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  1. #include <stdio.h>
  2.  
  3. int main(void) {
  4. 	int a[9]={0,};
  5. 	int result[19]={0,};
  6. 	double per[19]={0,};
  7. 	double sum_p=0;
  8. 	int sum=0;
  9. 	int k=0;
  10. 	for(a[0]=-1;a[0]<=1;a[0]++){
  11. 		for(a[1]=-1;a[1]<=1;a[1]++){
  12. 			for(a[2]=-1;a[2]<=1;a[2]++){
  13. 				for(a[3]=-1;a[3]<=1;a[3]++){
  14. 					for(a[4]=-1;a[4]<=1;a[4]++){
  15. 						for(a[5]=-1;a[5]<=1;a[5]++){
  16. 							for(a[6]=-1;a[6]<=1;a[6]++){
  17. 								for(a[7]=-1;a[7]<=1;a[7]++){
  18. 									for(a[8]=-1;a[8]<=1;a[8]++){
  19. 										sum=0;
  20. 										for(k=0;k<9;k++) sum+=a[k];
  21. 										sum=sum+9;
  22. 										result[sum]++;
  23. 									}
  24. 								}
  25. 							}
  26. 						}
  27. 					}
  28. 				}
  29. 			}
  30. 		}
  31. 	}
  32.  
  33. 	//3^9=19683
  34.  
  35. 	for(k=0;k<19;k++){
  36. 		per[k]=(double)result[k]/(19683);
  37. 	}
  38. 	sum_p=0;
  39. 	for(k=0;k<19;k++){//self=1
  40. 		if(k-9<-1) sum_p+=per[k];
  41. 	}
  42. 	printf("you=1, win=%f\n",1-sum_p-per[8]/2);
  43. 	sum_p=0;
  44. 	for(k=0;k<19;k++){//self=0
  45. 		if(k-9<0) sum_p+=per[k];
  46. 	}
  47. 	printf("you=0, win=%f\n",1-sum_p-per[9]/2);
  48. 	sum_p=0;
  49. 	for(k=0;k<19;k++){//self=-1
  50. 		if(k-9<1) sum_p+=per[k];
  51. 	}
  52. 	printf("you=-1, win=%f\n",1-sum_p-per[10]/2);
  53.  
  54. 	return 0;
  55. }
  56.  
Success #stdin #stdout 0s 2248KB
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stdin
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Standard input is empty
stdout
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you=1, win=0.653584
you=0, win=0.500000
you=-1, win=0.346416
https://ideone.com/KkL6Oj
language:
C (gcc 8.3)
created:
10 years ago
visibility:
public

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