Kuchiguse (20)简单字符串处理，输入坑

Kuchiguse (20)

时间限制 1000 ms 内存限制 65536 KB 代码长度限制 100 KB 判断程序 Standard (来自 小小)

题目描述

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:
Itai nyan~ (It hurts, nyan~)
Ninjin wa iyada nyan~ (I hate carrots, nyan~)
Now given a few lines spoken by the same character, can you find her Kuchiguse?

输入描述:

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.

输出描述:

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".

输入例子:

3
Itai nyan~
Ninjin wa iyadanyan~
uhhh nyan~

输出例子:

nyan~

分析:

找出最长公共后缀，用gets读时先getchar读掉n后的换行符，还有加上这两行读入优化会出错。。。。。（字符串题中慎用）

//std::ios::sync_with_stdio(false);
//std::cin.tie(0);

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<bits/stdc++.h>
#define ll long long
const int inf=0x3f3f3f3f;
using namespace std;

int main()
{
    int n; cin>>n;
    char a[106][300];
    int minn=inf;
    getchar();
    for(int i=1;i<=n;i++){
        char s[300];
        gets(s); int len=strlen(s);
        int t=0;
        for(int j=len-1;j>=0;j--){
            a[i][++t]=s[j];
        }
        minn=min(minn,len);
        //cout<<"len"<<len<<endl;

    }
    int ans=0,f=0;
    for(int i=1;i<=minn;i++){
        for(int j=2;j<=n;j++){
            if(a[j][i]!=a[1][i]){
                f=1; break;
            }
        }
        if(f) break;
        ans++;
    }
    if(ans==0) cout<<"nai"<<endl;
    else{
        for(int i=ans;i>=1;i--) cout<<a[1][i];
        cout<<endl;
    }
    return 0;
}