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今天刷了小白书的字符串专题，各种WA以及PE。UVaOJ中有时候会把PE判成WA，这样会导致很难查错。
这道题目有个坑，只有表格中列出的才是镜像字母，没有列出了的表示没有镜像字母，在这上WA了一次。
#include <iostream>
#include <string>
using namespace std;
const char pAlphabet[] = {
‘A‘, ‘*‘, ‘*‘, ‘*‘, ‘3‘, ‘*‘, ‘*‘, ‘H‘, ‘I‘,
‘L‘, ‘*‘, ‘J‘, ‘M‘, ‘*‘, ‘O‘, ‘*‘, ‘*‘, ‘*‘,
‘2‘, ‘T‘, ‘U‘, ‘V‘, ‘W‘, ‘X‘, ‘Y‘, ‘5‘, ‘1‘,
‘S‘, ‘E‘, ‘*‘, ‘Z‘, ‘*‘, ‘*‘, ‘8‘, ‘*‘ };
bool Palindrome(string x);
bool Mirrored(string x);
int main()
{
string x;
while(cin >> x)
{
if(!Palindrome(x))
{
if(Mirrored(x)) { cout << x << " -- is a mirrored string." << endl; }
else { cout << x << " -- is not a palindrome." << endl; }
}
else
{
if(Mirrored(x)) { cout << x << " -- is a mirrored palindrome." << endl; }
else { cout << x << " -- is a regular palindrome." << endl; }
}
cout << endl;
}
return 0;
}
bool Palindrome(string x)
{
for(int i = 0; i < x.length() / 2; i++)
{
if(x[i] != x[x.length() - i - 1]) { return false; }
}
return true;
}
bool Mirrored(string x)
{
if(x.length() == 1)
{
if(x[0] >= ‘A‘ && x[0] <= ‘Z‘)
{
if(pAlphabet[x[0] - ‘A‘] != x[0]) { return false; }
}
else
{
if(pAlphabet[x[0] - ‘0‘ + 25] != x[0]) { return false; }
}
}
for(int i = 0; i < x.length() / 2; i++)
{
if(x[i] >= ‘A‘ && x[i] <= ‘Z‘)
{
if(pAlphabet[x[i] - ‘A‘] != x[x.length() - i - 1]) { return false; }
}
else
{
if(pAlphabet[x[i] - ‘0‘ + 25] != x[x.length() - i - 1]) { return false; }
}
}
return true;
}
这道题目要求八个方向都要搜一遍，一开始准备写八个函数，后来发现用dx[], dy[]数组就搞定了。
#include <iostream>
#include <string>
#include <memory.h>
using namespace std;
const int MAX = 128;
const int dx[] = { 0, 0, 1, -1, -1, 1, -1, 1 };
const int dy[] = { 1, -1, 0, 0, 1, 1, -1, -1 };
int T, N, M, Q;
char pMap[MAX][MAX];
char ToLower(char x);
void Find(string x);
bool Search(string s, int x, int y);
int main()
{
string x;
cin >> T;
for(int i = 1; i <= T; i++)
{
memset(pMap, 0, sizeof(pMap));
cin >> N >> M;
for(int j = 1; j <= N; j++)
{
for(int k = 1; k <= M; k++)
{
cin >> pMap[j][k];
pMap[j][k] = ToLower(pMap[j][k]);
}
}
cin >> Q;
for(int j = 1; j <= Q; j++)
{
cin >> x;
Find(x);
}
if(i != T) { cout << endl; }
}
return 0;
}
char ToLower(char x)
{
if(x >= ‘a‘ && x <= ‘z‘) { return x; }
else { return x + 32; }
}
void Find(string x)
{
bool bFlag = false;
for(int i = 0; i < x.length(); i++)
{ x[i] = ToLower(x[i]); }
for(int i = 1; i <= N; i++)
{
for(int j = 1; j <= M; j++)
{
if(Search(x, i, j))
{ cout << i << " " << j << endl; bFlag = true; break; }
}
if(bFlag) { break; }
}
}
bool Search(string s, int x, int y)
{
for(int i = 0; i < 8; i++)
{
bool bFlag = true;
for(int j = 0; j < s.length(); j++)
{
int nx = x + dx[i] * j;
int ny = y + dy[i] * j;
if(pMap[nx][ny] != s[j]) { bFlag = false; break; }
}
if(bFlag) { return true; }
}
return false;
}
这道题目卡了好久，一直不知道为什么我的代码会各种WA以及PE。最后还是采用了网上一种C语言的写法才AC了。但还是不知道为什么我的错了。
#include <stdio.h>
#include <string.h>
const int MAX = 128;
char s1[MAX], s2[MAX], s3[MAX], s4[MAX], s5[MAX], line[MAX];
void getss(char s[]);
int main()
{
int N;
scanf("%d", &N);
getchar();
for(int i = 1; i <= N; i++)
{
getss(s1);
getss(s2);
getss(s3);
getss(s4);
getss(s5);
gets(line);
line[strlen(line) - 3] = ‘\0‘;
printf("%s%s%s%s%s\n", s1, s2, s3, s4, s5);
printf("%s%s%s%s%s\n", line, s4, s3, s2, s5);
}
return 0;
}
void getss(char s[])
{
for(int i = 0; i < MAX; i++)
{
if((s[i] = getchar()) ==‘<‘ || s[i] == ‘>‘ || s[i] == ‘\n‘)
{ s[i] = ‘\0‘; break; }
}
}
这道题目一下就AC了，处理的时候注意一下坑数据。
这里还有一个知识点，除了atoi可以实现string到int的转换，还有atof，可以实现string到double的转换。
#include <iostream>
#include <iomanip>
#include <string>
#include <stdlib.h>
using namespace std;
int main()
{
int T;
cin >> T;
cin.ignore();
for(int j = 1; j <= T; j++)
{
cout << "Problem #" << j << endl;
double u = 0, i = 0, p = 0;
string x, U = "", I = "", P = "";
getline(cin, x);
if(x.find("U=") != string::npos) { U = x.substr(x.find("U=") + 2, x.find(‘V‘, x.find("U=")) - x.find("U=") - 2); }
if(x.find("I=") != string::npos) { I = x.substr(x.find("I=") + 2, x.find(‘A‘, x.find("I=")) - x.find("I=") - 2); }
if(x.find("P=") != string::npos) { P = x.substr(x.find("P=") + 2, x.find(‘W‘, x.find("P=")) - x.find("P=") - 2); }
if(U != "")
{
if(U[U.length() - 1] >= ‘0‘ && U[U.length() - 1] <= ‘9‘) { u = atof(U.c_str()); }
else
{
u = atof(U.substr(0, U.length() - 1).c_str());
if(U[U.length() - 1] == ‘m‘) { u /= 1000.0; }
if(U[U.length() - 1] == ‘k‘) { u *= 1000.0; }
if(U[U.length() - 1] == ‘M‘) { u *= 1000000.0; }
}
}
if(I != "")
{
if(I[I.length() - 1] >= ‘0‘ && I[I.length() - 1] <= ‘9‘) { i = atof(I.c_str()); }
else
{
i = atof(I.substr(0, I.length() - 1).c_str());
if(I[I.length() - 1] == ‘m‘) { i /= 1000.0; }
if(I[I.length() - 1] == ‘k‘) { i *= 1000.0; }
if(I[I.length() - 1] == ‘M‘) { i *= 1000000.0; }
}
}
if(P != "")
{
if(P[P.length() - 1] >= ‘0‘ && P[P.length() - 1] <= ‘9‘) { p = atof(P.c_str()); }
else
{
p = atof(P.substr(0, P.length() - 1).c_str());
if(P[P.length() - 1] == ‘m‘) { p /= 1000.0; }
if(P[P.length() - 1] == ‘k‘) { p *= 1000.0; }
if(P[P.length() - 1] == ‘M‘) { p *= 1000000.0; }
}
}
if(U != "" && I != "")
{ cout << "P=" << fixed << setprecision(2) << u * i << "W" << endl; }
if(U != "" && P != "")
{ cout << "I=" << fixed << setprecision(2) << p / u << "A" << endl; }
if(I != "" && P != "")
{ cout << "U=" << fixed << setprecision(2) << p / i << "V" << endl; }
cout << endl;
}
return 0;
}
只用运用string的find函数就可以解决了。
#include <iostream>
#include <memory.h>
using namespace std;
const int MAX = 32;
int pCnt[MAX];
string pKeyword[MAX], pExcuse[MAX], pTmp[MAX];
int main()
{
int N, K, nCase = 0;
while(cin >> N >> K)
{
cout << "Excuse Set #" << ++nCase << endl;
memset(pCnt, 0, sizeof(pCnt));
for(int i = 1; i <= N; i++)
{
cin >> pKeyword[i];
pKeyword[i] += ‘ ‘;
}
cin.ignore();
for(int i = 1; i <= K; i++)
{
getline(cin, pTmp[i]);
pExcuse[i] = pTmp[i];
for(int j = 0; j < pExcuse[i].length(); j++)
{
if(pExcuse[i][j] >= ‘A‘ && pExcuse[i][j] <= ‘Z‘)
{ pExcuse[i][j] += 32; }
if(!(pExcuse[i][j] >= ‘a‘ && pExcuse[i][j] <= ‘z‘))
{ pExcuse[i][j] = ‘ ‘; }
}
pExcuse[i] += ‘ ‘;
}
for(int i = 1; i <= K; i++)
{
for(int j = 1; j <= N; j++)
{
int nPos = 0;
while((nPos = pExcuse[i].find(pKeyword[j], nPos)) && nPos != string::npos)
{ pCnt[i]++; nPos += pKeyword[j].length() + 1; }
}
}
int nMax = 0;
for(int i = 1; i <= K; i++)
{ nMax = max(nMax, pCnt[i]); }
for(int i = 1; i <= K;i ++)
{
if(pCnt[i] == nMax) { cout << pTmp[i] << endl; }
}
cout << endl;
}
return 0;
}
一看就知道和二进制有关系，翻译过来果然如此（忽略字符‘.‘）。解法中用了秦九韶算法来进制转换。
#include <iostream>
#include <string>
using namespace std;
int main()
{
int nCnt = 0;
string x;
while(getline(cin, x))
{
int nTmp = 0;
if(x == "___________") { nCnt++; }
if(nCnt == 2) { break; }
if(x != "___________")
{
x = x.substr(1, x.length() - 2);
for(int i = 0; i < x.length(); i++)
{
if(x[i] != ‘.‘) { nTmp *= 2; }
if(x[i] == ‘o‘) { nTmp += 1; }
}
cout << (char)(nTmp);
}
}
return 0;
}
C++的STL中的set可以水过。
#include <iostream>
#include <string>
#include <set>
using namespace std;
const int MAX = 5120;
set<string> pSet;
bool IsAlpha(char x);
char ToLower(char x);
int main()
{
string x;
while(getline(cin, x))
{
for(int i = 0; i < x.length(); i++)
{
if(!IsAlpha(x[i]))
{ continue; }
string strTmp;
while(i < x.length() && IsAlpha(x[i]))
{ strTmp += ToLower(x[i++]); }
pSet.insert(strTmp);
}
}
for(set<string>::iterator it = pSet.begin(); it != pSet.end(); it++)
{ cout << *it << endl; }
return 0;
}
bool IsAlpha(char x)
{ return x >= ‘A‘ && x <= ‘Z‘ || x >= ‘a‘ && x <= ‘z‘; }
char ToLower(char x)
{
if(x >= ‘A‘ && x <= ‘Z‘) { x += 32; }
return x;
}
一开始TLE，还以为算法不行，后来发现是读入出了问题。
#include <iostream>
using namespace std;
const int MAX = 10240;
string x[MAX];
int main()
{
int nCase = 0, nPos = 0;
while(cin >> x[++nPos])
{
if(x[nPos] != "9") { continue; }
else
{
nPos--;
bool bFlag = true;
for(int i = 1; i <= nPos; i++)
{
for(int j = 1; j <= nPos; j++)
{
if(x[i].length() < x[j].length())
{ if(x[i] == x[j].substr(0, x[i].length())) { bFlag = false; break; } }
}
if(!bFlag) { break; }
}
if(bFlag) { cout << "Set " << ++nCase << " is immediately decodable" << endl; }
else { cout << "Set " << ++nCase << " is not immediately decodable" << endl; }
nPos = 0;
}
}
}
自己写一个replace函数就可以AC了。
#include <iostream>
using namespace std;
const int MAX = 16;
string strRules[MAX], strReplace[MAX];
string& ReplaceAll(string& str, const string& strOld, const string& strNew);
int main()
{
int N;
while(cin >> N)
{
string x;
if(N == 0) { break; }
cin.ignore();
for(int i = 1; i <= N; i++)
{
getline(cin, strRules[i]);
getline(cin, strReplace[i]);
}
getline(cin, x);
for(int i = 1; i <= N; i++)
{ x = ReplaceAll(x, strRules[i], strReplace[i]); }
cout << x << endl;
}
return 0;
}
string& ReplaceAll(string& str, const string& strOld, const string& strNew)
{
while(true)
{
string::size_type pos(0);
if((pos = str.find(strOld)) != string::npos)
{ str.replace(pos, strOld.length(), strNew); }
else { break; }
}
return str;
}
最后总结一个知识点，如何在C++中自己写基于string的relpace函数。
在这里，replace分为两种，一种是repalce_all，一种是replace_all_distinct。
/* replace 12 with 21
12212 -> 22211 */
string& replace_all(string& str,const string& old_value,const string& new_value)
{
while(true) {
string::size_type pos(0);
if( (pos=str.find(old_value))!=string::npos )
str.replace(pos,old_value.length(),new_value);
else break;
}
return str;
}
/* replace 12 with 21
12212 -> 21221 */
string& replace_all_distinct(string& str,const string& old_value,const string& new_value)
{
for(string::size_type pos(0); pos!=string::npos; pos+=new_value.length()) {
if( (pos=str.find(old_value,pos))!=string::npos )
str.replace(pos,old_value.length(),new_value);
else break;
}
return str;
}
还有很深的感触就是UVaOJ上面关于WA和PE分的不是很清楚。多了空行算WA，行内少了或多了空格算PE。
一套字符串题目刷下来，感到非常吃力，被各种WA以及PE。
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原文地址：http://www.cnblogs.com/Ivy-End/p/4279216.html
