序列只包含1和2，q次操作，两种类型：（1）询问是否存在区间和为s（2）使a[x] = y

题目

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int a[maxn];
void solve(){
	int n, q, i, j;
	cin >> n >> q;
	int sum = 0;
	set<int> S;//1的下标集合
	for(i = 1; i <= n; i++){
		cin >> a[i];
		if(a[i] == 1) S.insert(i);
		sum += a[i];
	}
	while(q--){
		int op, x, y;
		cin >> op;
		if(op == 1){
			cin >> x;
			int lst = 0, rst = 0, cnt, maxx = 0;
			if(S.size()){
				lst = *S.begin(), rst = *S.rbegin();//找最左端的1和最右端的1，看哪个离端点最近，该1的位置到端点的区间和为maxx，那么[1, maxx]的询问都能满足
			  cnt = min(lst - 1, n - rst);
			  maxx = sum - 2 * cnt;
			} 
			if(x <= maxx) cout << "Yes\n";
			else{
				if((x + sum) % 2 == 0 && x <= sum){//大于maxx的话，得小于等于sum且与sum奇偶性相同（只能-2，-2...）
					cout << "Yes\n";
				}
				else cout << "No\n";
			}
		}
		else{
			cin >> x >> y;
			if(a[x] != y){
				a[x] = y;
				if(y == 1){
					S.insert(x);
					sum--;
				}
				else{
					S.erase(x);
					sum++;
				}
			}
		}
	}
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int T;
	cin >> T;
	while(T--){
		solve();
	}
	return 0;
}