20 分 PAT:B1018 锤子剪刀布( 二 )

< N; i++) {a = in.next().charAt(0);b = in.next().charAt(0);if(a == 'B' && b == 'C') {aHand[0]++;} else if(a == 'B' && b == 'J') {bHand[2]++;} else if(a == 'C' && b == 'B') {bHand[0]++;} else if(a == 'C' && b == 'J') {aHand[1]++;} else if(a == 'J' && b == 'B') {aHand[2]++;} else if(a == 'J' && b == 'C') {bHand[1]++;}}in.close();aWin = aHand[0] + aHand[1] + aHand[2];bWin = bHand[0] + bHand[1] + bHand[2];count = N - aWin - bWin;System.out.printf("%d %d %d\n", aWin, count, bWin);System.out.printf("%d %d %d\n", bWin, count, aWin);int aC = 0, bC = 0;for(int i = 0; i < 3; i++) {if(aMax < aHand[i]) {aMax = aHand[i];aC = i;}if(bMax < bHand[i]) {bMax = bHand[i];bC = i;}}System.out.printf("%c %c", hand[aC], hand[bC]);}}