50%的数据满足：1 <= k <= 5, 1 <= ci <= 3
100%的数据满足：1 <= k <= 15, 1 <= ci <= 5
【题解】
本题中ci很小，因此可以直接5维保存可以涂i块的油漆有多少种颜色。然后利用乘法原理进行DP
1 #include<cstdio> 2 #include<algorithm> 3 #define LL long long 4 #define rg register 5 #define N 16 6 #define Mod (1000000007) 7 using namespace std; 8 LL f[N][N][N][N][N][6]; 9 bool v[N][N][N][N][N][6]; 10 int n,cnt[6]; 11 inline int read(){ 12 int k=0,f=1; char c=getchar(); 13 while(c<'0'||c>'9')c=='-'&&(f=-1),c=getchar(); 14 while('0'<=c&&c<='9')k=k*10+c-'0',c=getchar(); 15 return k*f; 16 } 17 LL dfs(int a,int b,int c,int d,int e,int last){ 18 if(v[a][b][c][d][e][last]) return f[a][b][c][d][e][last]; 19 if(a+b+c+d+e==0) return 1; 20 LL tmp=0; 21 if(a) tmp+=(a-(last==2))*dfs(a-1,b,c,d,e,1); 22 if(b) tmp+=(b-(last==3))*dfs(a+1,b-1,c,d,e,2); 23 if(c) tmp+=(c-(last==4))*dfs(a,b+1,c-1,d,e,3); 24 if(d) tmp+=(d-(last==5))*dfs(a,b,c+1,d-1,e,4); 25 if(e) tmp+=e*dfs(a,b,c,d+1,e-1,5); 26 v[a][b][c][d][e][last]=1; 27 return f[a][b][c][d][e][last]=(tmp%Mod); 28 } 29 int main(){ 30 n=read(); 31 for(rg int i=1;i<=n;i++){ 32 int x=read(); 33 cnt[x]++; 34 } 35 printf("%lld ",dfs(cnt[1],cnt[2],cnt[3],cnt[4],cnt[5],0)); 36 return 0; 37 }