題解

計算對手和自身已有牌的和並對M取模 計算過程中統計剩餘牌的數量
嘗試所有剩餘的牌 如果可以贏則贏 不能贏能平就平

AC程式碼

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;

int toVal(char *c) //將牌的字串轉為數值
{
	if (strlen(c) == 2)
		return 10;
	if (c[0] == 'A')
		return 1;
	if
 
 (c[0] == 'J')
		return 11;
	if (c[0] == 'Q')
		return 12;
	if (c[0] == 'K')
		return 13;
	return c[0] - '0';
}
int main()
{
#ifdef LOCAL
	freopen("C:/input.txt", "r", stdin);
#endif
	int T;
	cin >> T;
	while (T--)
	{
		int a[14] = { 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 }; //剩餘牌數量
		char s[10];
		scanf
 
("%s", s);
		int M = toVal(s);
		a[M]--; //抽出的牌數量需要減少
		int x = 0, y = 0, z = 0, v;
		for (int i = 0; i < 3; i++) //倆對手求和並取模
			scanf("%s", s), v = toVal(s), a[v]--, x = (x + v) % M; 
		for (int i = 0; i < 3; i++)
			scanf("%s", s), v = toVal(s), a[v]--, y = (y + v) % M;
		for (int i = 0; i < 2; i++)
 

			scanf("%s", s), v = toVal(s), a[v]--, z = (z + v) % M;
		scanf("%s", s);
		int flag = -1; //-1輸 0不會輸 1贏
		for (int i = 1; i <= 13; i++)
			if (a[i]) //還有牌
			{
				if ((z + i) % M > max(x, y)) 
					flag = max(flag, 1); //儘量取最優
				if ((z + i) % M == max(x, y))
					flag = max(flag, 0);
			}
		if (flag == -1)
			cout << "I'm a loser" << endl;
		else if (flag == 0)
			cout << "I may not lose" << endl;
		else
			cout << "I can win" << endl;
	}

	return 0;
}