LOJ6283 數列分塊入門7

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LOJ6283 數列分塊入門 7

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標籤

• 分塊入門

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前言

• 我的csdn和博客園是同步的，歡迎來訪danzh-博客園~

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簡明題意

• 維護序列，需要支持三種操作：
  1. 區間加
  2. 區間乘
  3. 單點查

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思路

• 學過線段樹的同學應該在洛谷上做過這一題，難點在於有多種標記該怎麼處理。我在線段樹分類下的一篇文章中講過如何處理多種標記。
• 這裏分塊和線段樹是一樣的，設一個tag_plus[]和tag_mult[]兩種標記。該如何更新標記？實際上我們第一步應該定義標記運算的法則，也就是，如果同時存在兩種標記，應該先計算哪一個呢？定義好了法則，然後就很容易可以更新標記了。

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注意事項

• 還是要注意最後一塊是不完整的。特別是重構塊時，特判一下最後一塊。

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總結

• 重構塊操作可以單獨寫出來~

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AC代碼

#pragma GCC optimize(2)
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;

const int maxn = 1e5 + 10;
const int mod = 10007;

int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}

int n, a[maxn];
int pos[maxn], len, tag_plus[maxn], tag_mult[maxn];

void reset(int id)
{
	for (int i = id * len - len + 1; i <= min(id * len, n); i++)
		a[i] = (a[i] * tag_mult[id] + tag_plus[id]) % mod;
	tag_mult[id] = 1, tag_plus[id] = 0;
}

void change_plus(int l, int r, int c) 
{
	reset(pos[l]);
	if (pos[l] != pos[r]) reset(pos[r]);

	for (int i = l; i <= min(pos[l] * len, r); i++)
		a[i] = (a[i] + c) % mod;
	if (pos[l] != pos[r])
		for (int i = pos[r] * len - len + 1; i <= r; i++)
			a[i] = (a[i] + c) % mod;
	for (int i = pos[l] + 1; i <= pos[r] - 1; i++) tag_plus[i] = (tag_plus[i] + c) % mod;
}

void change_mult(int l, int r, int c) 
{
	reset(pos[l]);
	if (pos[l] != pos[r]) reset(pos[r]);

	for (int i = l; i <= min(pos[l] * len, r); i++) 
		a[i] = a[i] * c % mod;
		
	if (pos[l] != pos[r])
		for (int i = pos[r] * len - len + 1; i <= r; i++) 
			a[i] = a[i] * c % mod;

	for (int i = pos[l] + 1; i <= pos[r] - 1; i++) {
		tag_plus[i] *= c;
		tag_mult[i] *= c;
		tag_plus[i] %= mod, tag_mult[i] %= mod;
	}
}

int ask(int l, int r, int c) {
	return (a[r] * tag_mult[pos[r]] + tag_plus[pos[r]]) % mod;
}

void solve() {
	fill(tag_mult + 1, tag_mult + 1 + maxn - 10, 1);
	scanf("%d", &n);
	len = sqrt(n);
	for (int i = 1; i <= n; i++)
	a[i] = read(), pos[i] = (i - 1) / len + 1;

	for (int i = 1; i <= n; i++) {
		int opt, l, r, c;
		opt = read(), l = read(), r = read(), c = read();
		if (opt == 0)
			change_plus(l, r, c);
		else if (opt == 1)
			change_mult(l, r, c);
		else
			printf("%d\n", (a[r] * tag_mult[pos[r]] + tag_plus[pos[r]]) % mod);
	}
}

int main() {
	freopen("Testin.txt", "r", stdin);
	freopen("Testout.txt", "w", stdout);
	solve();
	return 0;
}