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