CodeForces 607B：Zuma（区间DP）

Zuma

Time limit:1000 ms Memory limit:524288 kB

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Problem Description

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 ci. 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 ci (1 ≤ ci ≤ 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.

Example

Input 3 1 2 1 Output 1

Input 3 1 2 3 Output 3

Input 7 1 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.

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题意：

给出n个数字，我们来逐步删除，如果一个子串是回文数，就可以直接删除，问删除完最少的次数。

解题思路：

设dp[i][j]为删除区间[i,j]所需最小次数。。算了讲不清楚的

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Code：

#include <iostream>#include <cstdio>using namespace std;const int maxn=505;int a[maxn];int dp[maxn][maxn];int dfs(int i,int j){    if(i>=j)        return 1;    if(dp[i][j])        return dp[i][j];    int ans=1e9;    if(a[i]==a[j])        ans=min(ans,dfs(i+1,j-1));    for(int k=i;k<j;k++)        ans=min(ans,dfs(i,k)+dfs(k+1,j));    return dp[i][j]=ans;}int main(){    int n;    scanf("%d",&n);    for(int i=0; i<n; i++)    {        scanf("%d",a+i);    }    printf("%d\n",dfs(0,n-1));    return 0;}