【PTA刷题】乙级 1086 To 1095

B1086.就不告诉你 3min
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/*B1086*/
#include 
#include 
#include 
#include 
#include 

using namespace std;

int main()
{	
	int a, b;
	cin >> a >> b;
	string str = to_string( a*b );
	reverse(str.begin(), str.end());
	cout << stoi(str);

	return 0;
}

B1087.有多少种不同的值 5min
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我用的map，柳神用的set。利用了set中每个元素唯一的特性。

/*B1087*/
#include 
#include 
#include 
#include 
using namespace std;

int getsum(int n)
{
	return (n / 2) + (n / 3) + (n / 5);
}

int main()
{	
	int N;
	cin >> N;
	map<int, int> m;

	for (int i = 1; i <= N; i++)
	{
		int sum = getsum(i);
		m[sum]++;
	}
	cout << m.size();
	return 0;
}

B1088.三人行 27min
有两个点卡了好久。
主要原因是一开始看错题，把NXY三个整数看成了甲乙丙必须都是整数。实际上丙可以是小数。
我一开始用的i从10到99正向查找遍历，这样会有几次甲的更新，这样要写好几个continue导致最后甲数字固定了，乙却自己会变动到99。这样的话，到最后还要写两行代码算乙和丙的值，很不划算。还不如用柳神的方法从99遍历到10.

/*B1088*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

void OutPut(int M,double n)
{
	if (M > n)
		cout << " Gai";
	else if (M < n)
		cout << " Cong";
	else
		cout << " Ping";
}


int main()
{	
	int M, X, Y;
	cin >> M >> X >> Y;

	int jia = 0, yi = 0;
	double bing;
	for (int i = 99; i >= 10; i--)
	{
		jia = i;
		yi = 10 * (i % 10) + i / 10;
		bing = abs(i - yi) * 1.0 / X;
		if (bing * Y == yi)
		{
			cout << jia;
			OutPut(M, jia);
			OutPut(M, yi);
			OutPut(M, bing);

			return 0;
		}
	}
		
	cout << "No Solution";
	
	return 0;
}

B1090.危险品装箱 40min
乍一看肯定会超时，其实并没有。和柳神逻辑一模一样。用MAP下面value挂一个vector即可。
1.注意不兼容是双向不兼容。要做两次puch_back。
2.push_back的时候直接 map[root].push_back(value) 即可。

/*B1090*/
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;


int main()
{	
	int N, M;
	cin >> N >> M;

	map<string, vector<string> > m;
	for (int i = 0; i < N; i++)
	{
		string root, info;
		cin >> root >> info;
		m[root].push_back(info);
		m[info].push_back(root);//互相不兼容
	}

	for (int cnt = 0; cnt < M; cnt++)
	{
		int K,flag = 1;
		cin >> K;
		vector<string> vec;//对应每个集装箱
		for (int i = 0; i < K; i++)
		{
			string check;
			cin >> check;
			vec.push_back(check);
		}

		for (int i = 0; i < K; i++)
		{
			for (int j = 0; j < m[vec[i]].size(); j++)
			{
				if (find(vec.begin(), vec.end(), m[vec[i]][j]) != vec.end())
				{
					cout << "No\n";
					flag = 0;
					break;
				}
			}
			if (flag == 0) break;
		}

		if(flag == 1) cout << "Yes\n";
	}

	return 0;
}

B1091.N-自守数 17min
注意读题。题目两点要读清
1.N是小于10的。
2.原本K是几位，就要到末尾找几位。
注意这两点就很容易了

/*B1091*/
#include 
#include 
#include 
#include 

using namespace std;

const int MaxRound = 10;

int IsZishou(int K)
{	
	string str = to_string(K);
	int len = str.length();
	int tmp = 1;
	for (int i = 0; i < len; i++)
		tmp = tmp * 10;

	for (int i = 1; i < MaxRound; i++)
	{
		int n = i * K * K;
		if (K == n % tmp)
			return i;
	
	}
	return -1;
}

int main()
{	
	int M;
	cin >> M;
	for (int i = 0; i < M; i++)
	{	
		int K,N;
		cin >> K;
		N = IsZishou(K);
		if (N != -1)
			cout << N << " " << N * K * K << "\n";
		else
			cout << "No\n";		
	}

	return 0;
}

B1092.最好吃的月饼 10min
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/*B1092*/
#include 
#include 
using namespace std;

int MoonCake[1005] = { 0 };

int main()
{
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < M; i++)
	{
		for (int j = 0; j < N; j++)
		{	
			int sale;
			cin >> sale;
			MoonCake[j] += sale;
		}
	}
	int Max = -1;
	for (int i = 0; i < N; i++)
	{
		if (MoonCake[i] > Max)
			Max = MoonCake[i];
	}
	cout << Max << "\n";
	int flag = 0;
	for (int i = 0; i < N; i++)
	{
		if (MoonCake[i] == Max)
		{
			if (flag == 0)
			{
				cout << i+1;
				flag = 1;
			}
			else
				cout << " " << i+1;
		}
	}
	return 0;
}

B1093.字符串A+B 10min
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/*B1093*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

int main()
{
	string A, B;
	getline(cin, A);
	getline(cin, B);

	vector<char> appear;
	for (int i = 0; i < A.length(); i++)
	{
		if (find(appear.begin(), appear.end(), A[i]) == appear.end())//在已出现里面没找到
		{
			cout << A[i];
			appear.push_back(A[i]);
		}
		else
			continue;
	}
	for (int i = 0; i < B.length(); i++)
	{
		if (find(appear.begin(), appear.end(), B[i]) == appear.end())
		{
			cout << B[i];
			appear.push_back(B[i]);
		}
		else
			continue;
	}

	return 0;
}

B1094.谷歌的招聘 16min
挂了两个点扣了4分，和柳神代码几乎一模一样，原因未知。
虽然一开始没有考虑进去1和0不是素数，但是加上了也没啥不同。

注意一点：str.substr(开始未知，你要截取的个数)，第二个参数直接给个个数就行，不用指针加上偏移量。

/*B1094*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 


using namespace std;

bool IsPrime(int n)
{	
	if (n == 0 || n == 1) return false;
	for (int i = 2; i * i < n; i++)
	{
		if (n % i == 0) return false;
	}
	return true;
}

int main()
{
	string ori;
	int L, K;
	cin >> L >> K >> ori;

	int flag = 0;
	for (int i = 0; i < L-K; i++)
	{
		string tmp = ori.substr(i, K);
		if (IsPrime(stoi(tmp)))
		{
			cout << tmp;
			flag = 1;
			break;
		}
		else
			continue;
	}
	if (flag == 0) cout << 404<<"\n";


	return 0;
}

B1095.解码PTA准考证 78min
非常繁琐。只过了一个点，拿了15分。答案错误两个点，超时两个点。
需要注意一点。map.find()里面的查找，是查找某个键值有没有出现过，然后返回一个迭代器。
所以写法是 map.find(你要查找的key) != map.end()；
此外强调，map的迭代只能用迭代器做。因为map自己加进去的时候就排序了。要想排序还是得用vector,如果要处理的数据复杂，就用vector包结构体。

/*B1095*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

struct Data 
{
	string level;
	string count;
	string testday;
	string id;
	int score;
};
map < string, Data> m;

struct CountInfo
{
	string id;
	int cnt;
};

bool cmp(string a,string b)
{
	if (m[a].score != m[b].score)
		return m[a].score > m[b].score ? true : false;//分数大的在前面
	else
		return a < b ? true : false;//准考证号小的在前面
}

bool cmp3(CountInfo a,CountInfo b)
{
	if (a.cnt != b.cnt)
		return a.cnt > b.cnt ? true : false;
	else
		return a.id < b.id ? true : false;
}

int main()
{
	int N,M;
	cin >> N >> M;
	
	for (int i = 0; i < N; i++)
	{
		string str;
		int Score;
		cin >> str >> Score;
		Data d;
		d.level = str[0];
		d.count = str.substr(1, 3);
		d.testday = str.substr(4, 6);
		d.id = str.substr(10, 3);
		d.score = Score;
		m[str] = d;
	}

	for (int i = 1; i <= M; i++)
	{
		int todo;
		cin >> todo;
		if (todo == 1)
		{	
			int flag = 0;
			string CheckLevel;
			cin >> CheckLevel;
			if (CheckLevel == "T" || CheckLevel == "A" || CheckLevel == "B")
				flag = 1;
			vector<string> vec;//存放map的key即可
			for (auto it = m.begin(); it != m.end(); it++)
			{
				if (it->second.level == CheckLevel)
					vec.push_back(it->first);
			}
			sort(vec.begin(), vec.end(), cmp);

			cout << "Case " << i << ": " << todo << " " << CheckLevel<<"\n";
			if (flag == 1) {
				for (int j = 0; j < vec.size(); j++)
					cout << vec[j] << " " << m[vec[j]].score << "\n";
			}
			else if (flag == 0)
				cout << "NA\n";
		}
		else if (todo == 2)
		{	
			int flag = 0;
			string CheckCount;
			cin >> CheckCount;
			int StuNum = 0, SumScore = 0;
			for (auto it = m.begin(); it != m.end(); it++)
			{
				if (it->second.count == CheckCount)
				{
					StuNum++;
					SumScore += it->second.score;
					flag = 1;
				}
			}
			cout << "Case " << i << ": " << todo << " " << CheckCount<< "\n";
			if (flag == 1)
				cout << StuNum << " " << SumScore << "\n";
			else if (flag == 0)
				cout << "NA\n";
		}
		else if (todo == 3)
		{	
			int flag = 0;
			string CheckDay;
			cin >> CheckDay;
			map<string, int> CountM;
			vector<CountInfo> res;
			for (auto it = m.begin(); it != m.end(); it++)
			{
				if (it->second.testday == CheckDay)
				{
					CountM[it->second.count]++;//在MAP中对应考场人数加一
					flag = 1;//这个日期是存在的
				}
			}
			for (auto it = CountM.begin(); it != CountM.end(); it++)//遍历中间map存入res中
			{
				res.push_back({ it->first,it->second });
			}
			sort(res.begin(), res.end(), cmp3);
			cout << "Case " << i << ": " << todo << " " << CheckDay << "\n";
			
			if(flag == 1)
				for (int j = 0; j < res.size(); j++)
					cout << res[j].id << " " << res[j].cnt << "\n";
			else if(flag == 0)
				cout << "NA\n";

		}

	}
	return 0;
}