TOJ 2926 Series

Description

An arithmetic series consists of a sequence of terms such that each term minus its immediate predecessor gives the same result. For example, the sequence 3, 7, 11, 15 is the terms of the arithmetic series 3+7+11+15; each term minus its predecessor equals 4. (Of course there is no requirement on the first term since it has no predecessor.)

Given a collection of integers, we want to find the longest arithmetic series that can be formed by choosing a sub-collection (possibly the entire collection).

Input

There are multiple cases, and each case contains 2 lines: the first line contains the count of integers (between 2 and 1000 inclusive), the following line contains all the integers (between -1,000,000,000 and 1,000,000,000 inclusive) separated by one or more spaces.

Output

Print a single number for each case in a single line.

Sample Input

7
3 8 4 5 6 2 2
4
-1 -5 1 3
4
-10 -20 -10 -10

Sample Output

5
3
3

Source

ZOJ Monthly, 2005.9

看了别人的解题报告说是DP+二分。试着写了一个后来发现与实际的结果相差2，不管三七二十一直接在原来的结果上加2。

竟然过了~o(╯□╰)o

cjx就在想这是为什么呢？后来发现因为我是从第三个数开始处理的，每个运算的结果都会少个2。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#define MAXN 1001
using namespace std;

int n,cnt;
int d[MAXN];
int a[MAXN];
int dp[MAXN][MAXN];

int find(int l, int r, int v){    
    while(l<=r){
        int m=(l+r)/2;
        if(v==a[m])return m;
        else if(v<a[m]){
            r=m-1;    
        }else if(v>a[m]){
            l=m+1;
        }
    }
    return -1;
}

int main()
{
    int ans,count;
    int temp;
     while( scanf("%d",&n)!=EOF ){
        for(int i=0; i<n; i++){
            scanf("%d",&d[i]);
        }
        sort(d,d+n);
        cnt=0;
        count=0;
        ans=0;
        temp=-1000000001;
        for(int i=0; i<n; i++){
            if(d[i]!=temp){
                a[cnt++]=d[i];
                if(count>ans)
                    ans=count;
                count=1;
            }else{
                count++;
            }
        }
        if(count>ans){
            ans=count;
        }
        memset(dp,0,sizeof(dp));
        for(int i=0; i<cnt; i++){
            for(int j=i+1; j<cnt; j++){
                int v=a[j]+a[j]-a[i];
                int k=-1;
                k=find(j+1,cnt-1,v);
                if(k!=-1){
                    dp[j][k]=dp[i][j]+1;
                    if(dp[j][k]>=ans){
                        ans=dp[j][k];
                    }
                }
            }
        }
        printf("%d
",ans+2);
     }
    return 0;
}