UVA10252 POJ2629 Common Permutation【字符串排序】

Common permutation

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5782		Accepted: 1749

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

pretty
women
walking
down
the
street

Sample Output

e
nw
et

Source

The UofA Local 1999.10.16

问题链接：UVA10252 POJ2629 Common Permutation。

问题描述：

　　两个小写字母构成的字符串a和b，求各自的置换的最长公共子串，按字母顺序输出。

问题分析：（略）。

程序说明：

　　字符串类型变量的排序也是可以用函数sort()实现的。

AC的C++语言程序如下：

/* UVA10252 POJ2629 Common Permutation */

#include 
#include 
#include 
#include 
#include 

using namespace std;

int main()
{
    string a, b;
    int lena, lenb;

    while(getline(cin, a) && getline(cin, b)) {
        // 计算字符串长度
        lena = a.size();
        lenb = b.size();

        // 字符串排序
        sort(a.begin(), a.end());
        sort(b.begin(), b.end());

        // 匹配求公共子串，输出结果
        for(int i=0, j=0; i b[j])
                j++;
            else if(a[i] < b[j])
                i++;
        }
        printf("\n");
    }

    return 0;
}