有一行 N 个弹子，每一个都有一个颜色。每次可以让超过 K 个的连续的同颜色的一段 弹子消失，剩下的会重新紧凑在一起。你有无限的所有颜色的弹子，要求在这行弹子中插入 最少的弹子，使得弹子全部消失。

The first line of input contains two integers N (1 ≤ N ≤ 100) and K (2 ≤ K ≤ 5) - the number of marbles in the initial sequence and the minimal number of consecutive marbles of the same color he could wish to vanish. The next line contains exactly N integers between 1 and 100 (inclusive), separated by one space. Those numbers represent colors of marbles in the sequence Mirko found.

The output should contain only one line with a single integer number - the minimal number of marbles Mirko has to insert to achive the desired effect.

10 4
3 3 3 3 2 3 1 1 1 3

4

---

 #include<bits/stdc++.h>

 const int maxn = ;

 int n,m,K,tmp[maxn];

 int a[maxn],c[maxn],f[maxn][maxn][maxn];

 bool vis[maxn][maxn][maxn];

 int read()

 {

     char ch = getchar();

     int num = ;

     for (; !isdigit(ch); ch=getchar());

     for (; isdigit(ch); ch=getchar())

         num = (num<<)+(num<<)+ch-;

     return num;

 }

 inline void Min(int &x, int y){x = x<y?x:y;}

 int dp(int l, int r, int t)

 {

     if (l > r) return ;

     if (vis[l][r][t]) return f[l][r][t];

     vis[l][r][t] = ;

     if (a[r]+t >= K) Min(f[l][r][t], dp(l, r-, ));

     else Min(f[l][r][t], dp(l, r, t+1)+1);　
　　　　//这里一开始想成dp(l, r, t-1)+1.实际上这里的状态表示的是逆推。也就是说f[]到完成消除的状态有多远。　

     for (int k=l; k<r; k++)

         if (c[k]==c[r])

             Min(f[l][r][t], dp(l, k, t+)+dp(k+, r-, ));

     return f[l][r][t];

 }

 int main()

 {

     memset(f, 0x3f3f3f, sizeof f);

     n = read(), K = read();

     for (int i=; i<=n; i++)

         c[i] = read(), a[i] = ;

     printf("%d\n",dp(, n, ));

     return ;

 }