#include<iostream>
#include<string>
#include<vector>
using namespace std;

string str1, str2;
typedef long long ll;
const ll maxn = 1000;
const ll MOD = 1000000007;
const ll P = 10000019;

vector<pair<int, int>> pr1, pr2;

ll H1[maxn], H2[maxn];

ll powp[maxn];

void init(int len)
{      // 為了得到powp陣列，省去後面操作的時間
    powp[0] = 1;
    for(int i = 1; i <= len; i++)
    {
        powp[i] = (powp[i-1] * P)%MOD;
    }
}

void calH(ll H[], string str)
{
    //計算每個字串的雜湊值
    H[0] = str[0];
    for(int i = 1; i < str.length(); i++)
    {
        H[i] = (H[i-1] * P + str[i]) % MOD;
    }
}

int calsinglesubH(ll H[], int i, int j)
{
    if(i == 0) return H[j];
    else return ((H[j] - H[i-1]*powp[j-i+1]) % MOD + MOD) % MOD;
}

void calsubH(ll H[], int len, vector<pair<int, int>> &pr)
{   //計算每個字串單個子串的雜湊值，並用vector儲存
    for(int i = 0; i < len; i++)
    {
        for(int j = 0; j < len; j++)
        {
            int hashvalue = calsinglesubH(H, i, j);
            pr.push_back(make_pair(hashvalue, j-i+1));
        }
    }
}

int getmax()
{    //計算兩個字串最長相同子串的長度
    int ans = 0;
    for(int i = 0; i < pr1.size(); i++)
    {
        for(int j = 0; j < pr2.size(); j++)
        {
            if(pr1[i].first == pr2[j].first)
            {
                ans = max(ans, pr1[i].second);
            }
        }
    }
    return ans;
}

int main()
{
    getline(cin, str1);
    getline(cin, str2);

    init(max(str1.length(), str2.length()));

    calH(H1, str1);
    calH(H2, str2);

    calsubH(H1, str1.length(), pr1);
    calsubH(H2, str2.length(), pr2);

    printf("ans = %d", getmax());

    return 0;
}
 

測試用例：

ILOVEYOU

YOUDONTLOVEME

執行結果：

ans = 4

具體分析步驟有空再補，程式碼先看著