UOJ228 简单数据结构练习题

Description

传送门

维护一个数列, 有以下操作:

1. 对[l,r]同时加上x

2. 把[l,r]开根后下取整.

3. 查询[l,r]之和

n,m \(\leq\)$ 100000, $$a_i,x \leq 10^5$

Solution

考虑一个简易的线段树,直接对一个区间进行开根. 如果这个区间数字不同就继续往下递归. 因为 开根开不了多少次就会变成1,所以复杂度相对较小.但考虑一种情况:8 9 8 9 8 9 \(\dots\) 这样变成1后再区间修改变成原来的样子.然后就会TLE.

怎么办呢. 考虑一个小小优化,如果\(Max - \sqrt{Max} = Min - \sqrt{Min}\), 即区间变化量相同, 就直接转换为一个减法.

然后就跑过去了, 还跑得飞快.

分析一下复杂度:

对于一个数\(y =a^{2^{k}}\), 我们对他开根是:\(k\)次的, 即\(log_{2}({log_{a}(y)})\). 对于一个数, 我们在线段树上跳是\(log{n}\)次的.

我们定义势能V为相邻两数的差. 那么越开根号,V就越小.

显然,我们一次加法最多让V增加到最大值, 总势能为mV, 所以复杂度为\(O((n + m)logN * loglogV)\)

Codes

#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for(int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define Debug(s) debug("The massage in line %d, Function %s: %s\n", __LINE__, __FUNCTION__, s)
typedef long long LL;
typedef long double LD;
const int BUF_SIZE = (int)1e6 + 10;
struct fastIO {
    char buf[BUF_SIZE], buf1[BUF_SIZE];
    int cur, cur1;
    FILE *in, *out;
    fastIO() {
        cur = BUF_SIZE, in = stdin, out = stdout;
		cur1 = 0;
    }
    inline char getchar() {
        if(cur == BUF_SIZE) fread(buf, BUF_SIZE, 1, in), cur = 0;
        return *(buf + (cur++));
    }
    inline void putchar(char ch) {
        *(buf1 + (cur1++)) = ch;
        if (cur1 == BUF_SIZE) fwrite(buf1, BUF_SIZE, 1, out), cur1 = 0;
    }
    inline int flush() {
        if (cur1 > 0) fwrite(buf1, cur1, 1, out);
        return cur1 = 0;
    }
}IO;
#define getchar IO.getchar
#define putchar IO.putchar
int read() {
	char ch = getchar();
	int x = 0, flag = 1;
	for(;!isdigit(ch); ch = getchar()) if(ch == '-') flag *= -1;
	for(;isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
	return x * flag;
}
void write(LL x) {
	if(x < 0) putchar('-'), x = -x;
	if(x >= 10) write(x / 10);
	putchar(x % 10 + 48);
}
void putString(char s[], char EndChar = '\n') {
	rep(i, 0, strlen(s) - 1) putchar(*(s + i));
	if(~EndChar) putchar(EndChar);
}

#define Maxn 100009
int n, m, a[Maxn];
namespace SGMT_tree {
	LL Max[Maxn << 2], Min[Maxn << 2], sum[Maxn << 2], tag[Maxn << 2];
#define lc(x) ((x) << 1)
#define rc(x) (((x) << 1) | 1)
	void pushup(int root) {
		Max[root] = max(Max[lc(root)], Max[rc(root)]);
		Min[root] = min(Min[lc(root)], Min[rc(root)]);
		sum[root] = sum[lc(root)] + sum[rc(root)];
	}
	void pushdown(int root, int l, int r) {
		LL v = tag[root];
		if(v) {
			int mid = (l + r) >> 1;
			Max[lc(root)] += v, Min[lc(root)] += v, tag[lc(root)] += v; sum[lc(root)] += 1ll * v * (mid - l + 1);
			Max[rc(root)] += v, Min[rc(root)] += v, tag[rc(root)] += v; sum[rc(root)] += 1ll * v * (r - mid);
			tag[root] = 0;
		}
	}
	void build(int root, int l, int r) {
		if(l == r) {
			Max[root] = Min[root] = sum[root] = a[l];
			tag[root] = 0;
			return ;
		}
		int mid = (l + r) >> 1;
		build(lc(root), l, mid);
		build(rc(root), mid + 1, r);
		pushup(root);
	}
	void modify(int root, int l, int r, int x, int y, LL v) {
		if(l > r || r < x || l > y) return ;
		if(x <= l && r <= y) {
			Max[root] += v, Min[root] += v, tag[root] += v;
			sum[root] += 1ll * v * (1ll * r - l + 1);
			return;
		}
		int mid = (l + r) >> 1;
		pushdown(root, l, r);
		modify(lc(root), l, mid, x, y, v);
		modify(rc(root), mid + 1, r, x, y, v);
		pushup(root);
	}
	void modify(int root, int l, int r, int x, int y) {
		if(l > r || r < x || l > y) return ;
		if(x <= l && r <= y) {
			if(Max[root] - ((int)sqrt(Max[root])) == Min[root] - ((int)sqrt(Min[root]))) {
				LL v = -Max[root] + ((int)sqrt(Max[root]));
				Max[root] += v, Min[root] += v, tag[root] += v, sum[root] += 1ll * v * (r - l + 1);
				return ;
			}
		}
		int mid = (l + r) >> 1;
		pushdown(root, l, r);
		modify(lc(root), l, mid, x, y);
		modify(rc(root), mid + 1, r, x, y);
		pushup(root);
	}
	LL query(int root, int l, int r, int x, int y) {
		if(l > r || r < x || l > y) return 0;
		if(x <= l && r <= y) return sum[root];
		int mid = (l + r) >> 1; LL res = 0;
		pushdown(root, l, r);
		res += query(lc(root), l, mid, x, y);
		res += query(rc(root), mid + 1, r, x, y);
		pushup(root);
		return res;
	}
}
namespace INIT {
	void Main() {
		n = read();	m = read();
		rep(i, 1, n) a[i] = read();
		SGMT_tree :: build(1, 1, n);
	}
}
namespace SOLVE {
	void Main() {
		rep(i, 1, m) {
			int opt = read();
			if(opt == 1) {
				int l = read(), r = read(), x = read();
				SGMT_tree :: modify(1, 1, n, l, r, x);
			}
			if(opt == 2) {
				int l = read(), r = read();
				SGMT_tree :: modify(1, 1, n, l, r);
			}
			if(opt == 3) {
				int l = read(), r = read();
				write(SGMT_tree :: query(1, 1, n, l, r)), putchar('\n');
			}
		}
	}
}
int main() {
#ifdef Qrsikno
	freopen("uoj228.in", "r", stdin);
	freopen("uoj228.out", "w", stdout);
#endif
	INIT :: Main();
	SOLVE :: Main();
#ifdef Qrsikno
	debug("\nRunning time: %.3lf(s)\n", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
	return IO.flush();
}