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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:Re:使用C++STL中的next_permutation()可以大大简化此题的难度 Posted by:qianguangpan at 2010-05-01 15:50:50 my_code: #include <iostream> #include <algorithm> using namespace std; char str[51]; int main(void) { while(scanf("%s",str)) { if(str[0]=='#') break; if(next_permutation(str,str+strlen(str))) { printf("%s\n",str); } else { printf("No Successor\n"); } } return 0; } Followed by: Post your reply here: |
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