一種典型的組合數DP 

f[i][j]表示前i種數值排列有j個不合法間隙的方案數

然後列舉第i個分成幾塊k 有l塊塞入j個不合法空隙中進行轉移

答案即為f[n][0]

不過還要乘以排列數

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#define cl(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long ll;

inline char nc(){
  static char buf[100000],*p1=buf,*p2=buf;
  if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
  return *p1++;
}

inline void read(int &x){
  char c=nc(),b=1;
  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=55;
const ll P=1e9+7;

int n;
int cnt[N],sum[N];
ll dp[N][555];
ll C[555][555],fac[555];

inline void Pre(){
  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-1]+C[i-1][j])%=P;
  }
  fac[0]=1;
  for (int i=1;i<=500;i++)
    fac[i]=fac[i-1]*i%P;
}

int main(){
  int T,test=0;
  freopen("t.in","r",stdin);
  freopen("t.out","w",stdout);
  Pre();
  read(T);
  while (T--){
    read(n);
    for (int i=1;i<=n;i++) read(cnt[i]),sum[i]=sum[i-1]+cnt[i];
    dp[1][cnt[1]-1]=1;
    for (int i=2;i<=n;i++)
      for (int j=0;j<=sum[i-1]+1;j++)
	if (dp[i-1][j])
	  for (int k=1;k<=cnt[i];k++)
	    for (int l=0;l<=k;l++)
	      (dp[i][j-l+cnt[i]-k]+=dp[i-1][j]*C[j][l]%P*C[sum[i-1]+1-j][k-l]%P*C[cnt[i]-1][k-1]%P)%=P;
    ll ret=dp[n][0];
    for (int i=1;i<=n;i++)
      ret*=fac[cnt[i]],ret%=P;
    printf("Case %d: %lld\n",++test,ret);
    cl(dp);
  }
  return 0;
}