记录编号	412478	评测结果	MMMMM
题目名称	[CCPC2015][HDU5548] 麻将 Mahjong	最终得分	0
用户昵称	liu_runda	是否通过	未通过
代码语言	C++	运行时间	0.000 s
提交时间	2017-06-09 15:57:56	内存使用	0.00 MiB

#include<cstdio>
const int mod=1000000007;
int K,M;
int f[330][330][5][5][5][5];
int main(){
  freopen("mahjong.in","r",stdin);
  freopen("mahjong.out","w",stdout);
  scanf("%d%d",&K,&M);
  f[0][0][0][0][4][4]=1;//(4,4):没有找出方案
  for(int i=0;i<K;++i){
    for(int j=0;j<=M;++j){
      for(int r1=0;r1<=4;++r1){
	for(int r2=r1;r2<=4;++r2){
	  if(r1==4&&r2==4)continue;
	  //f[i][j][r1][r2][4][4]
	  int x=f[i][j][r1][r2][4][4];
	  if(x==0)continue;
	  for(int k=r2;k<=4;++k){//第i种花色中使用的麻将数目
	    int res=k-r2;

	    if(res>=2){
	      if(res>=3){
		(f[i+1][j+k][r2-r1][r2-r1+res-3][r2-r1][r2-r1+res-2]+=x)%=mod;
	
	      }
	      else {

		(f[i+1][j+k][r2-r1][r2-r1+res][r2-r1][r2-r1+res-2]+=x)%=mod;
	      }
	    }
	    //碰?
	    
	    if(res>=3){

	      (f[i+1][j+k][r2-r1][r2-r1+res-3][4][4]+=x)%=mod;
	    }
	    else{

	      (f[i+1][j+k][r2-r1][r2-r1+res][4][4]+=x)%=mod;
	    }
	  }
	}
      }
      for(int r3=0;r3<=3;++r3){
	for(int r4=r3;r4<=4;++r4){
	  int x=f[i][j][4][4][r3][r4];
	  if(x==0)continue;

	  for(int k=r4;k<=4;++k){
	    
	    int res=k-r4;
	    if(res>=3){
	      
	      (f[i+1][j+k][4][4][r4-r3][r4-r3+res-3]+=x)%=mod;
	    }
	    else{

	      (f[i+1][j+k][4][4][r4-r3][r4-r3+res]+=x)%=mod;
	    }
	  }
	}
      }
      for(int r1=0;r1<4;++r1){
	for(int r2=r1;r2<=4;++r2){
	  if(r1==4&&r2==4)continue;
	  for(int r3=0;r3<=4;++r3){
	    for(int r4=r3;r4<=4;++r4){
	      if(r3==4&&r4==4)continue;
	      //f[i][j][r1][r2][r3][r4]
	      int x=f[i][j][r1][r2][r3][r4];
	      if(x==0)continue;
	      for(int k=0;k<=4;++k){
		if(k<r2){
		  if(k>=r4){
		    int res=k-r4;
		    if(res>=3){
		      
		      (f[i+1][j+k][4][4][r4-r3][r4-r3+res-3]+=x)%=mod;
		    }
		    else{
		      (f[i+1][j+k][4][4][r4-r3][r4-r3+res]+=x)%=mod;
		    }
		  }
		}else if(k<r4){
		  if(k>=r2){
		  }
		}else{
		  int res=k-r2;
		  if(res>=2){
		    int res2=k-r4;
		    if(res2>=3)res2-=3;
		    //if(res>=3)res-=3;
		    int nxt3=r2-r1,nxt4=r2-r1+res-2;
		    int NXT3=r4-r3,NXT4=r4-r3+res2;

		    if(nxt3!=NXT3||nxt4!=NXT4){
		      if(res>=3)res-=3;
		       (f[i+1][j+k][r2-r1][r2-r1+res][r4-r3][r4-r3+res2]+=x)%=mod;
		    }
		  }else{
		    int res2=k-r4;
		    if(res2>=3){
		      (f[i+1][j+k][r2-r1][r2-r1+res][r4-r3][r4-r3+res2-3]+=x)%=mod;
		    }
		    else{
		      (f[i+1][j+k][r2-r1][r2-r1+res][r4-r3][r4-r3+res2]+=x)%=mod;
		    }
		  }
		}
	      }
	    }
	  }
	}
      }
    }
  }
  int ans=0;
  for(int i=0;i<=4;++i){
    for(int j=0;j<=4;++j){
      ans=(ans+f[K][M][i][j][0][0])%mod;
    }
  }
  printf("%d\n",ans);
  return 0;
}
/*
 5 0 2 0 0 0
to 3 5 4 4 0 0
*/
/*
0 1 4 3
0 4 1 3 
3 1 1 3
1 2 2 3 
//f[3][5][1][1][4][4]
0 2 2 4
//f[3][4][2][2][4][4]
0 3 3 2
3 0 3 2 
3 3 0 2
1 1 4 2 
1 4 1 2 
4 1 1 2
2 2 2 2 
//f[3][6][0][0][4][4]

*/