题解 | #公共子串计算#

#include <iostream>
using namespace std;

int main() {
    // dp[i][j]: s1[0..i)以s[i-1]结尾, s2[0..j)以s[j-1]结尾的LCS
    int dp[151][151] = {0};
    int max_len = 0;
    string s1, s2;
    cin >> s1 >> s2;
    int len1 = s1.size(), len2 = s2.size();
    // dp[x][0]与dp[0][x]必然为0 (按照定义是空串)
    for(int i = 1; i <= len1; ++i)
    {
        for(int j = 1; j <= len2; ++j)
        {
            dp[i][j] = s1[i-1] == s2[j-1] ? dp[i-1][j-1] + 1 : 0; 
            max_len = max(max_len, dp[i][j]);
        }
    }
    printf("%d", max_len);
}