输入两个字符串，比如abdbcc和abc，输出第二个字符串在第一个字符串中的连接次序

解析：本题算法是先找到第二个字符串中各个字符分别出现的位置，利用其位置构造多叉树（构造规则为子节点必须比父节点大）。遍历构造完成的多叉树即相应的组合。
例子中搜索得到a, b, c在源字符串中出现的位置序列分别为{1}、{2，4}、{5，6}.
1
2 4
5 6 5 6
然后将算法组合成125，126，145，146。

#include<string.h>#include<stdlib.h>#include<stdio.h>#include<iostream>using namespace std;void PrintfArrary(char *p_str, char *s_str,int* print_arr, int p_len,     int s_len, int print_arr_num,int p_start_num, int s_start_num){    int pStartnum = p_start_num, sStartNum = s_start_num;    int printNum = print_arr_num;    if (printNum == s_len)    {        for (int i = 0;i < s_len;i++)        {            cout<< *(print_arr + i);            cout << " ";        }        cout << endl;        return;    }    for (int i = pStartnum;i < p_len;i++)    {        for (int j = sStartNum; j < s_len; j++)        {            if (*(p_str+i)==*(s_str+j))            {                print_arr[printNum] = i + 1;                pStartnum = i;                sStartNum = j;                PrintfArrary(p_str, s_str, print_arr, p_len, s_len, printNum + 1, pStartnum + 1,                    sStartNum + 1);            }        }    }}void ConnectSequence(char *p_str, char *s_str){    int p_len = strlen(p_str);    int s_len = strlen(s_str);    int *print_arr = new int[s_len];    unsigned int print_arr_num = 0;    if (NULL == p_str && NULL == s_str)    {        cout << "string erro" << endl;        return;    }    if (NULL == print_arr)    {        cout << "allocate error" <<endl;        return;    }    PrintfArrary(p_str, s_str, print_arr, p_len, s_len, 0, 0, 0);}int main(){    char ParString[] = "abdbcca";    char SonString[] = "abc";    ConnectSequence(ParString, SonString);    getchar();    return 0;}