[Codeforces Round #250 (Div. 1) -D] The Child and Sequence

题意翻译

给定数列，区间查询和，区间取模，单点修改。

$n,m\le 10^5$

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 mm operations. An operation can be one of the following:

Print operation $l,r$ . Picks should write down the value of .
Modulo operation $l,r,x$ . Picks should perform assignment $a[i]=a[i]\ \mathrm{mod}\ x$ for each $i$ $(l\le i\le r)$ .
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: $n,m$ ( $1\le n,m\le 10^{5}$ ) . The second line contains $n$ integers, separated by space: $a[1],a[2],...,a[n] (1\le a[i]\le 10^{9})$ — initial value of array elements.

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

If $type=1$ , there will be two integers more in the line: $l,r (1\le l\le r\le n)$ , which correspond the operation $1$ .
If $type=2$ , there will be three integers more in the line: $l,r,x (1\le l\le r\le n; 1\le x\le 10^{9})$ , which correspond the operation 2.
If $type=3$ , there will be two integers more in the line: $k,x (1\le k\le n; 1\le x\le 10^{9})$ , 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：

样例解释

对于样例#1：

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.

解题分析

考虑每次取模后，如果当前数值比模数大，则肯定取模后的值至少比原来少一半，因此不考虑单点修改的话每个数字最多修改 $log(N)$ 次。所以总复杂度为 $O(Nlog^2(N))$ 。

我们维护一个区间最大值，如果当前区间最大值 $<$ 模数，则直接 $return$ 。

代码如下：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <algorithm>
#define R register
#define IN inline
#define gc getchar()
#define W while
#define MX 100050
#define ll long long
#define ls (now << 1)
#define rs (now << 1 | 1)
template <class T>
IN void in(T &x)
{
    x = 0; R char c = gc;
    W (!isdigit(c)) c = gc;
    W (isdigit(c))
    x = (x << 1) + (x << 3) + c - 48, c = gc;
}
int dot, q;
ll dat[MX], MOD;
struct Node
{ll sum, mx;} tree[MX << 5];
namespace SGT
{
    IN void pushup(const int &now)
    {
        tree[now].mx = std::max(tree[ls].mx, tree[rs].mx);
        tree[now].sum = tree[ls].sum + tree[rs].sum;
    }
    void build(const int &now, const int &lef, const int &rig)
    {
        if(lef == rig) return tree[now].mx = tree[now].sum = dat[lef], void();
        int mid = lef + rig >> 1;
        build(ls, lef, mid), build(rs, mid + 1, rig);
        pushup(now);
    }
    void modify(const int &now, const int &lef, const int &rig, const int &tar, const ll &del)
    {
        if(lef == rig) return tree[now].mx = tree[now].sum = del, void();
        int mid = lef + rig >> 1;
        if(tar <= mid) modify(ls, lef, mid, tar, del);
        else modify(rs, mid + 1, rig, tar, del);
        pushup(now);
    }
    void mod(const int &now, const int &lef, const int &rig, const int &lb, const int &rb)
    {
        if(tree[now].mx < MOD) return;
        if(lef == rig) return tree[now].sum = tree[now].mx = tree[now].sum % MOD, void();
        int mid = lef + rig >> 1;
        if(lb <= mid) mod(ls, lef, mid, lb, rb);
        if(rb > mid) mod(rs, mid + 1, rig, lb, rb);
        pushup(now);
    }
    IN ll query(const int &now, const int &lef, const int &rig, const int &lb, const int &rb)
    {
        if(lef >= lb && rig <= rb) return tree[now].sum;
        int mid = lef + rig >> 1; ll ret = 0;
        if(lb <= mid) ret += query(ls, lef, mid, lb, rb);
        if(rb > mid) ret += query(rs, mid + 1, rig, lb, rb);
        return ret;
    }
}
int main(void)
{
    int typ, a, b; ll c;
    in(dot), in(q);
    for (R int i = 1; i <= dot; ++i) in(dat[i]);
    SGT::build(1, 1, dot);
    W (q--)
    {
        in(typ);
        if(typ == 1)
        {
            in(a), in(b);
            printf("%I64d\n", SGT::query(1, 1, dot, a, b));
        }
        else if(typ == 2)
        {
            in(a), in(b), in(MOD);
            SGT::mod(1, 1, dot, a, b);
        }
        else
        {
            in(a), in(c);
            SGT::modify(1, 1, dot, a, c);
        }
    }
}