• HDU 5328 Problem Killer(尺取法)


    You are a "Problem Killer", you want to solve many problems.
    Now you have nn problems, the ii-th problem's difficulty is represented by an integer aiai (1ai1091≤ai≤109).
    For some strange reason, you must choose some integer ll and rr (1lrn1≤l≤r≤n), and solve the problems between the ll-th and the rr-th, and these problems' difficulties must form an AP (Arithmetic Progression) or a GP (Geometric Progression).
    So how many problems can you solve at most?

    You can find the definitions of AP and GP by the following links:
    https://en.wikipedia.org/wiki/Arithmetic_progression
    https://en.wikipedia.org/wiki/Geometric_progression

    InputThe first line contains a single integer TT, indicating the number of cases.
    For each test case, the first line contains a single integer nn, the second line contains nn integers a1,a2,,ana1,a2,⋯,an.

    T104,n106T≤104,∑n≤106
    OutputFor each test case, output one line with a single integer, representing the answer.Sample Input

    2
    5
    1 2 3 4 6
    10
    1 1 1 1 1 1 2 3 4 5

    Sample Output

    4
    6

    分析:分别对等差数列和等比数列进行尺取,找其中的最大值
    代码如下:
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    typedef long long ll;
    const int MAXN=1e6+10;
    ll a[MAXN];
    ll Q[MAXN];
    ll t,n,l,r,mid,maxx;
    void check1()
    {
        ll l=0,r=0,k,num;
        Q[r++]=a[0];
        Q[r++]=a[1];
        num=0;
        while(1)
        {
            while(r<n&&a[r]-Q[r-1]==Q[r-1]-Q[r-2])
            {
                Q[r]=a[r];
                r++;
            }
            num=max(r-l,num);
            if(r==n)break;
            Q[r]=a[r];
            r++;
            l=r-2;
        }
        maxx=max(num,maxx);
    }
    
    void check2()
    {
        ll l=0,r=0,k,num;
        Q[r++]=a[0];
        Q[r++]=a[1];
        num=0;
        while(1)
        {
            while(r<n&&a[r]*Q[r-2]==Q[r-1]*Q[r-1])
            {
                Q[r]=a[r];
                r++;
            }
            num=max(r-l,num);
            if(r==n)break;
            Q[r]=a[r];
            r++;
            l=r-2;
        }
          maxx=max(maxx,num);
    }
    int main()
    {
        scanf("%d",&t);
        while(t--)
        {
          maxx=0;
          scanf("%d",&n);
          for(int i=0;i<n;i++)
          scanf("%d",&a[i]);
          if(n<=2)
          {
              printf("%d
    ",n);
              continue;
          }
       /*   l=2;
          r=n;
           while(l+1<r)
           {
              mid=(l+r)/2;
              if(check1(mid)||check2(mid))
               l=mid;
              else
               r=mid;
           }*/
           check1();
           check2();
           printf("%lld
    ",maxx);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/a249189046/p/8094294.html
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