("%d\n", solve(1, y, 1, n, x));
		}
	}
	return 0;
}