5-1 字符串替换 (10分)

将文本文件中指定的字符串替换成新字符串。 由于目前的OJ系统暂时不能支持用户读入文件，我们编写程序从键盘输入文件中的内容，当输入的一行为end时，表示结束。end后面有两个字符串，要求用第二个字符串替换文本中所有的第一个字符串。

输入格式:

Xi’an Institute of Posts and Telecommunications is co-designed and implemented by the People’s Government of Shaanxi Province and the Ministry of Industry and Information Technology. The Institute is located in Xi’an, a historic city in Northwest China, famous for its magnificent ancient culture.

end （表示结束）

Institute （第一个字符串，要求用第二个字符串替换）

University （第二个字符串）

输出格式:

Xi’an University of Posts and Telecommunications is co-designed and implemented by the People’s Government of Shaanxi Province and the Ministry of Industry and Information Technology.The University is located in Xi’an, a historic city in Northwest China, famous for its magnificent ancient culture.

输入样例:

Xi’an Institute of Posts and Telecommunications is co-designed and implemented by the People’s Government of Shaanxi Province and the Ministry of Industry and Information Technology.The Institute is located in Xi’an, a historic city in Northwest China, famous for its magnificent ancient culture.endInstituteUniversity

输出样例:

Xi’an University of Posts and Telecommunications is co-designed and implemented by the People’s Government of Shaanxi Province and the Ministry of Industry and Information Technology.

The University is located in Xi’an, a historic city in Northwest China, famous for its magnificent ancient culture.

解析：使用string容器中的find来查找，replace来替换。另外需注意换行，使用getline来确定哪里需要换行。

关于find函数的返回值：1、找到，返回找到字符的地址或字符串的首地址； 2、没找到，返回string::npos

#include <iostream>#include <string>#include <vector>using namespace std;int main () {int found;string s1, s2, s;vector<string> v;getline(cin, s); //接收一串子直到碰到换行符（不包括换行符） v.push_back(s + '\n');while (1) {getline(cin, s);if ( s.compare("end") == 0 )//相等返回0.小于返回负数，反之正数 break;else {s += '\n';v.push_back(s);}}cin >> s1 >> s2;for ( int i = 0; i < v.size(); i++ ) {found = v[i].find(s1);// std::string::npos = -1while ( found != std::string::npos ) {//从第found个元素开始，连续s1.lenth()个元素，替换s2 v[i].replace(found, s1.length(), s2);//找下一个s1  found = v[i].find(s1, found + 1);}}vector<string>::iterator it;for ( it = v.begin(); it != v.end(); it++ ) {cout << *it;}return 0;}