spoj periodni

题解：

dp

方程弄出来就好做了

代码：

#include<bits/stdc++.h> 
const int N=505,M=1000000007;
typedef int arr[N];
typedef long long ll;
int n,k,c[N][N],h[N];
arr g;
void add(int &x,int y){x+=y;if (x>=M)x-=M;}
void dfs(int l,int r,int H,arr &ans)
{
    memset(ans,0,sizeof(ans));
    ans[0]=1;
    int p=l,m=r-l+1;
    for (int i=l;i<=r;i++)
     if (h[i]<h[p]) p=i;
    arr L,R;
    L[0]=R[0]=1;
    int h0=H;H=h[p]-H;
    if (l<p)dfs(l,p-1,h[p],L);
    if (p<r)dfs(p+1,r,h[p],R);
    memset(g,0,sizeof(g));
    for (int i=0;i<=p-l;i++)
     for (int j=m;j>=i;j--)
      (g[j]+=(ll)ans[j-i]*L[i]%M)%=M;
    memcpy(ans,g,sizeof(ans));
    memset(g,0,sizeof(g));
    for (int i=0;i<=r-p;i++)
     for (int j=m;j>=i;j--)
      (g[j]+=(ll)ans[j-i]*R[i]%M)%=M;
    memcpy(ans,g,sizeof(ans));
    memset(g,0,sizeof(g));
    ll P=1;
    for (int i=0;i<=m&&i<=H;i++)
     {
        for (int j=m;j>=i;j--)
         (g[j]+=(ll)ans[j-i]*c[m-j+i][i]%M*P%M)%=M;
        P=P*(H-i)%M;
     }
    memcpy(ans,g,sizeof(ans));
}
int main()
{
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;i++)scanf("%d",&h[i]);
    c[0][0]=1;
    for (int i=1;i<=500;i++)
     {
        c[i][0]=1;
        for (int j=1;j<=i;j++)
         c[i][j]=(c[i-1][j]+c[i-1][j-1])%M;
     }
    arr ans;
    dfs(1,n,0,ans);
    printf("%d\n",ans[k]);
    return 0;
}