PAT BASIC 1003

2019-11-11 04:00:42

字體：大中小

來源：轉載

供稿：網友

注意點

1、其實就是要求P之前的A的數量乘以PT之間A的數量等于T之后A的數量 2、注意PT中間等于零的情況 3、注意PT兩個字母：必須P在T之前出現，各有且僅有一次

代碼

#include<iostream>#include<string>using namespace std;void checkPAT();void checkPAT(){ string str; int pos_P = -1, pos_T = -1, state = 0; int i, j, len; int num[3]; cin >> str; len = str.size(); num[0] = 0; num[1] = 0; num[2] = 0; for (i = 0; i < len; i++) { if (str[i] != 'P' && str[i] != 'A' && str[i] != 'T') { cout << "NO/n"; return; } if (str[i] == 'A') num[state]++; else if (str[i] == 'P') { if (pos_P != -1) { cout << "NO/n"; return; } pos_P = i; state++; } else if (str[i] == 'T') { if (pos_T != -1) { cout << "NO/n"; return; } pos_T = i; state++; } } if (pos_P == -1 || pos_T == -1) { cout << "NO/n"; return; } if (pos_P >= pos_T) { cout << "NO/n"; return; } if (num[1] == 0) { cout << "NO/n"; return; } if (num[0] * num[1] == num[2]) cout << "YES/n"; else cout << "NO/n"; return;}int main(){ int n, i; cin >> n; for (i = 0; i < n; i++) checkPAT(); //while (1) //{ } return 0;}

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