Sample Input

clinton
homer
riemann
marjorie

Sample Output

0
rie 3

看輸出才題意。。。
拓展kmp特征很明顯嘛。。。。
註意開始就匹配到尾的情況

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include 
 
<queue>
#include <algorithm>
#include <cmath>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define
 
 rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
 
const int maxn = 50010, INF = 0x7fffffff;
int nex[maxn], ex[maxn];

void get_next(char *s)
{
    int i=0, j, po, len = strlen(s);
    nex[0] = len;
    while(i+1 < len && s[i] == s[i+1])
        i++;
    nex[1] = i;
    po = 1;
    for(int i=2; i<len; i++)
    {
        if(i+nex[i-po] < po + nex[po])
            nex[i] = nex[i-po];
        else
        {
            j = po + nex[po] - i;
            if(j < 0) j = 0;
            while(i + j < len && s[i+j] == s[j])
                j++;
            nex[i] = j;
            po = i;
        }
    }
}

int get_ex(char *s1, char *s2)
{
    int i=0, j, po, len1 = strlen(s1), len2 = strlen(s2);
    get_next(s2);
    while(s1[i] == s2[i] && i < len1 && i < len2)
        i++;
    ex[0] = i;
    if(i == len1)
        return 1;
    po = 0;
    for(int i=1; i<len1; i++)
    {
        if(i + nex[i - po] < po + ex[po])
            ex[i] = nex[i-po];
        else
        {
            j = po + ex[po] - i;
            if(j < 0) j = 0;
            while(i + j < len1 && j < len2 && s1[i+j] == s2[j])
                j++;
            ex[i] = j;
            po = i;
            if(ex[i] == len1 - i)
                return i + 1;
        }
    }
    return 0;
}

char s1[maxn], s2[maxn];
int main()
{
    while(~rs(s1))
    {
        rs(s2);
        int flag = get_ex(s2, s1);
        int len = strlen(s2);
        if(flag)
        {
            rep(i, flag-1, len)
                cout<<s2[i];
            cout<<" ";
            cout<< ex[flag-1] <<endl;
        }
        else
            cout<< "0" <<endl;

    }


    return 0;
}

Simpsons’ Hidden Talents HDU - 2594（拓展kmp）