Solution

这题的解法非常巧妙啊~
我们发现只有一个箱子中的气球全部被踩完后, 才可能对答案产生贡献, 因此用数组维护每个箱子中剩余的气球个数.
考虑每次可能对处于哪些区间的熊孩子产生影响: 我们设当前箱子为\(p\), 通过链表查找到最小的\(l\)满足\([l, p]\)区间的所有箱子都不含有气球, 以及最大的\(r\)满足\([p, r]\)区间中的所有箱子都不含有气球. 则会变高兴的熊孩子满足其对应区间的左端点在\([l, p]\)中且右端点在\([p, r]\)中. 我们用一颗可持久化线段树来维护每个位置\(p\)有哪些熊孩子满足\(l \le p\), 且将其对应的\(r\)存入线段树中. 每次操作后, 在线段树中容斥即可.

#include <cstdio>
#include <cctype>
#include <algorithm>namespace Zeonfai
{inline int getInt(){int a = 0, sgn = 1;char c;while(! isdigit(c = getchar())) if(c == '-') sgn *= -1;while(isdigit(c)) a = a * 10 + c - '0', c = getchar();return a * sgn;}
}
const int N = (int)1e5, M = (int)1e5;
int n, m;
struct bearChild
{int L, R;inline int friend operator <(bearChild a, bearChild b){return a.L == b.L ? a.R < b.R : a.L < b.L;}
}chd[M];
struct segmentTrees
{struct node{node *suc[2];int cnt;inline node() {cnt = 0; suc[0] = suc[1] = NULL;}};node *rt[N + 1];node* build(int L, int R){node *u = new node;if(L == R) return u;u->suc[0] = build(L, L + R >> 1); u->suc[1] = build((L + R >> 1) + 1, R);return u;}inline void build() {rt[0] = build(1, n);}inline void newTree(int id) {rt[id] = rt[id - 1];}inline node* insert(node *_u, node *u, int L, int R, int pos){if(u == _u) u = new node, *u = *_u;++ u->cnt;if(L == R) return u;if(pos <= L + R >> 1) u->suc[0] = insert(_u->suc[0], u->suc[0], L, L + R >> 1, pos);else u->suc[1] = insert(_u->suc[1], u->suc[1], (L + R >> 1) + 1, R, pos);return u; }inline void insert(int id, int pos){rt[id] = insert(rt[id - 1], rt[id], 1, n, pos);}int query(node *u, int curL, int curR, int L, int R){if(curL >= L && curR <= R) return u->cnt;int mid = curL + curR >> 1;int res = 0;if(L <= mid) res += query(u->suc[0], curL, mid, L, R);if(R > mid) res += query(u->suc[1], mid + 1, curR, L, R);return res;}inline int query(int id, int L, int R){return query(rt[id], 1, n, L, R);}
}seg;
int main()
{#ifndef ONLINE_JUDGEfreopen("balloon.in", "r", stdin);freopen("balloon.out", "w", stdout);#endifusing namespace Zeonfai;n = getInt(), m = getInt();static int a[N + 1];for(int i = 1; i <= n; ++ i) a[i] = getInt();for(int i = 0; i < m; ++ i) chd[i].L = getInt(), chd[i].R = getInt();std::sort(chd, chd + m);seg.build();for(int i = 1, p = 0; i <= n; ++ i){seg.newTree(i);for(; p < m && chd[p].L == i; ++ p) seg.insert(i, chd[p].R);}static int pre[N + 1], nxt[N + 1];for(int i = 1; i <= n; ++ i) pre[i] = i - 1, nxt[i] = i + 1;pre[1] = nxt[n] = -1;int q = getInt(), ans = 0;for(int i = 0; i < q; ++ i){int pos = (getInt() + ans - 1) % n + 1; -- a[pos];if(! a[pos]){ans += seg.query(pos, pos, ~ nxt[pos] ? nxt[pos] - 1 : n);if(~ pre[pos]) ans -= seg.query(pre[pos], pos, ~ nxt[pos] ? nxt[pos] - 1 : n);if(~ nxt[pos]) pre[nxt[pos]] = pre[pos];if(~ pre[pos]) nxt[pre[pos]] = nxt[pos];}printf("%d\n", ans);}
}