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  • codeforces 438D. The Child and Sequence(线段树)

    D. The Child and Sequence
    time limit per test
    4 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    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.

    Examples
    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
    output
    8
    5
    input
    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
    output
    49
    15
    23
    1
    9
    Note

    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:区间求和 单点修改 区间取模

    /*
    标记没法维护
    但可以发现一个性质,一个数每次取模最大也会比原来的1/2小
    所以单点改,如果一个区间最大值都比模数小,就不用往后递归了。
    */
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    
    #define N 100007
    #define ll long long
    
    using namespace std;
    ll n,m,ans,cnt;
    struct tree
    {
        ll l,r,mx,sum;
    }tr[N<<2];
    
    inline ll read()
    {
        ll x=0,f=1;char c=getchar();
        while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
        return x*f;
    }
    
    void pushup(ll k)
    {
        tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum;
        tr[k].mx=max(tr[k<<1].mx,tr[k<<1|1].mx);
    }
    
    void build(ll k,ll l,ll r)
    {
        tr[k].l=l;tr[k].r=r;tr[k].sum=tr[k].mx=0;
        if(l==r)
        {
            tr[k].sum=read();
            tr[k].mx=tr[k].sum;
            return;
        }
        ll mid=l+r>>1;
        build(k<<1,l,mid);build(k<<1|1,mid+1,r);
        pushup(k);
    }
    
    void changemod(ll k,ll l,ll r,ll mod)
    {
        if(l>r) return;
        if(tr[k].mx<mod) return;
        if(tr[k].l==tr[k].r)
        {
            tr[k].mx%=mod;
            tr[k].sum=tr[k].mx;
            return;
        }
        ll mid=tr[k].l+tr[k].r>>1;
        if(r<=mid) changemod(k<<1,l,r,mod);
        else if(l>mid) changemod(k<<1|1,l,r,mod);
        else changemod(k<<1,l,mid,mod),changemod(k<<1|1,mid+1,r,mod);
        pushup(k);
    }
    
    void change(ll k,ll pos,ll x)
    {
        if(tr[k].l==tr[k].r && tr[k].l==pos)
        {
            tr[k].sum=tr[k].mx=x;
            return;
        }
        ll mid=tr[k].l+tr[k].r>>1;
        if(pos<=mid) change(k<<1,pos,x);
        if(pos>mid)  change(k<<1|1,pos,x);
        pushup(k);
    }
    
    ll query(ll k,ll l,ll r)
    {
        if(l>r) return 0;
        if(tr[k].l==l && tr[k].r==r)
        return tr[k].sum;
        ll mid=tr[k].l+tr[k].r>>1;
        if(r<=mid) return query(k<<1,l,r);
        else if(l>mid) return query(k<<1|1,l,r);
        else return query(k<<1,l,mid)+query(k<<1|1,mid+1,r);
    }
    
    int main()
    {
        n=read();m=read();
        build(1,1,n);ll opt,x,y,z;
        for(ll i=1;i<=m;i++)
        {
            opt=read();
            if(opt==1)
            {
                x=read();y=read();
                printf("%lld
    ",query(1,x,y));
            }
            if(opt==2)
            {
                x=read();y=read();z=read();
                changemod(1,x,y,z);
            }
            if(opt==3)
            {
                x=read();y=read();
                change(1,x,y);
            }
        }
        return 0;
    }
    折花枝,恨花枝,准拟花开人共卮,开时人去时。 怕相思,已相思,轮到相思没处辞,眉间露一丝。
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  • 原文地址:https://www.cnblogs.com/L-Memory/p/7623172.html
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