Question

我试图利用地板功能计算一下在脱氧核糖核酸序列中阵列的机体数量,而没有计算任何三倍。这应该有5起关贸总协定,但我的职责是给我9起,如果你重叠关贸总协定,那是你会得到的。我如何确定这一点,以便只计算每个单项合同,不计算两倍T? 我的方案如下:

/***************************************************************************************/
#include <iostream>    
#include <cstring>      
#include <iomanip>

using namespace std;

    //FUNCTION PROTOTYPES
     int overlap(char *ptr1, char *ptr2);

int main()
{

    //Declare and initialize objects
   int count(0); // For DNA sequence

        //DNA SEQUENCE
    char DNA_sequence[] = "ATGCTAGTATTTGGATAGATAGATAGATAGATAGATAGATAAAAAAATTTTTTTT";
    char thymine_group[] = "TT";
    char *ptr1(DNA_sequence), *ptr2(thymine_group);

//Send QUOTE to function
count = overlap(ptr1, ptr2);

   //Print number of occurences.
    cout << " TT  appears in DNA sequence " << count << " times" << endl;
    return 0;
}

//FUNCTION 1 USING CHAR ARRAYS AND POINTERS

int overlap(char *ptr1, char *ptr2)
{
    int count(0);
    //Count number of occurences of strg2 in strg1.
    //While function strstr does not return NULL
    //increment count and move ptr1 to next section
    //of strg1.
    while ((ptr1=strstr(ptr1,ptr2)) != NULL)
    {
        count++;
        ptr1++;
    }
    return count;
}

/**************************************************************************************************/

Answer 1

Just change ptr1++; in your loop to ptr1 += strlen(ptr2);

Answer 2

try

int overlap(char *ptr1, char *ptr2)
{
    int count(0);
    //Count number of occurences of strg2 in strg1.
    //While function strstr does not return NULL
    //increment count and move ptr1 to next section
    //of strg1.
    while ((ptr1=strstr(ptr1,ptr2)) != NULL)
    {
        count++;
        ptr1 += strlen (ptr2);
    }
    return count;
}

Answer 3

int count(char *haystack, char* needle)
{       int c = 0;
        for(;*haystack;haystack++){
            if(strcmp(haystack, needle)==0){
                 c++;
                 haystack+=strlen(needle)-1;
            }
        }
        return c;
}

友情链接