codeforces 607 B 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 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.

Examples

input

31 2 1

output

1

input

31 2 3

output

3

input

71 4 4 2 3 2 1

output

2

solution；

    我们用dp[i][j]代表从i到j消除需要的最小步数，那么我们可以得到递推公式，dp[i][j]=min(dp[i][k]+dp[k+1][j])(k>i&&k<j),但是当a[i]==a[j]时，这时候要特判dp[i][j]=min(dp[i][j],dp[i+1][j-1]);

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<map>#include<queue>#include<stack>#include<string>#include<iostream>#include<set>using namespace std;typedef long long ll;#define mem(a) (memset(a,0,sizeof(a)))#define f0(n) for(int i=0;i<n;i++)#define f1(n) for(int i=1;i<=n;i++)#define si(x) scanf("%d",&x)#define sl(x) scanf("%I64d",&x)#define sd(x) scanf("%lf",&x)#define sii(x,y) scanf("%d%d",&x,&y)#define siii(x,y,z) scanf("%d%d%d",&x,&y,&z)#define mp(x,y) make_pair(x,y)const int maxn = 1e3 + 500;int n, m;int dp[maxn][maxn], a[maxn];int main(){si(n);f1(n)si(a[i]);for (int p = 0; p <= n; p++){for (int i = 1; i <= n; i++){int j = i + p;if (j > n)break;if (i == j){dp[i][j] = 1;continue;}if (i + 1 == j){dp[i][j] = (a[i] == a[j] ? 1 : 2);continue;}dp[i][j] = 1e9;for (int k = i; k < j; k++)dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);if (a[i] == a[j])dp[i][j] = min(dp[i][j], dp[i + 1][j - 1]);}}printf("%d\n", dp[1][n]);return 0;}