寒假练习 04

断断续续的把排序和检索专题刷完了，感觉英语还是不够，题目太长以后看起来就会很吃力。

还有一点感触就是STL的广泛应用。学到了很多新东西。

当然，不能忍受的就是答案最后多输出一行空行，UVaOJ会判WA。

UVaOJ 340

简单模拟题，一开始没有看懂题目。百度以后才明白的题意。朴素模拟以后即可得到答案。

#include <iostream>
#include <memory.h>

using namespace std;

const int MAX = 1024;

int pCode[MAX], pGuess[MAX], pVisited[MAX];

int main()
{
	int N, nCase = 0;
	while(cin >> N)
	{
		if(N == 0) { break; }
		memset(pCode, 0, sizeof(pCode));
		for(int i = 1; i <= N; i++)
		{ cin >> pCode[i]; }
		cout << "Game " << ++nCase << ":" << endl; 
		while(1)
		{
			int x = 0, y = 0, nCnt = 0;
			memset(pGuess, 0, sizeof(pGuess));
			memset(pVisited, 0, sizeof(pVisited));
			for(int i = 1; i <= N; i++)
			{
				cin >> pGuess[i];
				if(pGuess[i] == 0) { nCnt++; } 
			}
			if(nCnt == N) { break; }
			for(int i = 1; i <= N; i++)
			{
				if(pCode[i] == pGuess[i])
				{
					x++;
					pVisited[i] = 2;
				}
			}
			for(int i = 1; i <= N; i++)
			{
				if(pVisited[i] == 2) { continue; }
				for(int j = 1; j <= N; j++)
				{
					if(pVisited[j] != 0) { continue; }
					if(pCode[i] == pGuess[j])
					{
						y++;
						pVisited[j] = 1;
						break;
					}
				}
			}
			cout << "    (" << x << "," << y << ")" << endl; 
		}
	}
	return 0;
} 

　　

UVaOJ 10420

STL中的set水过。

#include <iostream>
#include <string>
#include <map> 

using namespace std;

map<string, int> pMap;

int main()
{
	int N;
	while(cin >> N)
	{
		pMap.clear();
		string x, y;
		for(int i = 1; i <= N; i++)
		{
			cin >> x;
			getline(cin, y);
			pMap[x]++;
		}
		map<string, int>::iterator it;
		for(it = pMap.begin(); it != pMap.end(); it++)
		{ cout << (*it).first << " " << (*it).second << endl; }
	}
	return 0;
} 

　　

UVaOJ 10474

水题，一开始没看见“CASE”是大写的，WA了好几次。

#include <iostream>
#include <algorithm>

using namespace std;

const int MAX = 10240;

int pData[MAX];

int main()
{
	int nCase = 0; 
	int N, Q, x;
	while(cin >> N >> Q)
	{
		if(N == 0 && Q == 0) { break; }
		for(int i = 1; i <= N; i++)
	 	{ cin >> pData[i]; }
	 	cout << "CASE# " << ++nCase << ":" << endl; 
	 	sort(pData + 1, pData + N + 1);
	 	for(int i = 1; i <= Q; i++)
	 	{
			cin >> x;
			bool bFlag = false;
			for(int j = 1; j <= N; j++)
			{
				if(pData[j] == x)
				{ cout << x << " found at " << j << endl; bFlag = true; break; }
			}
			if(!bFlag) { cout << x << " not found" << endl; }
	 	}
	}
	return 0;
}

　　

UVaOJ 152

这道题目一开始也没有看懂题意，百度以后才明白的题意。

为了避免sqrt的误差，可以直接判断平方的大小关系。

#include <iostream>
#include <iomanip>
#include <memory.h> 

using namespace std;

const int MAX = 10240;
const int pTable[] = { 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 };

struct Point
{
	int x, y, z;
};

int pOut[MAX];
Point pData[MAX];

int Calc(int i, int j);

int main()
{
	int x, y, z, nCnt = 0;
	memset(pOut, 0, sizeof(pOut));
	while(cin >> x >> y >> z)
	{
		if(x == 0 && y == 0 && z == 0) { break; }
		pData[++nCnt] = {x, y, z};
	}
	for(int i = 1; i <= nCnt; i++)
	{
		int nMin = 2147483647;
		for(int j = 1; j <= nCnt; j++)
		{
			if(i == j) { continue; }
			int nTmp = Calc(i, j);
			if(nMin > nTmp) { nMin = nTmp; }
		}
		pOut[nMin]++;
	}
	for(int i = 1; i <= 10; i++)
	{ cout << setw(4) << pOut[i]; }
	cout << endl;
	return 0; 
} 

int Calc(int i, int j)
{
	int x = pData[i].x - pData[j].x;
	int y = pData[i].y - pData[j].y;
	int z = pData[i].z - pData[j].z;
	int nTmp = x * x + y * y + z * z;
	for(int k = 1; k <= 10; k++)
	{
		if(nTmp < pTable[k]) { return k; }
	}
}

　　

UVaOJ 299

学过线性代数以后直接想到了逆序数。

#include <iostream>
#include <memory.h> 

using namespace std;

const int MAX = 10240;

int pData[MAX];

int main()
{
	int T, N;
	cin >> T; 
	for(int i = 1; i <= T; i++)
	{
		cin >> N;
		int nCnt = 0;
		memset(pData, 0, sizeof(pData));
		for(int j = 1; j <= N; j++)
		{ cin >> pData[j]; }
		for(int j = 1; j <= N; j++)
		{
			for(int k = j + 1; k <= N; k++)
			{
				if(pData[j] > pData[k]) { nCnt++; }
			}
		}
		cout << "Optimal train swapping takes " << nCnt << " swaps." << endl;
	}
	return 0;
} 

　　

UVaOJ 120

简单模拟题，也是百度以后才知道的题意。

如果当前区间最大的不是第一个，那么先将它翻转到第一个，然后在翻转到当前区间的最后一个。

如果它恰好是第一个，直接翻转到最后一个。

#include <iostream>
#include <string>
#include <sstream>
#include <deque>
#include <algorithm>

using namespace std;

deque<int> Q;

int main()
{
	int nTmp;
	string x;
	while(getline(cin, x))
	{
		cout << x << endl;
		istringstream iss(x);
		Q.clear();
		while(iss >> nTmp) { Q.push_front(nTmp); }
		for(deque<int>::iterator it = Q.begin(); it != Q.end(); it++)
		{
			deque<int>::iterator iMax = max_element(it, Q.end());
			if(iMax != it)
			{
				if(iMax != Q.end() - 1)
				{
					reverse(iMax, Q.end());
					cout << iMax - Q.begin() + 1 << " ";
				}
				reverse(it, Q.end());
				cout << it - Q.begin() + 1 << " ";
			}
		}
		cout << "0" << endl;
	} 
	return 0;
} 

　　

UVaOJ 156

STL中的map水过。但是对于map内部的自动排序还不是非常的理解。

#include <iostream>
#include <algorithm>
#include <map>
#include <vector>

using namespace std;

const int MAX = 10240;

map<string, int> pMap;
vector<string> pVec;
string pData[MAX], pSorted[MAX];

string ToLower(string x);

int main()
{
	int nCnt = 0;
	string x;
	while(cin >> x)
	{
		if(x == "#") { break; }
		nCnt++;
		pData[nCnt] = x;
		pSorted[nCnt] = ToLower(x);
		sort(pSorted[nCnt].begin(), pSorted[nCnt].end());
		pMap[pSorted[nCnt]]++;
	}
	for(map<string, int>::iterator it = pMap.begin(); it != pMap.end(); it++)
	{
		if((*it).second != 1) { continue; }
		int nPos;
		for(nPos = 1; nPos <= nCnt; nPos++)
		{
			if(pSorted[nPos] == (*it).first) { break; }
		}
		pVec.push_back(pData[nPos]);
	}
	sort(pVec.begin(), pVec.end());
	for(int i = 0; i < pVec.size(); i++)
	{ cout << pVec[i] << endl; }
}

string ToLower(string x)
{
	for(int i = 0; i < x.length(); i++)
	{
		if(x[i] >= 'A' && x[i] <= 'Z') { x[i] += 32; }
	}
	return x;
}

　　

UVaOJ 400

一开始没有注意到除数为0的情况，导致RE了好多次。

#include <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

const int MAX = 60;

vector<string> pVec;

int main()
{
	int N;
	string x;
	while(cin >> N)
	{
		int nMax = 0;
		pVec.clear();
		for(int i = 1; i <= N; i++)
		{
			cin >> x;
			pVec.push_back(x);
			nMax = max(nMax, (int)x.length());
		}
		int nCnt = MAX / (nMax + 2);
		if(nCnt == 0) { nCnt = 1; }
		int nRow = N / nCnt + (N % nCnt != 0); 
		sort(pVec.begin(), pVec.end());
		cout << "------------------------------------------------------------" << endl;
		for(int i = 1; i <= nRow; i++)
		{
			for(int j = 1; j <= nCnt; j++)
			{
				int nPos = (j - 1) * nRow + i - 1;
				if(nPos >= N) { continue; }
				cout.setf(ios::left);
				if(nPos >= (nCnt - 1) * nRow)	
				{ cout << setw(nMax) << pVec[nPos]; }
				else { cout << setw(nMax + 2) << pVec[nPos]; }
 				
			}
			cout << endl;
		}
	}
	return 0;
}

　　

UVaOJ 123

这道题目要求按照原来的顺序，而sort是非稳定排序，在这上面WA了好几次。

#include <iostream>
#include <string>
#include <set> 
#include <vector>
#include <algorithm>
#include <sstream> 

using namespace std;

const int MAX = 10240;

int cmp(pair<string, pair<string, int> > x, pair<string, pair<string, int> > y)
{
	if(x.first != y.first) { return x.first < y.first; }
	else { return x.second.second < y.second.second; }
}

string pData[MAX];
set<string> pSet;
vector<pair<string, pair<string, int> > > pVec;

string ToLower(string x);
string ToUpper(string x);

int main()
{
	int nPos = 0;
	string x;
	while(cin >> x)
	{
		if(x == "::") { break; }
		pSet.insert(x);
	}
	cin.ignore();
	while(getline(cin, x))
	{
		string strLow = ToLower(x);
		istringstream iss(strLow);
		int nCnt = 0;
		while(iss >> x)
		{
			if(x != "") { pData[++nCnt] = x; } 
		}
		for(int i = 1; i <= nCnt; i++)
		{
			if(!pSet.count(pData[i]))
			{
				string strTmp = "";
				for(int j = 1; j <= nCnt; j++)
				{
					if(j != i) { strTmp += pData[j] + " "; }
					else { strTmp += ToUpper(pData[j]) + " "; } 
				}
				strTmp = strTmp.substr(0, strTmp.length() - 1);
				pVec.push_back(make_pair(ToUpper(pData[i]), make_pair(strTmp, ++nPos)));
			}
		}
	}
	sort(pVec.begin(), pVec.end(), cmp);
	for(int i = 0; i < pVec.size(); i++)
	{ cout << pVec[i].second.first << endl; }
	return 0;
}

string ToLower(string x)
{
	for(int i = 0; i < x.length(); i++)
	{
		if(x[i] >= 'A' && x[i] <= 'Z')
		{ x[i] += 32; }
	}
	return x;
}

string ToUpper(string x)
{
	for(int i = 0; i < x.length(); i++)
	{
		if(x[i] >= 'a' && x[i] <= 'z')
		{ x[i] -= 32; }
	}
	return x;
}

　　

UVaOJ 10194

水题，直接模拟即可。注意写好排序的cmp函数即可。

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

const int MAX = 1024;

struct Player
{
	Player()
	{ Clear(); }
	
	void Clear()
	{
		strName = "";
		nPoint = nGame = 0;
		nWin = nTie = nLose = 0;
		nDiff = nScored = nAgainst = 0;
	}
	
	string strName;
	string strTmp;
	int nPoint, nGame;
	int nWin, nTie, nLose;
	int nDiff, nScored, nAgainst;
};

Player pData[MAX];

int GetPos(string x);
int Trans(string x);
string ToLower(string x);

int cmp(Player x, Player y)
{
	if(x.nPoint != y.nPoint) { return x.nPoint > y.nPoint; }
	if(x.nWin != y.nWin) { return x.nWin > y.nWin; }
	if(x.nDiff != y.nDiff) { return x.nDiff > y.nDiff; }
	if(x.nScored != y.nScored) { return x.nScored > y.nScored; }
	if(x.nGame != y.nGame) { return x.nGame < y.nGame; }
	return ToLower(x.strName) < ToLower(y.strName); 
}

int main()
{
	int N, nPlayer, nGame;
	string x;
	cin >> N;
	cin.ignore();
	for(int i = 1; i <= N; i++)
	{
		getline(cin, x);
		cout << x << endl;
		int nCnt = 0;
		cin >> nPlayer;
		cin.ignore();
		for(int j = 1; j <= nPlayer; j++)
		{
			getline(cin, x);
			pData[++nCnt].Clear();
			pData[nCnt].strName = x;
			pData[nCnt].strTmp = ToLower(x);
		}
		cin >> nGame;
		cin.ignore();
		for(int j = 1; j <= nGame; j++)
		{
			getline(cin, x);
			int nLeft = x.find_first_of('#'), nRight = x.find_last_of('#'), nMid = x.find('@');
			string strLeft = x.substr(0, nLeft);
			string strRight = x.substr(nRight + 1, x.length() - nRight - 1);
			string strLeftPoint = x.substr(nLeft + 1, nMid - nLeft - 1);
			string strRightPoint = x.substr(nMid + 1, nRight - nMid - 1);
			int nLeftPos = GetPos(strLeft), nRightPos = GetPos(strRight);
			int nLeftPoint = Trans(strLeftPoint), nRightPoint = Trans(strRightPoint);
			pData[nLeftPos].nGame++; pData[nRightPos].nGame++;
			pData[nLeftPos].nScored += nLeftPoint; pData[nRightPos].nScored += nRightPoint;
			pData[nLeftPos].nAgainst += nRightPoint; pData[nRightPos].nAgainst += nLeftPoint;
			if(nLeftPoint > nRightPoint) { pData[nLeftPos].nWin++; pData[nRightPos].nLose++; }
			else if(nLeftPoint < nRightPoint) { pData[nLeftPos].nLose++; pData[nRightPos].nWin++; }
			else { pData[nLeftPos].nTie++; pData[nRightPos].nTie++; }
		}
		for(int i = 1; i <= nPlayer; i++)
		{
			pData[i].nPoint = pData[i].nWin * 3 + pData[i].nTie;
			pData[i].nDiff = pData[i].nScored - pData[i].nAgainst;
		}
		sort(pData + 1, pData + nPlayer + 1, cmp);
		for(int i = 1; i <= nPlayer; i++)
		{
			cout << i << ") " << pData[i].strName 
				<< " " << pData[i].nPoint << "p, " 
				<< pData[i].nGame << "g (" 
				<< pData[i].nWin << "-" 
				<< pData[i].nTie << "-" 
				<< pData[i].nLose << "), " 
				<< pData[i].nDiff << "gd (" 
				<< pData[i].nScored << "-" 
				<< pData[i].nAgainst << ")" 
				<< endl; 
		}
		if(i != N) { cout << endl; }
	} 
	return 0;
}

int GetPos(string x)
{
	for(int nPos = 1; nPos < MAX; nPos++)
	{
		if(pData[nPos].strTmp == ToLower(x))
		{ return nPos; }
	}
}

int Trans(string x)
{
	int nRet = 0;
	for(int i = 0; i < x.length(); i++)
	{
		nRet *= 10;
		nRet += x[i] - '0';
	}
	return nRet;
}

string ToLower(string x)
{
	for(int i = 0; i <x.length(); i++)
	{
		if(x[i] >= 'A' && x[i] <= 'Z')
		{ x[i] += 32; }
	} 
	return x;
}

　　

UVaOJ 755

直接map水过。多输出了一行空行，导致WA了一次。

#include <iostream>
#include <map>
#include <vector>
#include <algorithm>

using namespace std;

const char pTable[] = { 
	'2', '2', '2', '3', '3', '3', 
	'4', '4', '4', '5', '5', '5', 
	'6', '6', '6', '7', '*', '7', 
	'7', '8', '8', '8', '9', '9', 
	'9', '*' };
	
int cmp(pair<string, int> x, pair<string, int> y)
{ return x.first < y.first; }

vector<pair<string, int> > pVec;
map<string, int> pMap;

int main()
{
	int T, N;
	string x;
	cin >> T;
	for(int i = 1; i <= T; i++)
	{
		pVec.clear();
		pMap.clear();
		cin >> N;
		cin.ignore();
		for(int j = 1; j <= N; j++)
		{
			getline(cin, x);
			string strTmp = "";
			for(int k = 0; k < x.length(); k++)
			{
				if(x[k] >= '0' && x[k] <= '9') { strTmp += x[k]; }
				if(x[k] >= 'A' && x[k] <= 'Z') { strTmp += pTable[x[k] - 'A']; }
			}
			while(strTmp.length() != 7) { strTmp = '0' + strTmp; }
			pMap[strTmp]++;
		}
		for(map<string, int>::iterator it = pMap.begin(); it != pMap.end(); it++)
		{
			if((*it).second > 1) { pVec.push_back(*it); }
		}
		if(pVec.size() == 0) { cout << "No duplicates." << endl; }
		else
		{
			sort(pVec.begin(), pVec.end(), cmp);
			for(int j = 0; j < pVec.size(); j++)
			{
				cout << pVec[j].first.substr(0, 3) << "-" 
					<< pVec[j].first.substr(3, 4) << " "
					<< pVec[j].second << endl; 
			}
		}
		if(i != T) { cout << endl; }
	}
	return 0;
}

　　

UVaOJ 10785

类似归并的思想。一开始打错表了，WA了一次。

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

const char pOdd[] = { 'A', 'U', 'E', 'O', 'I' };
const char pEven[] = { 
	'J', 'S', 'B', 'K', 'T',
	'C', 'L', 'D', 'M', 'V', 
	'N', 'W', 'F', 'X', 'G', 
	'P', 'Y', 'H', 'Q', 'Z', 
	'R' };

string Solve(int x);

int main()
{
	int T, N;
	cin >> T;
	for(int i = 1; i <= T; i++)
	{
		cin >> N;
		cout << "Case " << i << ": "
			 << Solve(N) << endl;
	}
	return 0;
}

string Solve(int x)
{
	string strRet = "";
	string strOdd = "";
	string strEven = "";
	int nOdd = (x + 1) / 2;
	int nEven = x / 2;
	int nOddCnt = nOdd / 21 + (nOdd % 21 != 0);
	int nEvenCnt = nEven / 5 + (nEven % 5 != 0);
	for(int i = 1; i <= nOddCnt; i++)
	{
		for(int j = 1; j <= 21; j++)
		{ strOdd += pOdd[i - 1]; }
	}
	for(int i = 1; i <= nEvenCnt; i++)
	{
		for(int j = 1; j <= 5; j++)
		{ strEven += pEven[i - 1]; }
	}
	strOdd = strOdd.substr(0, nOdd);
	strEven = strEven.substr(0, nEven);
	sort(strOdd.begin(), strOdd.end());
	sort(strEven.begin(), strEven.end());
	int nOddPos = 0, nEvenPos = 0;
	for(int i = 1; i <= x; i++)
	{
		if(i & 1) { strRet += strOdd[nOddPos++]; }
		else { strRet += strEven[nEvenPos++]; }
	}
	return strRet;
	
}

　　

这一组题目做下来，感觉英语还是要提升，有时候题目一长，干扰信息一多，读起来就感觉很有难度了。