1077. Kuchiguse (20)

题目大意:给一些字符串 求最大公共后缀子串

思路:

设置变量Count初始化0 记录每两串的公共后缀子串的长度

后每输入一个字符串 根据最小原则 更新Count

最后根据Count是否为0 来确定最终公共后缀子串

在这里我多此一举 将字符串倒转 然而 从后逐一判断 就可以了 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<string>
#include<cmath>
#include<algorithm>
using namespace std;
const int Max=1<<30;

int getCommonLen(string S1, string S2)
{
    int CL=0;
    int size=(S1.size()<S2.size() ? S1.size():S2.size());
    for(int i=0; i<size; i++)
    {
        if(S1[i]==S2[i])
           CL++;
        else
            return CL;
    }
    return CL;
}

int main()
{
    int T,Count=Max,flag=0;
    string S1, S2;

    cin>>T;
    getchar();

    getline(cin, S1);
    getline(cin, S2);
    reverse(S1.begin(),S1.end());
    reverse(S2.begin(),S2.end());
    Count=(getCommonLen(S1,S2)<Count ? getCommonLen(S1,S2):Count);

    if(Count==0)
        flag=1;
    T-=2;
    while(T--)
    {
        getline(cin,S1);
        if(!flag){
            reverse(S1.begin(),S1.end());
            Count=(getCommonLen(S1,S2)<Count ? getCommonLen(S1,S2):Count);
            if(Count==0)
                flag=1;
            S2=S1;
        }
    }
    if(flag)
        cout<<"nai"<<endl;
    else
    {
        S2=S2.substr(0,Count);
        reverse(S2.begin(),S2.end());
        cout<<S2<<endl;
    }
    return 0;
}