PAT Advanced 1035

来源：互联网发布：程序员述职报告范文编辑：程序博客网时间：2024/06/05 07:41

题意：

给你一些字符串，对于一些特殊的字符要进行转换。

需要注意的地方：

If no account is modified, print in one line "There are N accounts and no account is modified" where N is the total number of accounts. However, if N is one, you must print "There is 1 account and no account is modified" instead.是这句话。。。。。如果总数是1且不用改的话，那么accounts要写成account。。。。。我在这没认真读题被坑了一下

#include<iostream>#include<string>#include<fstream>#include<map>using namespace std;//ifstream fin("fin.txt");//streambuf *old = cin.rdbuf(fin.rdbuf());int times;map<string,string>mymap;string temp;string m[1111];string modify[1111];int k = 0;string pos;void solve(){bool flag = true;for (int i = 0; i < times; i++){temp = m[i];pos = temp;temp = mymap[temp];flag = true;int length = temp.length();for (int j = 0; j < length; j++){if (temp[j] == '1'){temp[j] = '@';flag = false;}else if (temp[j] == '0'){temp[j] = '%';flag = false;}else if (temp[j] == 'l'){temp[j] = 'L';flag = false;}else if (temp[j] == 'O'){temp[j] = 'o';flag = false;}}if (!flag){modify[k++] = pos;mymap[pos] = temp;}}if (k == 0){if (times == 1)cout << "There is 1 account and no account is modified" << endl;elsecout << "There are " << times << " accounts and no account is modified" << endl;}else{cout << k << endl;for (int i = 0; i < k; i++){cout << modify[i] << " " << mymap[modify[i]] << endl;}}}void input(){cin >> times;string mt;for (int i = 0; i < times; i++){cin >> m[i] >> mt;;mymap[m[i]] = mt;}solve();}int main(){input();return 0;}

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