poj 2629 Common permutation
2011-11-22 12:55
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Common permutation
Description
Given two strings of lowercase letters, a and b, print the longest string x of lowercase letters such that there is a permutation of x that is a subsequence of a and there is a permutation of x that is a subsequence of b.
Input
Input consists of pairs of lines. The first line of a pair contains a and the second contains b. Each string is on a separate line and consists of at most 1,000 lowercase letters.
Output
For each subsequent pair of input lines, output a line containing x. If several x satisfy the criteria above, choose the first one in alphabetical order.
Sample Input
Sample Output
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 5059 | Accepted: 1519 |
Given two strings of lowercase letters, a and b, print the longest string x of lowercase letters such that there is a permutation of x that is a subsequence of a and there is a permutation of x that is a subsequence of b.
Input
Input consists of pairs of lines. The first line of a pair contains a and the second contains b. Each string is on a separate line and consists of at most 1,000 lowercase letters.
Output
For each subsequent pair of input lines, 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
#include<iostream> #include<algorithm> using namespace std; int comp(char &a,char &b) { return a<b; } int main() { char a[1001]; char b[1001]; int i,j; int lena,lenb; int pos; while(gets(a) && gets(b)) { lena=strlen(a); lenb=strlen(b); sort(a,a+lena,comp); sort(b,b+lenb,comp); pos=-1; for(i=0;i<lena;i++) { for(j=pos+1;j<lenb;j++) { if(a[i]==b[j]) { pos=j; cout<<a[i]; break; } } } cout<<endl; } return 0; }
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