「HNOI2010」弹飞绵羊

「HNOI2010」弹飞绵羊

传送门

考虑分块。

每一个位置 \(i\) ，记 \(to[i]\) 表示从这个位置一直往右跳回落在哪个位置。

然后修改的时候直接暴改，查询也是暴跳，复杂度 \(O(n \sqrt n)\)

参考代码：

#include <algorithm>
#include <cstdio>
#include <cmath>
#define rg register
#define file(x) freopen(x".in", "r", stdin), freopen(x".out", "w", stdout)
using namespace std;
template < class T > inline void read(T& s) {
	s = 0; int f = 0; char c = getchar();
	while ('0' > c || c > '9') f |= c == '-', c = getchar();
	while ('0' <= c && c <= '9') s = s * 10 + c - 48, c = getchar();
	s = f ? -s : s;
}

const int _ = 200010;

int n, f[_], k[_];
int gap, pos[_], to[_], l[_ >> 1];

int main() {
#ifndef ONLINE_JUDGE
	file("cpp");
#endif
	read(n), gap = sqrt(n * 1.0);
	for (rg int i = 1; i <= n; ++i) {
		read(k[i]), pos[i] = (i - 1) / gap + 1;
		if (pos[i - 1] != pos[i]) l[pos[i]] = i;
	}
	l[pos[n] + 1] = n + 1;
	for (rg int i = n; i >= 1; --i) {
		if (i + k[i] >= l[pos[i] + 1])
			f[i] = 1, to[i] = i + k[i];
		else
			f[i] = f[i + k[i]] + 1, to[i] = to[i + k[i]];
	}
	int q; read(q);
	for (rg int opt, x, o = 1; o <= q; ++o) {
		read(opt), read(x), ++x;
		if (opt == 1) {
			int res = 0;
			while (x <= n) res += f[x], x = to[x];
			printf("%d\n", res);
		} else {
			read(k[x]);
			for (rg int i = l[pos[x] + 1] - 1; i >= l[pos[x]]; --i) {
				if (i + k[i] >= l[pos[x] + 1])
					f[i] = 1, to[i] = i + k[i];
				else
					f[i] = f[i + k[i]] + 1, to[i] = to[i + k[i]];
			}
		}
	}
	return 0;
}