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Re:使用C++STL中的next_permutation()可以大大简化此题的难度In Reply To:使用C++STL中的next_permutation()可以大大简化此题的难度 Posted by:jourkliu at 2009-03-09 01:07:40 > 程序仅供参考: > #include <iostream> > #include <string> > #include <algorithm> > #include <vector> > using namespace std; > int main() > { > string s; > while(cin>>s&&s!="#"){ > if(next_permutation(s.begin(),s.end())) cout<<s<<endl; > else cout<<"No Successor"<<endl; > } > return 0; > } 能解释一下什么意思么? next_permutation(s.begin(),s.end()) Followed by: Post your reply here: |
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