平常水题

                                                             C - 怪文書 / Dubious Document
                                                      Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement
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.
Input
Input is given from Standard Input in the following format:

n
S1
…
Sn
Output
Print the lexicographically smallest string among the longest strings that satisfy the condition. If the answer is an empty string, print an empty line.

Sample Input 1
3
cbaa
daacc

acacac

Sample Output 1

aac

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.

Sample Input 2
3
a
aa

b

Sample Output 2

The answer is an empty string.

（打比赛的时候没看懂题意，很尴尬......）

题意 ：用n个字符串S1.......Sn 。每个字符串都由小写的a........z中的字母组成。找出每个字符串共有的字符并挑选最少的。如：样例： S1  ：cbaa

                            S2  ：daacc

                            S3  ：acacac

三个字符串共有的字符是 a 和 c ，其中

S1  ：有2个a       S2  ：有2个a     S3  ：有3个a     挑选最少的  --  aa

S1  ： 有1个c       S2  ：有2个c     S3 ： 有3个c     挑选最少的  --  c

最后把挑选出来的字符按字典序最小排列 -- aac

解题思路：就直接扫一遍，每次就取最少的就行了（主要是考对字符串的处理）

#include<bits/stdc++.h>using namespace std;char h[1000][1000]；//第一个方括号记录第i个字符串，第二个方括号记录第i个字符串的第j个字符。int s[1000],ans1[1000],ans2[1000];int main(){int n,flag = 0;scanf("%d",&n);for(int i = 1;i <= n;i++){scanf("%s",h[i]);//这里字符串的输入要注意一下s[i] = strlen(h[i]);}for(int i = 49;i <=74;i++){ans2[i] = 10000000;//初始化}for(int i = 1;i <= n;i++){for(int j = 0;j < s[i];j++){    ans1[h[i][j] - '0']++;}for(int k = 49;k <= 74;k++){ans2[k] = min(ans1[k],ans2[k]);}memset(ans1,0,sizeof(ans1));}char p;for(int i = 49;i <= 74;i++){if(ans2[i] != 0){for(int j = 1;j <= ans2[i];j++){flag = 1;printf("%c",i + '0');}}}if(flag == 0) printf("\n");}