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)/;
         if(v==a[m])return m;
         else if(v<a[m]){
             r=m-;
         }else if(v>a[m]){
             l=m+;
         }
     }
     return -;
 }

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