> 算法：硬币状态 -1：light, 0：normal, 1: heavy.
>       天平状态 -1：down, 0: even, 1: heavy.
>       枚举24中可能。
>       利用  ((coinStatus[input[0][1]] + coinStatus[input[0][2]] + coinStatus[input[0][3]]+ coinStatus[input[0][4]] - coinStatus[input[0][5]] - coinStatus[input[0][6]] - coinStatus[input[0][7]]-coinStatus[input[0][8]]  ==   input[0][0])  这种方法判断枚举是否正确。 
> 
> #include<iostream>
> using namespace std;
> 
> void main() {
> 	char line1[5] = {'0'};
> 	char line2[5] = {'0'};
> 	char line3[5] = {'0'};
> 	int input[3][9];
> 	int coinStatus[12] = { 0 };
> 	int nn = 0;
> 	cin >> nn;
> 	while (nn--){
> 		for(int i = 0; i < 3; i++) {
> 			cin >> line1 >> line2 >> line3;
> 			for(int j = 1; j < 5; j++) { input[i][j] = line1[j-1] - 65;}
> 			for(int k = 1; k < 5; k++) { input[i][k+4] = line2[k-1] - 65;}
> 			switch (*line3) {
> 			case 'e' : input[i][0] = 0; break;
> 			case 'd' : input[i][0] = -1; break;
> 			case 'u' : input[i][0] = 1; break;
> 			default: ;
> 			}
> 		}
> 		int index = 0;
> 		for(index = 0; index < 12; index ++) {
> 			coinStatus[index] = 1;
> 			if((coinStatus[input[0][1]]+coinStatus[input[0][2]]+coinStatus[input[0][3]]+coinStatus[input[0][4]]
> 				-coinStatus[input[0][5]]-coinStatus[input[0][6]]-coinStatus[input[0][7]]-coinStatus[input[0][8]]
> 				== input[0][0]) && (coinStatus[input[1][1]]+coinStatus[input[1][2]]+coinStatus[input[1][3]]+coinStatus[input[1][4]]
> 				-coinStatus[input[1][5]]-coinStatus[input[1][6]]-coinStatus[input[1][7]]-coinStatus[input[1][8]]
> 				== input[1][0]) && (coinStatus[input[2][1]]+coinStatus[input[2][2]]+coinStatus[input[2][3]]+coinStatus[input[2][4]]
> 				-coinStatus[input[2][5]]-coinStatus[input[2][6]]-coinStatus[input[2][7]]-coinStatus[input[2][8]]
> 				== input[2][0])) cout << (char)(index + 65) << " is the counterfeit coin and it is heavy. " << endl;
> 			coinStatus[index] = -1;
> 			if((coinStatus[input[0][1]]+coinStatus[input[0][2]]+coinStatus[input[0][3]]+coinStatus[input[0][4]]
> 				-coinStatus[input[0][5]]-coinStatus[input[0][6]]-coinStatus[input[0][7]]-coinStatus[input[0][8]]
> 				== input[0][0]) && (coinStatus[input[1][1]]+coinStatus[input[1][2]]+coinStatus[input[1][3]]+coinStatus[input[1][4]]
> 				-coinStatus[input[1][5]]-coinStatus[input[1][6]]-coinStatus[input[1][7]]-coinStatus[input[1][8]]
> 				== input[1][0]) && (coinStatus[input[2][1]]+coinStatus[input[2][2]]+coinStatus[input[2][3]]+coinStatus[input[2][4]]
> 				-coinStatus[input[2][5]]-coinStatus[input[2][6]]-coinStatus[input[2][7]]-coinStatus[input[2][8]]
> 				== input[2][0])) cout << (char)(index + 65) << " is the counterfeit coin and it is light. " << endl;
> 			coinStatus[index] = 0;
> 		}
> 	}
> }
> 
> 
> 
> 

• Re:没说两边一定是4枚硬币呀，只说数目相等 (0) C0400203025 2004-02-24 14:14:50
  • Re:没说两边一定是4枚硬币呀，只说数目相等 (3) C0400203025 2004-02-24 14:15:05