c - 如何在空 vector 中传递值？

Question

如何使用 for 循环将值传递到空 vector (因子)？

在这段代码中，我试图找到一个数字的因数个数，然后将这些因数传递到一个空 vector 中，但它一直给我零

#include <stdio.h>
#include <math.h>

int main(void)
{

    int facVar = 57;
    int S;
    double numFact = 0;
    int k;
    int factors1[4];

    for (S = 1; S <= facVar; S = S + 1)
    {
        if (facVar % S == 0)
        {
            numFact = numFact + 1;

        }
    }
    int factNum = numFact;
    int factors[factNum];
    int indices[factNum];
    int i;
    printf("the number of factors is  %.lf \n", numFact);

    for (k = 0; k < factNum; k++)
    {

        factors[k] = S;
    }

    for (k = 0; k < factNum; k++)
    {
        printf("factors are %.lf", factors[k]);
    }

    return 0;
}

Answer 1

首先，您不需要调用:

#include <math.h>

之后，在第一个循环中，您检查数字的因子，但不将其存储在数组中:

for(S=1 ; S <= facVar ; S = S + 1){
if(facVar % S == 0){
    numFact = numFact + 1;

}
}

因此您需要将因子存储在数组因子中。该程序可帮助您获得所需的结果:

#include <stdio.h>
#include <stdlib.h>

int main(void){

int facVar = 57;
int S, k;
int numFact = 0;

for(S = 1, k = 0 ; S <= facVar ; S++){
    if(facVar % S == 0){
      numFact++;

    }
}
int factors[numFact];
for(S = 1, k = 0 ; S <= facVar ; S++){
    if(facVar % S == 0){
      factors[k] = S;
      k++;

    }
}
printf("the number of factors is:  %d \n" , numFact);

for(k = 0; k < numFact ; k++){
    printf("factors are: %d\n" , factors[k]);
}

return 0;

}