神奇递归Gym

While I was working in the company, the internet broke. I didn't have anything to do without internet, so I decided to write this ACM problem.

We need to mix between two strings and also we should keep the same order for both strings.

Example: if we have s1 = "ab" and s2 = "cd", we can generate six strings:

abcd acbd

acdb cdab

cadb cabd

thank you for your help in ”Count Mix Strings” problem. now I can calculate the complexity and make input and output files. You will be given the two strings and you should print all the distinct strings that could be generated in the alphabetical order.

Input
Your program will be tested on one or more test cases. The first line of the input will be a single integer T, the number of test cases (1 ≤ T ≤ 100).

Followed by the test cases, each test case is on one line. it contains two strings s1 and s2, both strings consist of at least 1 and at most 8 lower case English letters (from ‘a’ to ‘z’).

Output
For each test case, print all the distinct strings that could be generated in the alphabetical order.

print a blank line after each test case.

Example
Input
2
a aa
ab cd
Output
aaa

abcd
acbd
acdb
cabd
cadb

cdab

给你两个字符串，混合在一起，按字典序输出。

#include <bits/stdc++.h>//#include <ext/pb_ds/tree_policy.hpp>//#include <ext/pb_ds/assoc_container.hpp>//using namespace __gnu_pbds;using namespace std;#define pi acos(-1)#define endl '\n'#define me(x) memset(x,0,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)typedef long long LL;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,-1,-1,1,1};const int dy[]={0,1,0,-1,1,-1,1,-1};const int maxn=1e3+5;const int maxx=2e6+100;const double EPS=1e-7;const int mod=1000000007;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}//typedef tree<pt,null_type,less< pt >,rb_tree_tag,tree_order_statistics_node_update> rbtree;/*lch[root] = build(L1,p-1,L2+1,L2+cnt);    rch[root] = build(p+1,R1,L2+cnt+1,R2);中前*//*lch[root] = build(L1,p-1,L2,L2+cnt-1);    rch[root] = build(p+1,R1,L2+cnt,R2-1);中后*/long long gcd(long long a , long long b){if(b==0) return a;a%=b;return gcd(b,a);}string s1,s2;set<string >s;int len1,len2;void dfs(int i,int j,string ans){    if(i>=len1&&j>=len2)    {        s.insert(ans);        return ;    }    if(i<len1) dfs(i+1,j,ans+s1[i]);    if(j<len2) dfs(i,j+1,ans+s2[j]);}int main(){    int n;    cin>>n;    while(n--)    {        s1.clear();s2.clear();        s.clear();        cin>>s1>>s2;        len1=s1.size();        len2=s2.size();        dfs(0,0,"");        foreach(it,s)        cout<<*it<<endl;        cout<<endl;    }}