[PAT B1018] hammer and scissors

We should be able to play "hammer and scissors" game: they both give gesture, as shown in Figure outcome of the rules:

Now given the two men clash records, statistics of both wins, flat, negative number, and the two sides are what give the greatest chance of winning gesture.
Input format:
Enter line 1 gives a positive integer N (<= 105), i.e., both the number of confrontation. Then N rows, each row gives information confrontation, i.e., the gesture and B sides at the same time is given. C stands for "hammer", the representative J "scissors", B stands for "cloth", the first letter represents a party, on behalf of the second party, there is an intermediate space.
Output Format:
output the first and second lines are given A, B wins, flat, negative number, separated by an inter-digital space. Line 3 shows two letters represent A, B wins the highest number of gestures, there is an intermediate space. If the solution is not unique, the minimum solution alphabetically output.
Sample input:
10
CJ
JB
CB
BB
the BC
the CC
CB
JB
the BC
JJ
Output Sample:
. 5. 3 2
2. 3. 5
BB

#include <stdio.h>

int num(char c) {
    if (c == 'B') return 0;
    if (c == 'C') return 1;
    if (c == 'J') return 2;
}

int main() {
    int maxa = 0, maxb = 0;
    char mp[3] = {'B', 'C', 'J'};
    int timeA[3] = {0}, timeB[3] = {0}, winHandA[3], winHandB[3]; //time存放 胜 负 平 的次数
    char a, b;
    int numA = 0, numB = 0, N;


    scanf("%d", &N);

    for (int i = 0; i < N; i++) {
        getchar();//吸收空格
        scanf("%c %c", &a, &b);
        numA = num(a);
        numB = num(b);
        //取胜顺序 0--> 1 --> 2 --> 0

        if ((numA + 1) % 3 == numB) {
            //a胜
            timeA[0]++;
            timeB[1]++;
            winHandA[numA]++;
        } else if (numA == numB) {
            //平
            timeA[2]++;
            timeB[2]++;
        } else if ((numB + 1) % 3 == numA) {
            //b胜
            timeA[1]++;
            timeB[0]++;
            winHandB[numB]++;
        }

    }
    for (int i = 0; i < 3; i++) {
        if (winHandA[i] > winHandA[maxa]) //比较字典序
            maxa = i;
        if (winHandB[i] > winHandB[maxb])
            maxb = i;
    }
    printf("%d %d %d\n", timeA[0], timeA[2], timeA[1]);
    printf("%d %d %d\n", timeB[0], timeB[2], timeB[1]);
    printf("%c %c", mp[maxa], mp[maxb]);
}
/*
10
C J
J B
C B
B B
B C
C C
C B
J B
B C
J J
 */

Test Results: