PATBasic——1003. 我要通过！(20)

1003. 我要通过！(20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

“答案正确”是自动判题系统给出的最令人欢喜的回复。本题属于PAT的“答案正确”大派送 —— 只要读入的字符串满足下列条件，系统就输出“答案正确”，否则输出“答案错误”。

得到“答案正确”的条件是：

1. 字符串中必须仅有P, A, T这三种字符，不可以包含其它字符；
2. 任意形如 xPATx 的字符串都可以获得“答案正确”，其中 x 或者是空字符串，或者是仅由字母 A 组成的字符串；
3. 如果 aPbTc 是正确的，那么 aPbATca 也是正确的，其中 a, b, c 均或者是空字符串，或者是仅由字母 A 组成的字符串。

现在就请你为PAT写一个自动裁判程序，判定哪些字符串是可以获得“答案正确”的。

输入格式： 每个测试输入包含1个测试用例。第1行给出一个自然数n (<10)，是需要检测的字符串个数。接下来每个字符串占一行，字符串长度不超过100，且不包含空格。

输出格式：每个字符串的检测结果占一行，如果该字符串可以获得“答案正确”，则输出YES，否则输出NO。

输入样例：

8PATPAATAAPATAAAAPAATAAAAxPATxPTWhateverAPAAATAA

输出样例：

YESYESYESYESNONONONO

这个我真的是被整晕了的，先把不完整的代码贴上来，请忽略我，我只是放在这里，过一会儿再来看，因为现在实在是太晕了。如有高人指点，不胜感激！菜鸟真的是伤不起啊

//1h 第一版本（想看正确答案者，请忽略我）

int main(){int n;cin >> n;vector<string> strings;for (int i = 0; i < n; i++){string input_s;cin >> input_s;strings.push_back(input_s);}for (int j = 0; j < strings.size(); j++){string each_string = strings[j];string::iterator iters;int k = 0;int p=0,t=0;for (iters = each_string.begin(); iters != each_string.end(); iters++){char each_char = *iters;if (each_char == 'P'){p = k;}if (each_char == 'T'){t = k;}k++;}if ((p != 0) && (t != (k - 1))){int pp = 0;for (int m = 0; m < p; m++){if (each_string.at(m) == 'A'){pp++;}}int tt = 0;for (int n = p + 1; n < t; n++){if (each_string.at(n) == 'A'){tt++;}}int hh = 0;for (int h = t + 1; h < k; h++){if (each_string.at(h) == 'A'){hh++;}}if ((pp == p) && (tt == (t - p - 1)) && (hh = k - t - 1)){cout << "YES" << endl;}else{cout << "NO" << endl;}}else{if ((p == 0) && (t == (k - 1))){int xx=0;for (int x = 1; x < t; x++){if (each_string.at(x) == 'A'){xx++;}}if (xx == (t - 1)){cout << "YES" << endl;}else{cout << "NO" << endl;}}}}return 0;}

第二版本代码，只能得六分。可能还是我没有找到合适的算法去解这道题！先坚持不看答案，自己再想！

#include<vector>#include <sstream>#include<cmath>#include<iomanip>#include<iostream>#include <ctype.h>#include <stdlib.h>#include <algorithm>using namespace std;bool IsAllA(string a, int m, int n)//判断数组a中，m到n位置的元素是否全为A{bool temp = true;for (int i = m; i <= n; i++){if (a[i] != 'A'){temp = false;}}return temp;}int main(){int n;cin >> n;string str[10];bool state = 0;vector<string> rightstrs;for (int i = 0; i < n; i++){cin >> str[i];}for (int i = 0; i < n; i++){string s = str[i];if (s[0] == 'P')//判断第一个元素是否为P{if (s[s.length() - 1] == 'T')//判断最后一位是否为T{//判断中间的元素是否全为Aif (s.length() == 2){state = false;}else{state = IsAllA(s, 1, s.length() - 2);}if (state == true){rightstrs.push_back(s);}}else{state = false;}}else if (s[0]=='A'){int pl = s.find('P');//找到第一个P出现的位置int tl = s.find('T');//找到T的位置if ((IsAllA(s, 1, pl - 1))&&(pl>0)&&(tl>0))//首A到首P全是A，且PT都存在{if (((tl - pl) == 2) && (s[tl - 1] == 'A'))//判断P和T中间是否只有一个A元素{state = IsAllA(s, tl + 1, s.length() - 1);if (state == true){rightstrs.push_back(s);}}else if ((IsAllA(s,pl+1,tl-1))&&(IsAllA(s,tl+1,s.length()-1)))//判断PT中间不止一个A的情况,但满足PT两边全是A{string a = s.substr(0,pl);string b = s.substr(pl + 2, tl-pl-2);string ca = s.substr(tl+1,s.length()-1-tl);string c;for (int x = 0; x < (ca.length() - a.length()); x++){c += "A";}string temp = a + "P" + b + "T" + c;for (int y = 0; y < rightstrs.size(); y++){if (temp == rightstrs[y]){state = true;break;}}}else{state = false;}}else{state = false;}}else{state = false;}if (state==1)cout << "YES" << endl;elsecout << "NO" << endl;}return 0;}

第三版本答案，17分，不知什么临界点没考虑到

#include<vector>#include <sstream>#include<cmath>#include<iomanip>#include<iostream>#include <ctype.h>#include <stdlib.h>#include <algorithm>using namespace std;bool IsAllA(string a, int m, int n)//判断数组a中，m到n位置的元素是否全为A{bool temp = true;for (int i = m; i <= n; i++){if (a[i] != 'A'){temp = false;}}return temp;}bool IfCheck(string s){int  count = 0;for (int i = 0; i < s.length(); i++){if ((s[i] == 'A') || (s[i] == 'P') || (s[i] == 'T')){count++;}}if (count == s.length()){return true;}else{return false;}}int main(){int n;cin >> n;string str[10];vector<string> rightstrs;for (int i = 0; i < n; i++){cin >> str[i];}for (int i = 0; i < n; i++){string s = str[i];int slen = s.length();bool state = 0;if (IfCheck(s)){int pl = s.find('P');int tl = s.find('T');int al = s.find('A');if ((pl >= 0) && (tl >= 0) && (al >= 0))//如果三种元素都存在{if (pl == 0) //第一个元素是否为P{if (tl == (slen - 1))//最后元素是否为T{if (s=="PAT"){state = true;rightstrs.push_back(s);}else if (IsAllA(s, 1, slen - 2))//中间元素全是A{string temp = "P" + s.substr(2,slen-2);for (int j = 0; j < rightstrs.size(); j++){if (temp == rightstrs[j]){state = true;rightstrs.push_back(s);}}}}}else if (al==0)//第一个元素为A的情况{if (((tl - pl) == 2) && (s[pl + 1] == 'A'))//中间含有PAT，判断两边A的个数是否相同{int cnta1 = pl;int cnta2 = slen - 1 - tl;if ((cnta1 == cnta2) && (IsAllA(s, 0, pl - 1)) && (IsAllA(s, tl+1, slen - 1)))//注意判断是否全是A{state = true;rightstrs.push_back(s);}}else {string a = s.substr(0, pl);//这里是通过aPbATca 组合原本的aPbTc，然后判断它是否在原来的正确字符串集合中出现过string b = s.substr(pl + 2, tl-pl-2);string ca = s.substr(tl+1,slen-1-tl);string c;if ((a != "") && (b != "") && (ca != "")){for (int x = 0; x < (ca.length() - a.length()); x++){c += "A";}if (IsAllA(a, 0, a.length() - 1) && IsAllA(b, 0, b.length() - 1) && IsAllA(c, 0, c.length() - 1)){string temp = a + "P" + b + "T" + c;for (int y = 0; y < rightstrs.size(); y++){if (temp == rightstrs[y]){state = true;rightstrs.push_back(s);break;}}}}}}}}if (state==1)cout << "YES" << endl;elsecout << "NO" << endl;}return 0;}

第四版本的代码，18分，自己都被自己进步的速度惊呆了，不过不要放弃，坚持就是胜利！！！！（先放在这里，改天再来继续做）

#include<vector>#include <sstream>#include<cmath>#include<iomanip>#include<iostream>#include <ctype.h>#include <stdlib.h>#include <algorithm>using namespace std;bool IsAllA(string a){bool state = true;for (int i = 0; i < a.length(); i++){if (a[i] != 'A'){state = false;break;}}return state;}bool IfDealed(string s){int num = 0;for (int i = 0; i < s.length(); i++){if ((s[i] == 'P') || (s[i] == 'A') || (s[i] == 'T')){num++;}}if (num == s.length()){return true;}else{return false;}}int main(){int n;cin >> n;string strs[10];vector<string> IsTrue;for (int i = 0; i < n; i++){cin >> strs[i];}for (int i = 0; i < n; i++){string s = strs[i];if (IfDealed(s)){if (s.length() < 3){cout << "NO" << endl;}else{int pl = s.find('P');int tl = s.find('T');//这次版本的思路是，首先判断P为首字母时的情况，只有PAT，基于条件的PAAAT才正确，其他错误if (pl == 0) {if (tl == (s.length() - 1)){if ((tl - pl) == 2){if (s[pl + 1] == 'A'){cout << "YES" << endl;IsTrue.push_back(s);}else{cout << "NO" << endl;}}else{//这种情况下也应该判断一下，字符串PT中间是否都为Aint numa = tl - pl - 1;string temp;for (int k = 0; k < numa - 1; k++){temp += "A";}temp = "P" + temp + "T";bool iftrue = false;for (int j = 0; j < IsTrue.size(); j++){if (temp == IsTrue[j]){iftrue = true;IsTrue.push_back(s);cout << "YES" << endl;}}if (!iftrue)cout << "NO" << endl;}}else{cout << "NO" << endl;}}//首字母不是P，利用aPbTcelse{int pa = pl;int pta = tl - pl - 1;int ta = s.length() - tl - 1;if ((ta / pa) == pta){string a = s.substr(0, pl);//这里是通过aPbATca 组合原本的aPbTc，然后判断它是否在原来的正确字符串集合中出现过string b = s.substr(pl + 1, tl - pl - 2);string c = s.substr(tl + 1, s.length() - 1 - tl);if ((IsAllA(a)) && (IsAllA(b)) && (IsAllA(c))){//感觉是这里的问题，满足上述条件，还要判断原来的aPbTc是否正确啊，有点晕cout << "YES" << endl;IsTrue.push_back(s);}else{cout << "NO" << endl;}}else{cout << "NO" << endl;}}}}else{cout << "NO" << endl;}}return 0;}