LibreOJ #6283

题目链接:#6283. 数列分块入门 7

题目大意

给出一个长为 \(n\) 的数列，以及 \(n\) 个操作，操作涉及区间乘法，区间加法，单点询问。

solution

我们这题可以借鉴线段树的懒标记, 然后我们的操作就变成了:

修改操作: 加和: 对于整块直接加和,对于散块,我们懒标记下放,然后暴力加和
乘: 对于整块,我们更新懒标记,对于散块, 我们下放,然后暴力乘
查询操作: 我们直接求就好了

Code:

/**
*    Author: Alieme
*    Data: 2020.9.8
*    Problem: LiberOJ #6283
*    Time: O()
*/
#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>

#define int long long
#define rr register

#define inf 1e9
#define MAXN 100010

using namespace std;

inline int read() {
	int s = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= ch == '-', ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

const int mod = 10007;

void print(int x) {
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) print(x / 10);
	putchar(x % 10 + 48);
}

int n, len;

int sum[MAXN], a[MAXN], id[MAXN], mul[MAXN];

inline void pushdown(int x) {
	if (mul[x] ^ 1) 
		for (rr int i = (x - 1) * len + 1; i <= min(x * len, n); i++)
			a[i] = (a[i] * mul[x]) % mod;
	mul[x] = 1;
	if (sum[x])
		for (rr int i = (x - 1) * len + 1; i <= min(x * len, n); i++)
			a[i] = (a[i] + sum[x]) % mod;
	sum[x] = 0;
}

inline void add_sum(int l, int r, int x) {
	int start = id[l], end = id[r];
	if (id[l] == id[r]) {
		pushdown(start);
		for (rr int i = l; i <= r; i++) a[i] = (a[i] + x) % mod;
		return ;
	}
	pushdown(start), pushdown(end);
	for (rr int i = l; id[i] == start; i++) a[i] = (a[i] + x) % mod;
	for (rr int i = start + 1; i < end; i++) sum[i] = (sum[i] + x) % mod;
	for (rr int i = r; id[i] == end; i--) a[i] = (a[i] + x) % mod;
}

inline void add_mul(int l, int r, int x) {
	int start = id[l], end = id[r];
	if (id[l] == id[r]) {
		pushdown(start);
		for (rr int i = l; i <= r; i++) a[i] = (a[i] * x) % mod;
		return ;
	}
	pushdown(start), pushdown(end);
	for (rr int i = l; id[i] == start; i++) a[i] = (a[i] * x) % mod;
	for (rr int i = start + 1; i < end; i++) sum[i] = (sum[i] * x) % mod, mul[i] = (mul[i] * x) % mod;
	for (rr int i = r; id[i] == end; i--) a[i] = (a[i] * x) % mod;
}

inline int query(int x) {
	return (a[x] * mul[id[x]] % mod + sum[id[x]]) % mod;
}

signed main() {
	n = read();
	len = sqrt(n);
	for (rr int i = 1; i <= n; i++) a[i] = read(), a[i] %= mod, id[i] = (i - 1) / len + 1, mul[id[i]] = 1;
	for (rr int i = 1; i <= n; i++) {
		int opt = read(), l = read(), r = read(), c = read();
		if (opt == 0) add_sum(l, r, c);
		if (opt == 1) add_mul(l, r, c);
		if (opt == 2) cout << query(r) << "\n";
	}
}