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
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
}