Zuma （区间DP）

Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color c_i. The goal of the game is to destroy all the gemstones in the line as quickly as possible.

In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?

Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.

Input

The first line of input contains a single integer n (1?≤?n?≤?500) — the number of gemstones.

The second line contains n space-separated integers, the i-th of which is c_i (1?≤?c_i?≤?n) — the color of the i-th gemstone in a line.

Output

Print a single integer — the minimum number of seconds needed to destroy the entire line.

Examples

Input

31 2 1

Output

1

Input

31 2 3

Output

3

Input

71 4 4 2 3 2 1

Output

2

Note

In the first sample, Genos can destroy the entire line in one second.

In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.

In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.

题目大意：

先输入n，然后输入n个数字，每次操作可以将该序列的某一个回文串去掉，求最少需要多少次才能将该序列消完。

#include <bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
int a[505],dp[505][505];
int main()
{
    memset(dp,INF,sizeof dp);
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>a[i];
    for(int i=1;i<=n;i++)
        dp[i][i]=1;
    for(int l=1;l<n;l++)
        for(int i=1;i+l<=n;i++)
        {
            int j=i+l;
            if(l==1)
            {
                if(a[i]==a[j]) dp[i][j]=1;
                else dp[i][j]=2;
            }
            else
            {
                if(a[i]==a[j]) dp[i][j]=dp[i+1][j-1];
                for(int k=i;k<j;k++)
                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
            }
        }
    cout<<dp[1][n]<<‘\n‘;
    return 0;
}

原文地址：https://www.cnblogs.com/zdragon1104/p/9319914.html