题解 | #公共子串计算#笨办法硬算#C语言

#define    _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()

{
    char in1[200] = { 0 }, in2[200] = { 0 };
    gets(in1);
    gets(in2);
    int len1, len2;
    int i, j, t1, t2, k = 0;
    int max;
    int out[2000];
    len1 = strlen(in1);
    len2 = strlen(in2);
    for (i = 0; i < len1; i++)
    {
        for (j = 0; j < len2; j++)
        {
            if (in1[i] == in2[j])
            {
                int h = 0;
                for (t1 = i, t2 = j; in1[t1] == in2[t2] && t1 < len1 && t2 < len2; t1++, t2++)
                {
                    h++;
                }

                out[k] = h;
                k++;
            }
        }
    }
    for (i = 0; i < k - 1; i++)
    {
        if (out[i] > out[i + 1])
        {
            max = out[i];
            out[i] = out[i + 1];
            out[i + 1] = max;
        }
    }
    printf("%d", out[k - 1]);
    return 0;
}