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  • Codeforces Round #307 (Div. 2) E. GukiZ and GukiZiana 分块

    E. GukiZ and GukiZiana
    time limit per test
    10 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a, indexed with integers from 1 to n, and number yGukiZiana(a, y) represents maximum value of j - i, such that aj = ai = y. If there is no y as an element in a, then GukiZiana(a, y) is equal to  - 1. GukiZ also prepared a problem for you. This time, you have two types of queries:

    1. First type has form l r x and asks you to increase values of all ai such that l ≤ i ≤ r by the non-negative integer x.
    2. Second type has form y and asks you to find value of GukiZiana(a, y).

    For each query of type 2, print the answer and make GukiZ happy!

    Input

    The first line contains two integers nq (1 ≤ n ≤ 5 * 105, 1 ≤ q ≤ 5 * 104), size of array a, and the number of queries.

    The second line contains n integers a1, a2, ... an (1 ≤ ai ≤ 109), forming an array a.

    Each of next q lines contain either four or two numbers, as described in statement:

    If line starts with 1, then the query looks like l r x (1 ≤ l ≤ r ≤ n0 ≤ x ≤ 109), first type query.

    If line starts with 2, then th query looks like y (1 ≤ y ≤ 109), second type query.

    Output

    For each query of type 2, print the value of GukiZiana(a, y), for y value for that query.

    Examples
    input
    4 3
    1 2 3 4
    1 1 2 1
    1 1 1 1
    2 3
    output
    2
    input
    2 3
    1 2
    1 2 2 1
    2 3
    2 4
    output
    0
    -1

     思路:分块,map超时,MLE,vector+二分即可;

    #include<bits/stdc++.h>
    using namespace std;
    #define ll long long
    #define pi (4*atan(1.0))
    #define eps 1e-14
    #define bug(x,y) cout<<"bug"<<x<<" "<<y<<endl;
    #define bug(x) cout<<"xxx "<<x<<endl;
    const int N=5e5+10,M=1e6+10,inf=2e9+10,mod=6;
    const ll INF=1e18+10;
    ll a[N],add[N];
    int n,q,ka,pos[N];
    vector<pair<ll,int> >vec[710];
    int cmp(pair<ll,int> a,pair<ll,int> b)
    {
        if(a.first==b.first)
            return a.second<b.second;
        return a.first<b.first;
    }
    int ansl,ansr;
    void brute(int l,int r,int pos)
    {
        vec[pos].clear();
        for(int i=l; i<=r; i++)
            vec[pos].push_back(make_pair(a[i],i));
        sort(vec[pos].begin(),vec[pos].end(),cmp);
    }
    void brutef(int l,int r,int pos,ll x)
    {
        for(int i=l; i<=r; i++)
        {
            if(add[pos]+a[i]==x)
            {
                ansl=min(ansl,i);
                ansr=max(ansr,i);
            }
        }
    }
    void update(int l,int r,ll x)
    {
        if(pos[l]==pos[r])
        {
            for(int i=l;i<=r;i++)
                a[i]+=x;
            brute((pos[l]-1)*ka+1,pos[l]*ka,pos[l]);
            return;
        }
        for(int i=pos[l]+1; i<=pos[r]-1; i++)
            add[i]+=x;
        for(int i=l; i<=min(r,pos[l]*ka); i++)
            a[i]+=x;
        brute((pos[l]-1)*ka+1,pos[l]*ka,pos[l]);
        for(int i=max(l,(pos[r]-1)*ka+1); i<=r; i++)
            a[i]+=x;
        if(pos[r]<=n/ka)
            brute((pos[r]-1)*ka+1,pos[r]*ka,pos[r]);
    }
    void slove(ll x)
    {
        ansl=n+1;
        ansr=0;
        for(int i=1; i<=n/ka; i++)
        {
            int l1=0,l2=0,l=-1;
            int r1=ka-1,r2=ka-1,r=-1;
            while(l1<=r1)
            {
                int mid=(l1+r1)>>1;
                if(vec[i][mid].first<x-add[i])
                    l1=mid+1;
                else if(vec[i][mid].first==x-add[i])
                {
                    l=mid;
                    r1=mid-1;
                }
                else
                    r1=mid-1;
            }
            while(l2<=r2)
            {
                int mid=(l2+r2)>>1;
                if(vec[i][mid].first<x-add[i])
                    l2=mid+1;
                else if(vec[i][mid].first==x-add[i])
                {
                    r=mid;
                    l2=mid+1;
                }
                else
                    r2=mid-1;
            }
            if(l!=-1)
            {
                ansl=min(ansl,vec[i][l].second);
                ansr=max(ansr,vec[i][r].second);
            }
        }
        brutef((n/ka)*ka+1,n,n/ka+1,x);
        //cout<<ansr<<" "<<ansl<<endl;
        if(ansl==n+1)
            puts("-1");
        else
            printf("%d
    ",ansr-ansl);
    }
    int main()
    {
        scanf("%d%d",&n,&q);
        for(int i=1; i<=n; i++)
            scanf("%lld",&a[i]);
        ka=sqrt(n);
        for(int i=1; i<=n; i++)
            pos[i]=(i-1)/ka+1;
        for(int i=1; i<=n/ka; i++)
            brute(ka*(i-1)+1,ka*i,i);
        while(q--)
        {
            int flag,l,r;
            ll x;
            scanf("%d",&flag);
            if(flag==1)
            {
                scanf("%d%d%lld",&l,&r,&x);
                update(l,r,x);
            }
            else
            {
                scanf("%lld",&x);
                slove(x);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jhz033/p/6435249.html
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