• Codeforces Round #250 (Div. 1) D. The Child and Sequence 线段树 区间取摸


    D. The Child and Sequence

    Time Limit: 20 Sec

    Memory Limit: 256 MB

    题目连接

    http://codeforces.com/contest/438/problem/D

    Description

    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 .
    2. Modulo operation l, r, x. Picks should perform assignment a[i] = a[imod 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?

    Input

    The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — 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 ≤ 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 ≤ 109), which correspond the operation 2.
    • If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3.

    Output

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

    Sample Input

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

    Sample Output

    8
    5

    HINT

    题意

    给你n个数,三个操作

    1.输出[l,r]的和

    2.将[l,r]中的数,对v取摸

    3.把a[x]变成v

    题解:

    线段树

    区间定值和区间和很简单

    区间取摸的话,需要维护一个区间最大值,如果这个区间的最大值小于要取摸的数,那么就直接break就好了

    代码

    #include<iostream>
    #include<stdio.h>
    using namespace std;
    #define maxn 100005
    struct node
    {
        int l,r;
        long long mx,sum;
    }a[maxn*4];
    int d[maxn];
    void build(int x,int l,int r)
    {
        a[x].l = l,a[x].r = r;
        if(l==r)
        {
            a[x].mx = a[x].sum = d[l];
            return;
        }
        int mid = (l+r)/2;
        build(x<<1,l,mid);
        build(x<<1|1,mid+1,r);
        a[x].mx = max(a[x<<1].mx , a[x<<1|1].mx);
        a[x].sum = a[x<<1|1].sum + a[x<<1].sum;
    }
    void change(int x,int pos,long long val)
    {
        int l = a[x].l,r = a[x].r;
        if(l==r)
        {
            a[x].mx = a[x].sum = val;
            return;
        }
        int mid = (l+r)/2;
        if(pos<=mid)
            change(x<<1,pos,val);
        else
            change(x<<1|1,pos,val);
        a[x].mx = max(a[x<<1].mx,a[x<<1|1].mx);
        a[x].sum = a[x<<1].sum + a[x<<1|1].sum;
    }
    void mod(int x,int l,int r,long long val)
    {
        int L = a[x].l,R = a[x].r;
        if(a[x].mx<val)return;
        if(L==R)
        {
            a[x].sum%=val;
            a[x].mx%=val;
            return;
        }
        int mid = (L+R)/2;
        if(r<=mid)
            mod(x<<1,l,r,val);
        else if(l>mid)
            mod(x<<1|1,l,r,val);
        else mod(x<<1,l,mid,val),mod(x<<1|1,mid+1,r,val);
        a[x].sum = a[x<<1].sum + a[x<<1|1].sum;
        a[x].mx = max(a[x<<1].mx,a[x<<1|1].mx);
    }
    long long get(int x,int l,int r)
    {
        int L = a[x].l,R = a[x].r;
        if(L>=l&&R<=r)
        {
            return a[x].sum;
        }
        int mid = (L+R)/2;
        long long sum1 = 0,sum2 = 0;
        if(r<=mid)
            sum1 = get(x<<1,l,r);
        else if(l>mid)
            sum2 = get(x<<1|1,l,r);
        else sum1 = get(x<<1,l,mid),sum2 = get(x<<1|1,mid+1,r);
        return sum1 + sum2;
    }
    int main()
    {
        int n,q;
        scanf("%d%d",&n,&q);
        for(int i=1;i<=n;i++)
            scanf("%d",&d[i]);
        build(1,1,n);
        while(q--)
        {
            int op;
            scanf("%d",&op);
            if(op==1)
            {
                int x,y;scanf("%d%d",&x,&y);
                printf("%lld
    ",get(1,x,y));
            }
            if(op==2)
            {
                int x,y,z;scanf("%d%d%d",&x,&y,&z);
                mod(1,x,y,z);
            }
            if(op==3)
            {
                int x,y;scanf("%d%d",&x,&y);
                change(1,x,y);
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4917116.html
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