PAT 1077 Kuchiguse [一般]
1077 Kuchiguse (20 分)
The Japanese language is notorious for its sentence ending particles. Personal preference of such particles can be considered as a reflection of the speaker's personality. Such a preference is called "Kuchiguse" and is often exaggerated artistically in Anime and Manga. For example, the artificial sentence ending particle "nyan~" is often used as a stereotype for characters with a cat-like personality:
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Itai nyan~ (It hurts, nyan~)
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Ninjin wa iyada nyan~ (I hate carrots, nyan~)
Now given a few lines spoken by the same character, can you find her Kuchiguse?
Input Specification:
Each input file contains one test case. For each case, the first line is an integer N (2≤N≤100). Following are N file lines of 0~256 (inclusive) characters in length, each representing a character's spoken line. The spoken lines are case sensitive.
Output Specification:
For each test case, print in one line the kuchiguse of the character, i.e., the longest common suffix of all N lines. If there is no such suffix, write nai
.
Sample Input 1:
3
Itai nyan~
Ninjin wa iyadanyan~
uhhh nyan~
Sample Output 1:
nyan~
Sample Input 2:
3
Itai!
Ninjinnwaiyada T_T
T_T
Sample Output 2:
nai
题目大意:给出m个句子,判断所有句子的最长后缀。
//有空格怎么办?那就不是一个character了。。。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include <algorithm> #include<cstdio> #include <vector> #include<cstring> #include<string.h> #include<string> using namespace std; int main() { int n; cin>>n; string pe="",now;//那就以第一个作为标准,剩下的去和它作比较。 int j=258,k=258,mins=j; for(int i=0;i<n;i++){//如何读取一行,真的是不会。。。。 getline(cin,now); if(pe==""){ pe=now;continue; } for(j=pe.size()-1,k=now.size()-1;k>=0,j>=0;j--,k--){ if(pe[j]!=now[k]){ break; } } if(j<mins){ mins=j; } } if(mins==258){ cout<<"nai"; }else{ for(int i=mins;i<pe.size();i++){ cout<<pe[i]; } } return 0; }
//本来是写成这个样子,根本就不行啊,不过整行读取字符串的函数写的是对的。
根据柳神的修改了之后的,可以AC了
#include <iostream> #include <algorithm> #include<cstdio> #include <vector> using namespace std; int main() { int n; cin>>n; string pe="",now,ans;//那就以第一个作为标准,剩下的去和它作比较。 getline(cin,now); for(int i=0;i<n;i++){//如何读取一行,真的是不会。。。。 getline(cin,now); reverse(now.begin(),now.end()); if(i==0){ ans=now; }else{ int j=0; for(j=0;j<ans.size()&&j<now.size();j++){ if(ans[j]!=now[j])break; } ans=ans.substr(0,j); } } reverse(ans.begin(),ans.end()); if(ans=="") cout<<"nai"; else cout<<ans; return 0; }
1.使用reverse进行倒转,因为是比较后缀,不太好比较,最后的答案需要再翻转过来。
2.关于这个getline(cin,now)读取一行的问题,在输入了n之后,其实就有一个\n被getline给读取进来了,所以这样就浪费了一个,需要单独加一个getline来消灭\n才对!
3.使用substr进行截取答案。