根据输入的字符串，判断并输出有效的字符串的长度，和该字符串。vc++

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题目的大概描述是这样的：

有一段字符串设置为密码，要求输入的字符串最长的对称的字符串为有效密码。

比如输入 aba, aa, abba, w2gtttg1, abaaab 等，其有效密码分别为：aba, aa, abba, gtttg, baaab.

下面是我写的实现代码，工作环境是VC6.0. （希望高手多多指教～）

// test.cpp : Defines the entry point for the console application.

//#include "stdafx.h"#include <iostream>#include <string>#include <stdlib.h>using namespace std;void getValidString(const std::string inputStr, std::string& outputStr, int& length);int main(int argc, char* argv[]){//printf("Hello World!\n");    std::string inputStr;cin >> inputStr;std::string outputStr;int length = 0;getValidString(inputStr, outputStr, length);cout << outputStr << " - " << length << endl;return 0;}void getValidString(const std::string inputStr, std::string& outputStr, int& length){int inLen = inputStr.length();length = 0;if (inLen <= 0){return;}for (int i=0; i<inLen; ++i){for (int j=0; j<(inLen-i)/2; ++j){bool isCode1 = false;bool isCode2 = false;// check if the current check subString is the valid code stringfor (int k=0; k<=j; k++){// check second mode: ababaif (i+j+k+2 >= inLen){isCode2 = false;break;}if(inputStr[i+j-k] == inputStr[i+j+k+2]){isCode2 = true;}else{isCode2 = false;break;}}// get the valid string and lengthif (isCode2 && 2*(j+1) + 1 > length){length = 2*(j+1) + 1;outputStr = inputStr.substr(i, length);}else {for (int k=0; k<=j; k++){// check first mode：abbaif(inputStr[i+j-k] == inputStr[i+j+k+1]){isCode1 = true;}else{isCode1 = false;break;}}if(isCode1 && 2*(j+1) > length){length = 2*(j+1);outputStr = inputStr.substr(i, length);}} //end of esle}//end of for j}//end of for i}

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