[HDOJ 1039] Easier Done Than Said? （基础，字符串） .

先贴上在VC6.0上运行无误但在HDOJ后台海量测试数据通之下无法通过的代码，以便以后分析错误原因。

#include <iostream>
using namespace std;
bool isVowel(char letter)
{
switch(letter)
{
case 'a':case 'e':case 'i':case 'o':case 'u':
return true;
default:
return false;
}
}

bool isSame(char ch_1,char ch_2)
{
if(ch_1 == ch_2 && (ch_1 != 'e' && ch_1 != 'o'))
return true;
else
return false;
}

bool allSame(char ch_1,char ch_2,char ch_3)
{
if( isVowel(ch_1) && isVowel(ch_2) && isVowel(ch_3) \
|| ((!isVowel(ch_1)) && (!isVowel(ch_2)) && (!isVowel(ch_3)) ))
return true;

else
return false;
}

bool isAcceptable(char *pwd)
{
char *pBegin,*pCur,*pNext;
int vNum;
int len = strlen(pwd);

if(len < 3)
{
if(len == 1 && isVowel((char)pwd[0]))
return true;

if(len == 2 && isSame(pwd[0],pwd[1]))
return true;
else
return false;
}
else
{
pBegin = pwd;
pCur = pwd + 1;
pNext = pwd + 2;
char *pEnd = pwd + len;//最后一个字符串的下一位为结束点
vNum = 0;
while(pNext != pEnd)
{
if(isVowel(*pBegin))
++vNum;
if(allSame(*pBegin,*pCur,*pNext))//三个连续的字母
return false;

if(isSame(*pBegin,*pCur) || isSame(*pCur,*pNext)){ //两个连续的字母相同
return false;
}
else{
pBegin++;
pCur++;
pNext++;
}
}
if(vNum == 0) //一个元音都没有
return false;
else
return true;
}

}
int main()
{
char pwd[1000];
while(cin >> pwd && (strcmp(pwd,"end") != 0))
{
if(isAcceptable(pwd))
{
cout << "<" << pwd << "> is acceptable." << endl;
}
else
{
cout << "<" << pwd << "> is not acceptable." << endl;
}
}
}

版本2:更为简洁，少了非常多的函数调用开销，效率较之前版本有所提升。不过仍旧不能通过OJ的测试。

#include <iostream>
using namespace std;
bool isVowel(char letter)
{
switch(letter)
{
case 'a':case 'e':case 'i':case 'o':case 'u':
return true;
default:
return false;
}
}

bool notVowel(char ch)
{
if(!isVowel(ch))
return true;
else
return false;
}
bool isAcceptable(char *pwd)
{
int len = strlen(pwd);
int n_vowel = 0;
if(1 == len && isVowel(pwd[0]))
return true;
for(int i = 1; i < len; ++i)
{
if(isVowel(pwd[i]))
++n_vowel;
if(pwd[i-1] == pwd[i] && (pwd[i-1] != 'e') && (pwd[i-1] != 'o')) //两个连续并且相同的字母
return false;
if(isVowel(pwd[i-1]) && isVowel(pwd[i]) && isVowel(pwd[i+1])) //三个全是元音
return false;
if(notVowel(pwd[i-1]) && notVowel(pwd[i]) && notVowel(pwd[i+1])) //三个全是辅音
return false;
}
if(!n_vowel)
return false;
else
return true;

}
int main()
{
char pwd[1000];
while(cin >> pwd && (strcmp(pwd,"end") != 0))
{
if(isAcceptable(pwd))
{
cout << "<" << pwd << "> is acceptable." << endl;
}
else
{
cout << "<" << pwd << "> is not acceptable." << endl;
}
}
}

网友答案：

5440699

2012-03-01 10:48:22

Accepted

1039

15MS

256K

936 B

C++

ajioy

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

int isVowel(char *pwd,int i)
{return (pwd[i] == 'a' || pwd[i] == 'e' || pwd[i] == 'i' || pwd[i] == 'o' || pwd[i] == 'u');}
int main()
{
char pwd[1000];
char str[2][30] = {"not acceptable.","acceptable."};
while(cin >> pwd && ( strcmp(pwd,"end") != 0))
{
int len = strlen(pwd);
int flag1=0,flag2=1,flag3=1;
for(int i = 0; i < len; ++i)
{
if(isVowel(pwd,i))
{
flag1 = 1;
if(i < len - 2 && isVowel(pwd,i+1) && isVowel(pwd,i+2) ) flag2 = 0;
if(i < len - 1 && pwd[i] == pwd[i+1] && (pwd[i] != 'e' && pwd[i] != 'o')) flag3 = 0;
}
else
{
if(i < len - 2 && !isVowel(pwd,i+1) && !isVowel(pwd,i+2)) flag2 = 0;
if(i < len - 1 && pwd[i] == pwd[i+1]) flag3 = 0;
}
}
cout << "<" << pwd <<"> is " << str[flag1 && flag2 && flag3] << endl;
}
}