PAT乙級 1017 A除以B (20分) & 1018 錘子剪刀布 (20分) &1019 數字黑洞 (20分)

1017 A除以B (20分)

其實我覺得這題就不簡單，可能是我沒有刷過題目，使用正常的手算的方式來解題即可。這個位置是最關鍵的。把第i位或者第i-1位的餘數作爲被除數和除數相除。

int temp1 = temp / b;
aa[i] = temp1;//除數的結果保存到aa
temp = temp % b * 10 + (a[i + 1] - '0');

#include<iostream>
#include<vector>
#include<string>
using namespace std;

int main()
{
	string a;
	int b = 0;
	int aa[1001] = { 0 };
	cin >> a >> b;
	int jieguo=0;
	int temp = (a[0]-'0');//保留餘數，同時作爲被除數的第一位
	if (a.size() == 1)
	{
		jieguo = (a[0] - '0') / b;
		temp = (a[0] - '0') % b;
		cout << jieguo << " " << temp;
		return 0;
	}
	for (int i = 0; i < a.size(); i++)
	{
		if (i != a.size() - 1)//當到a.size（）時會數組越界需要執行else
		{
			int temp1 = temp / b;
			aa[i] = temp1;//除數的結果保存到aa
			temp = temp % b * 10 + (a[i + 1] - '0');
		}
		else
		{
			aa[i] = temp / b;
			temp = temp % b;
		}
	}

	if (aa[0] == 0)//使用上述方法輸出結果會存在第一位爲0的情況所以需要區分
	{
		for (int i = 1; i < a.size(); i++)
			cout << aa[i];
		cout << " " << temp;
	}
	else
	{
		for (int i = 0; i < a.size(); i++)
			cout << aa[i];
		cout << " " << temp;
	}
	return 0;
}

1018 錘子剪刀布 (20分)

如果你是測試點1沒有通過說明你最後的輸出獲勝次數最多的手勢有問題。我直接用最粗暴的方式確定手勢，題目的意思是要是有相同的勝利次數就按照，bcj的順序輸出

    if (b[2] >= b[0] && b[2] >= b[1])maxjia = 2;
	if (b[0] > b[2] && b[0] >= b[1])maxjia = 0;
	if (b[1] > b[0] && b[1] > b[2])maxjia = 1;
	if (b[5] >= b[3] && b[5] >= b[4])maxyi = 5;
	if (b[3] > b[5] && b[3] >= b[4])maxyi = 3;
	if (b[4] > b[3] && b[4] > b[5])maxyi = 4;

全部代碼：

#include<iostream>
#include<vector>
#include<string>
using namespace std;

int main()
{
	int sum = 0;
	cin >> sum;
	int a[6] = { 0 };//前三個數字表示甲贏平輸的個數012，後三個表示乙的345
	int b[6] = { 0 };//前三個表示甲用石頭剪刀布勝利的個數012，後三個表示乙的345
	for (int i = 0; i < sum; i++)//c>j>b>c
	{
		char jia, yi;
		cin >> jia >> yi;
		if (jia == yi)
		{
			a[1]++; a[4]++; continue;
		}
		if (jia == 'C')
		{
			if(yi=='J')
			{
				a[0]++; a[5]++; b[0]++; continue;
			}
			else
			{
				a[2]++; a[3]++; b[5]++; continue;
			}
		}
		if (jia == 'J')
		{
			if (yi == 'C')
			{
				a[3]++; a[2]++; b[3]++; continue;
			}
			else
			{
				a[0]++; a[5]++; b[1]++; continue;
			}
		}
		if (jia == 'B')
		{
			if (yi == 'C')
			{
				a[0]++; a[5]++; b[2]++; continue;
			}
			else
			{
				a[3]++; a[2]++; b[4]++; continue;
			}
		}
	}
	int maxjia=2,maxyi=5;
	cout << a[0] << " " << a[1] << " " << a[2] << "\n" << a[3] << " " << a[4] << " " << a[5] << "\n";
	if (b[2] >= b[0] && b[2] >= b[1])maxjia = 2;
	if (b[0] > b[2] && b[0] >= b[1])maxjia = 0;
	if (b[1] > b[0] && b[1] > b[2])maxjia = 1;
	if (b[5] >= b[3] && b[5] >= b[4])maxyi = 5;
	if (b[3] > b[5] && b[3] >= b[4])maxyi = 3;
	if (b[4] > b[3] && b[4] > b[5])maxyi = 4;


	if (maxjia == 0)
		cout << "C" << " ";
	if (maxjia == 1)
		cout << "J" << " ";
	if (maxjia == 2)
		cout << "B" << " ";
	if (maxyi == 3)
		cout << "C";
	if (maxyi == 4)
		cout << "J";
	if (maxyi == 5)
		cout << "B";
	return 0;
}

1019 數字黑洞 (20分)

測試點5需要輸入是6174的時候輸出一行，否則出錯即：

如果忽略前綴爲0的條件則測試點234錯誤即：0001,0023,0456

#include<iostream>
#include<vector>
#include<string>
using namespace std;
void swap(char* a, char* b) //交換兩個變量
{
	char temp = *a;
	*a = *b;
	*b = temp;
}

string up(string a)//從小到大
{
	for (int i = 0; i < 3; i++)
	{
		int min = i;
		for (int j = i + 1; j < 4; j++)
			if (a[j] < a[min])
				min = j;
		swap(&a[min], &a[i]);
	}
	return a;
}
string down(string a)//從大到小
{
	for (int i = 0; i < 3; i++)
	{
		int min = i;
		for (int j = i + 1; j < 4; j++)
			if (a[j] > a[min])
				min = j;
		swap(&a[min], &a[i]);
	}
	return a;
}

int main()
{
	string a;
	cin >> a;
	if (a.length() == 3) {//出現輸入只有123位數時補充位數，否則數組越界
		a = a.insert(0, "0");
	}
	else if (a.length() == 2) {
		a = a.insert(0, "00");
	}
	else if (a.length() == 1) {
		a = a.insert(0, "000");
	}
	while (true)
	{
		if (a[0] == a[1] && a[1] == a[2] && a[2] == a[3])
		{
			printf("%04d - %04d = 0000", stoi(a), stoi(a));
			//cout << stoi(a) << " - " << stoi(a) << " = " << "0000";
			break;
		}
		else
		{
			int max = stoi(down(a)),min=stoi(up(a));
			printf("%04d - %04d = %04d\n", max,min,max-min);
			//cout << max << " - " << min << " = " << max-min<<"\n";
			a = to_string(max - min);
			if (a.length() == 3) {//出現輸入只有123位數時補充位數，否則數組越界
				a = a.insert(0, "0");
			}
			else if (a.length() == 2) {
				a = a.insert(0, "00");
			}
			else if (a.length() == 1) {
				a = a.insert(0, "000");
			}
			if (a == "6174")break;
		}
	}
	

	return 0;
}