Dubious Document

最新推荐文章于 2024-02-06 18:28:22 发布

转载最新推荐文章于 2024-02-06 18:28:22 发布 · 261 阅读

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原文链接：http://www.cnblogs.com/hao-tian/p/9152544.html

本文介绍了一个有趣的编程挑战——纸艺重构。参与者需从给定的新闻标题中剪裁字符并重新排列组合成新的字符串。任务的目标是在不知道确切标题的情况下找出能从任一给定字符串中构造出的最长且字典序最小的字符串。

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问题 K: Dubious Document

时间限制: 1 Sec 内存限制: 128 MB
提交: 34 解决: 28
[提交][状态][讨论版][命题人:admin]

题目描述

Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string.
He will receive a headline which contains one of the strings S1,…,Sn tomorrow. He is excited and already thinking of what string he will create. Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains.
Find the longest string that can be created regardless of which string among S1,…,Sn the headline contains. If there are multiple such strings, find the lexicographically smallest one among them.

Constraints
1≤n≤50
1≤|Si|≤50 for every i=1,…,n.
Si consists of lowercase English letters (a - z) for every i=1,…,n.

输入

Input is given from Standard Input in the following format:
n
S1
…
Sn

输出

Print the lexicographically smallest string among the longest strings that satisfy the condition. If the answer is an empty string, print an empty line.

样例输入

3
cbaa
daacc
acacac

样例输出

aac

提示

The strings that can be created from each of cbaa, daacc and acacac, are aa, aac, aca, caa and so forth. Among them, aac, aca and caa are the longest, and the lexicographically smallest of these three is aac.

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int main()
{
    int n;
    char str[55];
    cin>>n;
    cin>>str;
    int len = strlen(str);
    int ans[50]={0};
    for(int i=0;i<len;i++)
    {
        ans[str[i]-'a']++;
    }
    for(int i=0;i<n-1;i++)
    {
        cin>>str;
        len = strlen(str);
        int temp[55]={0};
        for(int j=0;j<len;j++)
        {
            temp[str[j]-'a']++;
        }
        for(int i=0;i<26;i++)
        {
            ans[i] = min(ans[i],temp[i]);
        }
    }
    for(int i=0;i<26;i++)
    {
        while(ans[i])
        {
            printf("%c",i+'a');
            ans[i]--;
        }
    }
}

转载于:https://www.cnblogs.com/hao-tian/p/9152544.html