洛谷P1328生活大爆炸版石頭剪刀布

題目
這道題只給了一部分表格，剩下的需要自己去推，我最開始以爲灰色部分屬於不存在的組合，不進行判斷。

我的做法

#include <iostream>
#include <cmath>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAX_SIZE = 200;

int Awin(int a, int b)
{
    if (a == 0) {
        switch (b) {
            case 0: return 0;
            case 1: return -1;
            case 2: return 1;
            case 3: return 1;
            case 4: return -1; 
        }
    }
    else if (a == 1) {
        switch (b) {
            case 0: return 1;
            case 1: return 0;
            case 2: return -1;
            case 3: return 1;
            case 4: return -1;
        }
    }
    else if (a == 2) {
        switch (b) {
            case 0: return -1;
            case 1: return 1;
            case 2: return 0;
            case 3: return -1;
            case 4: return 1;
        }
    }
    else if (a == 3) {
        switch (b) {
            case 0: return -1;
            case 1: return -1;
            case 2: return 1;
            case 3: return 0;
            case 4: return 1;
        }
    }
    else {
        switch (b) {
            case 0: return 1;
            case 1: return 1;
            case 2: return -1;
            case 3: return -1;
            case 4: return 0;
        }
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int N, NA, NB;
    cin >> N >> NA >> NB;
    int A[MAX_SIZE], B[MAX_SIZE];
    memset(A, -1, sizeof(A));
    memset(B, -1, sizeof(B));
    for (int i = 0; i < NA; i++) {
        cin >> A[i];
    }
    for (int i = 0; i < NB; i++) {
        cin >> B[i];
    }
    int cnt_A = 0, cnt_B = 0, idx_A = 0, idx_B = 0;
    for (int i =  0; i < N; i++, idx_A++, idx_B++) {
        if (A[idx_A] == -1) {
            idx_A = 0;
        }
        if (B[idx_B] == -1) {
            idx_B = 0;
        }
        if (Awin(A[idx_A], B[idx_B]) == 1) {
            cnt_A++;
        } 
        else if (Awin(A[idx_A], B[idx_B]) == -1 ) {
            cnt_B++;
        } 
        else if (Awin(A[idx_A], B[idx_B]) == 2) {
            i--;
        }
    }
    cout << cnt_A << ' ' << cnt_B << endl;
    return 0;
}

題解裏看到的簡易做法

#include <iostream>
#include <cmath>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAX_SIZE = 200;
const int pk[5][5] = {{0, -1, 1, 1, -1},//利用二維數組保存狀態
                      {1, 0, -1, 1, -1},
                      {-1, 1, 0, -1, 1},
                      {-1, -1, 1, 0, 1},
                      {1, 1, -1, -1, 0}};
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int N, NA, NB;
    cin >> N >> NA >> NB;
    int A[MAX_SIZE], B[MAX_SIZE];
    for (int i = 0; i < NA; i++) {
        cin >> A[i];
    }
    for (int i = 0; i < NB; i++) {
        cin >> B[i];
    }
    int cnt_A = 0, cnt_B = 0, idx_A = 0, idx_B = 0;
    for (int i =  0; i < N; i++, idx_A++, idx_B++) {
        if (pk[A[idx_A % NA]][B[idx_B % NB]] == 1) {
            cnt_A++;
        }
        else if (pk[A[idx_A % NA]][B[idx_B % NB]] == -1) {
            cnt_B++;
        }
    }
    cout << cnt_A << ' ' << cnt_B << endl;
    return 0;
}