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啊 忘记不要涂鸦了 不好意思啊。。。

Posted by oyjpart at 2007-02-16 15:59:01 on Problem 1013
In Reply To:枚举的思路 Posted by:oyjpart at 2007-02-16 15:03:58
> #include <stdio.h>
> int main() {
> 	int ntc, i, j, k, d[3], l;
> 	char a[3][7], b[3][7], tmp[5];
> 	scanf("%d", &ntc);
> 	while(ntc--) {
> 		for(i = 0; i<3; i++) {
> 			scanf("%s %s %s", a[i], b[i], tmp);
> 			if(tmp[0] == 'd') d[i] = -1;
> 			else d[i] = tmp[0] == 'u';
> 		}
> 		for(i = 0; i<12; i++) 
> 			for(l = -1; l<=1; l+= 2) {
> 				for(j = 0; j<3; j++) {
> 					int wa = 0, wb = 0;
> 					for(k = 0; a[j][k]; k++) wa += (a[j][k]-'A'==i)*l;
> 					for(k = 0; b[j][k]; k++) wb += (b[j][k]-'A'==i)*l;
> 					if(wa - wb != d[j])		break;
> 				}
> 				if(j != 3) continue;
> 				printf("%c is the counterfeit coin and it is ", (char)(i+'A'));
> 				if(l == 1) printf("heavy.\n");
> 				else printf("light.\n");
> 			}
> 	}
> 	return 0;
> }

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