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:

Print operation $l,r$. Picks should write down the value of $\sum_{i=1}^{r} a[i]$.
Modulo operation $l,r,x$. Picks should perform assignment $a[i]=a[i]$ $\%$ $x$ for each $i$ $(l<=i<=r)$.
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

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

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

}