Common Permutation

Given two strings a and b, print the longest string x of letters such that there is a permutation of x that isa subsequence of a and there is a permutation of x that is a subsequence of b.

 

Input

The input file contains several cases, each case consisting oftwo consecutive lines.This means that lines 1 and 2 are a test case, lines 3 and 4 are another test case, and so on.Each line contains one string of lowercase characters, withfirst line of a pair denoting a and the second denoting b.Each string consists of at most 1,000 characters.

 

Output

For each set of input, output a line containing x. If several x satisfy the criteria above, choose the first one in alphabetical order.

 

Sample Input

 

pretty
women
walking
down
the
street

 

Sample Output

 

e
nw
et

my code:
#include <stdio.h>#include <stdlib.h>#include <string.h>int cmp(const void *a,const void *b){if((*(char*)a  - *(char*)b)>0)return 1;return -1;}int main(){char str1[1005];char str2[1005];int len1,len2,i,k,x,y;while(gets(str1)){//if(gets(str1)==NULL)//break;gets(str2);len1 = strlen(str1);len2 = strlen(str2);qsort(str1,strlen(str1),sizeof(str1[0]),cmp);qsort(str2,strlen(str2),sizeof(str2[0]),cmp);//k = (len1-len2)?len1:len2;k = (len1>len2)?len1:len2;//for(x=0,y=0,i = 0;i <k;i++ ) 这样写 会提前跳出循环 一定要加强逻辑的训练 找出那些条件才是循环的控制点x=0,y=0;while((x<len1)&&(y<len2)){if(str1[x]==str2[y]){printf("%c",str1[x]);x++;y++;}else if(str1[x]>str2[y])y++;elsex++;//if((x>=len1)||(y>=len2))//if((x>len1)&&(y>len2))//break;}printf("/n");}}