Rabin-Karp算法代码中的负散列值

Question

我是从这个网站了解拉宾-卡普算法的：https://www.geeksforgeeks.org/rabin-karp-algorithm-for-pattern-searching/

他们在C++中为算法编写了以下代码：

#include <bits/stdc++.h> 
using namespace std; 
  
// d is the number of characters in the input alphabet  
#define d 256  
  
/* pat -> pattern  
    txt -> text  
    q -> A prime number  
*/
void search(char pat[], char txt[], int q)  
{  
    int M = strlen(pat);  
    int N = strlen(txt);  
    int i, j;  
    int p = 0; // hash value for pattern  
    int t = 0; // hash value for txt  
    int h = 1;  
  
    // The value of h would be "pow(d, M-1)%q"  
    for (i = 0; i < M - 1; i++)  
        h = (h * d) % q;  
  
    // Calculate the hash value of pattern and first  
    // window of text  
    for (i = 0; i < M; i++)  
    {  
        p = (d * p + pat[i]) % q;  
        t = (d * t + txt[i]) % q;  
    }  
  
    // Slide the pattern over text one by one  
    for (i = 0; i <= N - M; i++)  
    {  
  
        // Check the hash values of current window of text  
        // and pattern. If the hash values match then only  
        // check for characters on by one  
        if ( p == t )  
        {  
            /* Check for characters one by one */
            for (j = 0; j < M; j++)  
            {  
                if (txt[i+j] != pat[j])  
                    break;  
            }  
  
            // if p == t and pat[0...M-1] = txt[i, i+1, ...i+M-1]  
            if (j == M)  
                cout<<"Pattern found at index "<< i<<endl;  
        }  
  
        // Calculate hash value for next window of text: Remove  
        // leading digit, add trailing digit  
        if ( i < N-M )  
        {  
            t = (d*(t - txt[i]*h) + txt[i+M])%q;  
  
            // We might get negative value of t, converting it  
            // to positive  
            if (t < 0)  
            t = (t + q);  
        }  
    }  
}  
  
/* Driver code */
int main()  
{  
    char txt[] = "GEEKS FOR GEEKS";  
    char pat[] = "GEEK"; 
        
      // A prime number  
    int q = 101;  
      
      // Function Call 
      search(pat, txt, q);  
    return 0;  
}  

我不理解的是这段代码：

            // We might get negative value of t, converting it  
            // to positive  
            if (t < 0)  
            t = (t + q);  

t怎么可能是负的呢？我们从t中减去的东西总是小于t，然后我们再加上一些东西，那么t为负的可能性从何而来？

我在没有这个if语句的情况下测试了代码，它不能正常工作。exepected输出为：

Pattern found at index 0
Pattern found at index 10

但是我得到了：

Pattern found at index 0

Answer 1

Aki Suihkonen有它；对于正的模数，结果要么是零，要么与被除数具有相同的符号，而Rabin-Karp假设结果总是非负的。

例如，如果我们这样做

t = 3
t = (t + 5) % 7
t = (t - 5) % 7

则这些值为

(3 + 5) % 7 == 1
(1 - 5) % 7 == -4

如果我们加7，就会得到3。