字符串匹配的RabinKarp算法的c语言实现

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</pre><pre name="code" class="cpp">#include<string.h>int check( char *s1,char *s2,int n );int main(){char s1[10000],s2[1000000];int n,i,j,k;int T = 563;scanf( "%d",&n );for( i = 0;i < n;++i ){int key1 = 0,key2 = 0,temp = 1;  int len1,len2;int count = 0;scanf( "%s %s",s1,s2 );len1 = strlen(s1),len2 = strlen(s2);if( len1 > len2 ){printf( "0\n" );break;}for( j = 0;j < len1;++j ){key1 = ( key1 * 26 + s1[j] - 'A' ) % T;key2 = ( key2 * 26 + s2[j] - 'A' ) % T;}for( j = 0;j < len1 - 1;++j )temp = (temp % T * 26 % T) % T;   // printf( "%d\n",temp );// printf( "key1: %d  key2 : %d\n",key1,key2 );if( key1 == key2 && check(s1,s2,len1) )++count;for( j = 1;j <= len2 - len1;++j ){int t = key2 - (s2[j-1] - 'A') * temp;if( t < 0 )while( t < 0 )t += T;key2 = ( t * 26 + s2[j+len1-1] - 'A' ) % T;// printf( "ignore: %c,add %c ",s2[j-1],s2[j+len1-1] );// printf( "key2 : %d\n",key2 );if( key2 != key1 )continue;else {if( check( s1,s2 + j,len1 ) )++count;}}printf( "%d\n",count );}return 0;}int check( char *a,char *b,int n ){int i = 0;for( i = 0;i < n;++i )if( a[i] != b[i] )return 0;return 1;}

字符串匹配的RabinKarp算法就是将模式转化为数字形式，然后在母版上寻找其数值与模式数值相同的片段，如果模式长度较长，则选用取模后所得数相同的值在进行精确验证其是否相等。如果在文本搜索中能够匹配的次数很少，则其事件复杂度可以看做是O(n+m);下面程序是将字符集a-z看做是26位进制的数值，然后对其进行求模寻找相等值，如果有在使用check函数精确验证其正确性。

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