【BZOJ4631】踩氣球【暴力】【線段樹優化】

【題目鏈接】

記錄每個盒子被哪些熊孩子的區間覆蓋，用鏈表存起來。

又因爲熊孩子的區間是連續的，所以用線段樹優化一下就行了。

注意空間大小，不然容易RE WA

/* Think Thank Thunk */
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int maxn = 100005, maxm = 5000005;

int n, m, head[maxn << 2], cnt, tot[maxn], ans;
LL tr[maxn << 2], sum[maxn];

struct data {
	int v, next;
} g[maxm];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline void add(int u, int v) {
	g[cnt] = (data){v, head[u]};
	head[u] = cnt++;
}

inline void update(int p) {
	for(int i = head[p]; ~i; i = g[i].next) {
		tot[g[i].v]--;
		if(!tot[g[i].v]) ans++;
	}
}

inline void build(int p, int l, int r) {
	tr[p] = sum[r] - sum[l - 1];
	head[p] = -1;
	if(l == r) return;
	int mid = l + r >> 1;
	build(p << 1, l, mid); build(p << 1 | 1, mid + 1, r);
}

inline void insert(int p, int l, int r, int x, int y, int c) {
	if(x <= l && r <= y) {
		add(p, c); tot[c]++;
		return;
	}
	int mid = l + r >> 1;
	if(x <= mid) insert(p << 1, l, mid, x, y, c);
	if(y > mid) insert(p << 1 | 1, mid + 1, r, x, y, c);
}

inline void change(int p, int l, int r, int x) {
	tr[p]--;
	if(!tr[p]) update(p);
	if(l == r) return;
	int mid = l + r >> 1;
	if(x <= mid) change(p << 1, l, mid, x);
	else change(p << 1 | 1, mid + 1, r, x);
}

int main() {
	n = iread(); m = iread();
	for(int i = 1; i <= n; i++) sum[i] = sum[i - 1] + iread();
	build(1, 1, n);
	for(int i = 1; i <= m; i++) {
		int l = iread(), r = iread();
		insert(1, 1, n, l, r, i);
	}
	for(m = iread(); m; m--) {
		int x = (iread() + ans - 1) % n + 1;
		change(1, 1, n, x);
		printf("%d\n", ans);
	}
	return 0;
}