POJ_1256_Anagram

POJ_1256_Anagram
题意：字符串排列，大写字母较小写字母对应小（A < b < C）
思路：先写一个比较函数在sort一下，再利用STL中next_permutation输出所有从小到大的排列。

#include <iostream>#include <iomanip>#include <cstring>#include <cstdio>#include <cstdlib>#include <cmath>#include <set>#include <map>#include <list>#include <stack>#include <deque>#include <queue>#include <vector>#include <algorithm>#include <functional>#define debug(x) cout << "--------------> " << x << endlusing namespace std;const double PI = acos(-1.0);const double eps = 1e-10;const long long INF = 0x7fffffff;const long long MOD = 1000000007;bool cmp(const char &a, const char &b){    if(a <= 'Z' && a >= 'A' && b <= 'Z' && b >= 'A')        return a < b;    if(a <= 'z' && a >= 'a' && b <= 'z' && b >= 'a')        return a < b;    if(a <= 'Z' && a >= 'A' && b <= 'z' && b >= 'a')        return a + 32 <= b;    if(a <= 'z' && a >= 'a' && b <= 'Z' && b >= 'A')        return a - 32 < b;}int main(){    int n;    scanf("%d", &n);    while(n--)    {        char str[27];         cin >> str;         int len = strlen(str);         sort(str, str+len, cmp);         cout << str << endl;         while(next_permutation(str, str+len, cmp))         {            cout << str << endl;         }    }     return 0;}