Bzoj1079: [scoi2008] coloring scheme

1079: [scoi2008] coloring scheme time limit: 10 sec memory limit: 162 MB
Submit: 822 solved: 525
[Submit] [Status] Description

N blocks are arranged in one row, numbered from left to right as 1 ~ N. You have k colors of paint, where the I-color paint is enough to paint CI blocks. All paint is enough to fill all blocks, that is, C1 + C2 +... + ck = n. It is difficult to paint two adjacent blocks in the same color, so you want to calculate the color of any two adjacent blocks.

Input

The first row is a positive integer k, and the second row contains k integers C1, C2,..., CK.

Output

Output an integer, that is, the result of the total number of Modulo 1,000,000,007.

Sample input3
1 2 3
Sample output10
Hint

 

[Example 2] Input 5 2 2 2 2 2 output 39480 [Example 3] Input 10 1 1 1 2 2 3 3 4 5 5 output 85937576 data scale] 50% of data meet: 1 <= k <= 5, 1 <= CI <= 3 100% of data satisfied: 1 <= k <= 15, 1 <= CI <= 5

 

Source

Question:

I thought for a morning that I could not do anything...

The difficulty of this question is that, if the State is represented by 15 numbers ranging from 1 to 5, t is inevitable... Zookeeper did not expect the remaining several identical paints to be treated equally. It would only take one time...

If I have been violent, maybe I can think of it?

Code:

 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set>10 #include<queue>11 #define inf 1<<3012 #define maxn 500+10013 #define maxm 500+10014 #define eps 1e-1015 #define ll long long16 #define pa pair<int,int>17 #define mod 100000000718 using namespace std;19 inline int read()20 {21     int x=0,f=1;char ch=getchar();22     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}23     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}24     return x*f;25 }26 ll f[16][16][16][16][16][6];27 bool mark[16][16][16][16][16][6];28 int n,x,a[6];29 ll dp(int a,int b,int c,int d,int e,int k)30 {31    if(mark[a][b][c][d][e][k])return f[a][b][c][d][e][k];32    if(a+b+c+d+e==0)return 1;33    ll t=0;34    if(a)t+=(a-(k==2))*dp(a-1,b,c,d,e,1);35    if(b)t+=(b-(k==3))*dp(a+1,b-1,c,d,e,2);    36    if(c)t+=(c-(k==4))*dp(a,b+1,c-1,d,e,3);37    if(d)t+=(d-(k==5))*dp(a,b,c+1,d-1,e,4);38    if(e)t+=e*dp(a,b,c,d+1,e-1,5);39    mark[a][b][c][d][e][k]=1;40    return f[a][b][c][d][e][k]=(t%mod);41 }42 int main()43 {44     freopen("input.txt","r",stdin);45     freopen("output.txt","w",stdout);46     n=read();47     while(n--)x=read(),a[x]++;48     printf("%lld\n",dp(a[1],a[2],a[3],a[4],a[5],0));49     return 0;50 }

View code

 

Bzoj1079: [scoi2008] coloring scheme