题解【Codeforces438D】The Child and Sequence

题目描述

At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.

Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array \(a[1],a[2],...,a[n]\). Then he should perform a sequence of \(m\) operations. An operation can be one of the following:

1. Print operation \(l,r\). Picks should write down the value of \(\sum_{i=1}^{r} a[i]\).

2. Modulo operation \(l,r,x\). Picks should perform assignment \(a[i]=a[i]\) \(\%\) \(x\) for each \(i\) \((l<=i<=r)\).

3. Set operation \(k,x\). Picks should set the value of \(a[k]\) to \(x\) (in other words perform an assignment \(a[k]=x\)).

Can you help Picks to perform the whole sequence of operations?

输入输出格式

输入格式

The first line of input contains two integer: \(n,m\) \((1<=n,m<=10^{5})\). The second line contains \(n\) integers, separated by space: \(a[1],a[2],...,a[n] (1<=a[i]<=10^{9})\) — initial value of array elements.

Each of the next m m m lines begins with a number type type type .

• If \(type=1\), there will be two integers more in the line: \(l,r (1<=l<=r<=n)\) , which correspond the operation \(1\).

• If \(type=2\), there will be three integers more in the line: \(l,r,x (1<=l<=r<=n; 1<=x<=10^{9})\) , which correspond the operation \(2\).

• If \(type=3\), there will be two integers more in the line: \(k,x (1<=k<=n; 1<=x<=10^{9})\) , which correspond the operation \(3\).

输出格式

For each operation \(1\), please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.

输入输出样例

输入样例#1

5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3

输出样例#1

8
5

输入样例#2

10 10
6 9 6 7 6 1 10 10 9 5
1 3 9
2 7 10 9
2 5 10 8
1 4 7
3 3 7
2 7 9 9
1 2 4
1 6 6
1 5 9
3 1 10

输出样例#2

49
15
23
1
9

说明

Consider the first testcase:

• At first, \(a={1,2,3,4,5}\).

• After operation \(1\), \(a={1,2,3,0,1}\).

• After operation \(2\), \(a={1,2,5,0,1}\).

• At operation \(3\), \(2+5+0+1=8\).

• After operation \(4\), \(a={1,2,2,0,1}\).

• At operation \(5\), \(1+2+2=5\).

题意翻译

给定数列，区间查询和，区间取模，单点修改。

\(n,m\)小于\(10^5\)

题解

线段树基础题。

区间查询和、单点修改很简单，也很基础，这里就不在赘述。

重点来看一下区间取模。

首先，我们不难知道，当一个数\(a \% b\)时，如果\(a < b\)，那么这个取模是没有什么意义的（\(*\)）。

如果，我们执行区间取模时，一个一个数去取模，那么复杂度会非常高，达到\(\Theta (n \times m)\)，绝对会\(TLE\)。

因此考虑一种类似搜索“剪枝”的方式来优化区间取模。

这时，我们就要用到上面的（\(*\)）了。

用一个数组\(mx[]\)来记录区间内的最大值，如果这个最大值都小于我们要取模的那个数了，就直接\(return\)返回掉，因为对这个区间取模就已经没有意义了。

很容易就可以写出\(AC\)代码。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define int long long

using namespace std;

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

int n/*数的个数*/, m/*操作个数*/, tr[100003 << 2]/*区间和*/, mx[100003 << 2]/*区间最大值*/, a[100003]/*每个数的值*/;

inline void pushup(int p)//上传节点操作
{
	mx[p] = max(mx[p << 1], mx[(p << 1) | 1]);//更新区间最大值
	tr[p] = tr[p << 1] + tr[(p << 1) | 1];//加上区间和
}

void build(int s, int t, int p)//建树操作
{
	if (s == t)//已经是叶子节点了
	{
		mx[p] = tr[p] = a[s];//更新节点的参数
		return;//返回
	}
	int mid = (s + t) >> 1;//计算中间值
	build(s, mid, p << 1); //递归左子树
	build(mid + 1, t, (p << 1) | 1);//递归右子树
	pushup(p);//上传当前节点
}

void modify(int l/*要修改的数的编号,即目标节点*/, int r/*要更新的值*/, int s, int t, int p)//单点修改操作
{
	if (s == t)//已经到了目标节点
	{
		mx[p] = tr[p] = r; //更新节点参数
		return;//直接返回
	}
	int mid = (s + t) >> 1;//计算中间值
	if (l <= mid) //目标节点在左区间
		modify(l, r, s, mid, p << 1);//递归左子树寻找
	else //目标节点在右区间
		modify(l, r, mid + 1, t, (p << 1) | 1);//递归右区间查找
	pushup(p);//上传当前节点
}

void getmod(int l/*区间左界*/, int r/*区间右界*/, int mod/*要取模的值*/, int s, int t, int p)//区间取模操作
{
	if (mx[p] < mod) return;//"剪枝"操作
	if (s == t)//已经到了叶子节点
	{
		tr[p] = tr[p] % mod; //取模
		mx[p] = tr[p];//更新最大值
		return;//返回
	}
	int mid = (s + t) >> 1;//计算中间值
	if (l <= mid) getmod(l, r, mod, s, mid, p << 1);//查找中点左边的区间进行取模
	if (r > mid) getmod(l, r, mod, mid + 1, t, (p << 1) | 1);//查找中点右边的区间进行取模
	pushup(p);//上传当前节点
}

int getans(int l, int r, int s, int t, int p)//查询区间和操作
{
	if (l <= s && t <= r) return tr[p];//当前区间完全包含于目标区间,就直接返回当前区间的和
	int mid = (s + t) >> 1, ans = 0;//计算中间值及初始化答案
	if (l <= mid) ans = ans + getans(l, r, s, mid, p << 1);//加上中点左边的区间进行求和
	if (r > mid) ans = ans + getans(l, r, mid + 1, t, (p << 1) | 1);//加上中点右边的区间进行求和
	return ans;//返回答案
}

signed main()
{
	n = gi(), m = gi();
	for (int i = 1; i <= n; i++) a[i] = gi();
	//以上为输入
	build(1, n, 1);//建树
	while (m--)
	{
		int fl = gi(), x = gi(), y = gi();
		if (fl == 1)//是输出区间和操作
		{
			printf("%lld\n", getans(x, y, 1, n, 1));//就输出区间和
		}
		else if (fl == 2)//区间取模操作
		{
			int md = gi();//输入模数
			getmod(x, y, md, 1, n, 1);//进行取模
		}
		else
		{
			modify(x, y, 1, n, 1);//否则就进行单点修改,注意是把点x的值修改为y
		}
	}
	return 0;//结束
}