Submission #853222

#TimeUsernameProblemLanguageResultExecution timeMemory
853222alexddFishing Game (RMI19_fishing)C++17
0 / 100
417 ms218160 KiB
#include<bits/stdc++.h> using namespace std; #define int long long const int MOD = 1e9+7; int dp[305][305][305]; int cardA[205]; int cardB[205]; int cardC[205]; int n; signed main() { ios_base::sync_with_stdio(0);cin.tie(0); int t; cin>>n>>t; while(t--) { for(int i=1;i<=2*n;i++) cin>>cardA[i]; for(int i=1;i<=2*n;i++) cin>>cardB[i]; for(int i=1;i<=2*n;i++) cin>>cardC[i]; int cntab=0,cntbc=0,cntca=0; for(int i=1;i<=2*n;i++) { for(int j=1;j<=2*n;j++) { if(cardA[i]==cardB[j]) cntab++; if(cardB[i]==cardC[j]) cntbc++; if(cardC[i]==cardA[j]) cntca++; } } for(int i=0;i<=3*n;i++) for(int j=0;j<=3*n;j++) for(int k=0;k<=3*n;k++) dp[i][j][k]=0; dp[cntab][cntbc][cntca]=1; for(int tot=cntab+cntbc+cntca;tot>0;tot--) { for(int ab=0;ab<=tot;ab++) { for(int bc=0;bc+ab<=tot;bc++) { int ca = tot-ab-bc; if(dp[ab][bc][ca]==0) continue; int prod=1; int init=dp[ab][bc][ca]; for(int a=0;a<2;a++) { if((a==0 && ab==0) || (a==1 && ca==0)) continue; if(a==0)///scoatem ab { prod*=ab; ab--; } else///scoatem ca { prod*=ca; ca--; bc++; } for(int b=0;b<2;b++) { if((b==0 && bc==0) || (b==1 && ab==0)) continue; if(b==0)///scoatem bc { prod*=bc; bc--; } else///scoatem ab { prod*=ab; ab--; ca++; } for(int c=0;c<2;c++) { if((c==0 && ca==0) || (c==1 && bc==0)) continue; if(c==0)///scoatem ca { prod*=ca; ca--; } else///scoatem bc { prod*=bc; bc--; ab++; } dp[ab][bc][ca]+=(init*prod)%MOD; dp[ab][bc][ca]%=MOD; if(c==0)///scoatem ca { ca++; prod*=ca; } else///scoatem bc { bc++; ab--; prod*=bc; } } if(b==0)///scoatem bc { bc++; prod/=bc; } else///scoatem ab { ab++; ca--; prod/=ab; } } if(a==0)///scoatem ab { ab++; prod/=ab; } else///scoatem ca { ca++; bc--; prod/=ca; } } } } } cout<<dp[0][0][0]<<"\n"; } return 0; } /** dp[ab][bc][ca] = numarul de moduri de a ajunge la o configuratie in care exista ab perechi care au un element in A si celalalt in B dp[ab-1][bc-1][ca-1] += dp[ab][bc][ca] * ab * bc * ca daca luam din A un ab, luam din B un bc, luam din C un ca dp[ab-1-1][bc][ca-1+1] += dp[ab][bc][ca] daca luam din A un ab, luam din B un ab, luam din C un ca */
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