• 转换成一个数在(0,X + Y)的加减问题
• 考虑一种使用线段树处理的方法， 维护前缀最大值， 前缀最小值， 前缀和， 然后查询的时候先询问右区间是否会同时碰到上下界， 会的话左区间无用直接递归右区间， 否则的话递归左区间， 然后右区间只会碰到上边界或者下边界， 分两种情况讨论即可

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define M 500010
#define ls now << 1
#define rs now << 1 | 1
#define lson l, mid, now << 1
#define rson mid + 1, r, now << 1 | 1
#define ll long long
#define mmp make_pair
using namespace std;
int read() {int nm = 0, f = 1;char c = getchar();for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';return nm * f;
}
int n, q, x, y, z;
ll sum[M << 2], minn[M << 2], maxx[M << 2];void pushup(int now) {sum[now] = sum[ls] + sum[rs];minn[now] = min(minn[ls], sum[ls] + minn[rs]);maxx[now] = max(maxx[ls], sum[ls] + maxx[rs]);
}void modify(int l, int r, int now, int pl, int v) {if(l > pl || r < pl) return;if(l == r) {sum[now] = v;minn[now] = min(v, 0);maxx[now] = max(v, 0);return;}int mid = (l + r) >> 1;modify(lson, pl, v), modify(rson, pl, v);pushup(now);
}
ll s;ll merge(ll v, int now) {if(v + maxx[now] > s) return s - maxx[now] + sum[now];if(v + minn[now] < 0) return sum[now] - minn[now];return v + sum[now];
}ll query(int l, int r, int now, ll v) {if(l == r) return max(min(s, sum[now] + v), 0ll);int mid = (l + r) >> 1;if(maxx[rs] - minn[rs] > s) return query(rson, 0);else return merge(query(lson, v), rs);
}int main() {
//  freopen("stone1.in", "r", stdin);n = read(), q = read();x = read(), y = read();for(int i = 1; i <= n; i++) {z = read();if(i % 2 == 0) z = -z;modify(1, n, 1, i, z);}while(q--) {int op = read();if(op == 1) x = read();else if(op == 2) y = read();else {int pl = read(), v = read();if(pl % 2 == 0) v = -v;modify(1, n, 1, pl, v);}s = x + y;cout << query(1, n, 1, x) << "\n";}return 0;
}