POJ 1002 解题报告 487-3279
典型的字符串处理问题。
刚开始用的vector，果断超时。题目本身没什么难度，只在选择容器上要考虑。
AC咯。
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#include <iostream>#include <string>#include <iomanip>#include <map>using namespace std;int nicknameToNumber(string&);int main (){int iPhoneTotal;cin >> iPhoneTotal;map<int, int> mPhone;while (iPhoneTotal--){string strTemp;cin >> strTemp;++mPhone[nicknameToNumber(strTemp)];}int iFlag = 0;map<int, int>::iterator iter = mPhone.begin();while (iter != mPhone.end()){if (iter->second > 1){iFlag = 1;cout << setw(3) << setfill('0') << iter->first/10000 << "-" << setw(4) << iter->first%10000 << " " << iter->second << endl;}iter++;}if (!iFlag){cout << "No duplicates." << endl;}return 0;}int nicknameToNumber(string& strPhone){int r = 0;int i = 0;while (i < strPhone.size()){if ('-' == strPhone[i] || 'Q' == strPhone[i] || 'Z' == strPhone[i]){}else if (strPhone[i] >= 'A' && strPhone[i] <= 'P'){r = r * 10 + 2 + (strPhone[i] - 'A')/3;}else if (strPhone[i] >= 'R' && strPhone[i] <= 'Y'){r = r * 10 + 2 + (strPhone[i] - 'A' -1)/3;}else if (strPhone[i] <= '9' && strPhone[i] >= '0'){r = r * 10 + (strPhone[i] - '0');}i++;}return r;}