poj 1016

该题主要是字符串的处理，没有什么难度，但是不细心的话很容易出错！我的代码还是比较繁琐，以后敲代码之前还是要先构思一下，不要遗漏逻辑分支以及局域变量的reset！对string的理解还不够透彻，不能灵活应用，以前一直以为+号在string里面只能用于string与string之间，现在竟然发现string s = 1 + ‘3’；等于“4”！

code：#include <iostream>#include <string>#include <vector>#include <map>using namespace std;int main(){vector<int> state;map<char, int> result;string number, temp, former;cin>>number;vector<string> sumstring, tempstring;vector<bool> type;int count = 0;while(number != "-1"){sumstring.push_back(number);while(true){tempstring.push_back(number);for(int ix = 0; ix != 10; ++ix){result.insert(make_pair('0'+ix, 0));}++count;for(string::iterator iter = number.begin(); iter != number.end(); ++iter){++result[*iter];}for(map<char, int>::iterator iter = result.begin(); iter != result.end(); ++iter){if(iter->second != 0){char seco[5];_itoa_s(iter->second,seco,10);string sec(seco);temp += sec;temp.push_back(iter->first);}}size_t ix = 0;for(ix = 0; ix != tempstring.size(); ++ix){if(temp == tempstring[ix]){if(ix == 0 && tempstring.size() == 1){type.push_back(1);state.push_back(count);}else if(ix == tempstring.size()-1 && tempstring.size() > 1){type.push_back(0);state.push_back(count);}else{type.push_back(1);state.push_back(tempstring.size()-ix);}break;}}if(ix != tempstring.size()){temp.clear();result.clear();break;}if(count == 15){type.push_back(0);state.push_back(16);temp.clear();result.clear();break;}number = temp;temp.clear();result.clear();}tempstring.clear();count = 0;cin>>number;}for(size_t ix = 0; ix != type.size(); ++ix){if(type[ix] == 1){if(state[ix] == 1){cout<<sumstring[ix]<<" is self-inventorying/n";}else{cout<<sumstring[ix]<<" enters an inventory loop of length "<<state[ix]<<"/n";}}else if(type[ix] == 0){if(state[ix] == 16){cout<<sumstring[ix]<<" can not be classified after 15 iterations/n";}else{cout<<sumstring[ix]<<" is self-inventorying after "<<state[ix]-1<<" steps/n";}}}return 0;}

看了一下网上的做法，简洁多了！

#include"iostream"#include"string"#include"algorithm"using namespace std;string str[18];void change(int i){ int w[10]={0}; for(unsigned int j=0;j<str[i].size();j++) w[str[i][j]-'0']++; string s=""; for(int j=0;j<=9;j++){ if(w[j]>=1&&w[j]<=9){ s+=w[j]+'0'; s+=j+'0'; } else if(w[j]>=10){ s+=w[j]/10+'0'; s+=w[j]%10+'0'; s+=j+'0'; } } str[i+1]=s;}int main(){ while(cin>>str[0],str[0]!="-1") { int j=0,flag=0,i; for(j;j<15;j++) { change(j); for(i=0;i<=j;i++) if(str[i]==str[j+1]){ flag=1; break; } if(flag) break; } if(flag){ if(j==0&&i==0){ cout<<str[0]<<" is self-inventorying"<<endl; } else if(i==j) cout<<str[0]<<" is self-inventorying after "<<j<<" steps"<<endl; else cout<<str[0]<<" enters an inventory loop of length "<<j-i+1<<endl; } else cout<<str[0]<<" can not be classified after 15 iterations"<<endl; } return 0;}