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  1. #include <bits/stdc++.h>
  2.  
  3. using namespace std;
  4. typedef long long ll;
  5.  
  6. const int MX=43;
  7.  
  8. void fnc(){
  9. 	array<array<ll, MX+5>, MX+5> cmb;
  10. 	for(int i=0; i<MX+5; i++)  cmb[i][0] = 1;
  11. 	for(int i=1; i<MX+5; i++) for(int j=1; j<=i;j++) 
  12. 		cmb[i][j] = cmb[i-1][j] + cmb[i-1][j-1];
  13.  
  14.  
  15. 	ll sm = 0;
  16. 	for(int i=3 ; i<=6; i++){
  17. 		ll w = cmb[6][i] * cmb[MX-6][6-i]; //当たりがi個の組み合わせ数
  18. 		sm += w;
  19. 		printf("%d : %6lld\n", i, w);
  20. 	}
  21.  
  22. 	printf("\n%lld / %lld --> %.15f  by comb\n", 
  23. 		sm, cmb[MX][6], 1.0*sm/cmb[MX][6]);
  24.  
  25. }
  26.  
  27. void fnc2(){
  28. 	vector<double> dp(8); //dp[あたり数] = 確率
  29. 	dp[0] =1.0;
  30. 	for(int i=0; i<6; i++){
  31. 		vector<double> np(8);
  32. 		for(int j=0; j<6; j++){
  33. 			int atr = 6-j; //残り当たり数
  34. 			np[j+1] += dp[j] * atr/(MX-i);
  35. 			np[j]   += dp[j] * (MX-i-atr)/(MX-i);
  36. 		}
  37. 		dp.swap(np);
  38. 	}
  39. 	double ans =0;
  40. 	for(int i=3; i<7; i++) ans +=dp[i];
  41. 	printf("\n%.15f  by dp\n",  ans);
  42. }
  43.  
  44.  
  45. int main()
  46. {
  47. 	fnc();  //コンビネーションで求める
  48. 	fnc2(); //確率DPで求める
  49. }
  50.  
Success #stdin #stdout 0s 4296KB
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stdin
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Standard input is empty
stdout
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3 : 155400
4 :   9990
5 :    222
6 :      1

165613 / 6096454 --> 0.027165463726947  by comb

0.027165463726947  by dp
https://ideone.com/gCsPec
ロト６
language:
C++14 (gcc 8.3)
created:
7 years ago
visibility:
secret

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