c - 得到结果多 1，除数计数 程序 - C

Question

Write a program which reads in an integer k, and prints out the number of positive integers between 1 and 100000 (inclusive) which have exactly k divisors.

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

    int main(void){
    int k=0, N=0, i=0, r=0, n_divisors=0, result=0;
    printf("Enter the number of divisors: ");

/*scanning in an integer value*/
    scanf(" %d",&k);
/*loop to count the no. of divisors in an integer from 1 to 100000*/ 
    for(N=1; N<=100000; N++){
            n_divisors=0;
/*loop for incrementing the no. divisors in an integer by 2 */
      for(i=1; i<sqrt(N); i++ ){
/*Testing if i is a divisor of the integer.*/
                r=(N%i);
                if( r==0)
                    n_divisors+=2;
      }
/* If the no. of divisors is equal to k, then increment result*/
                if(n_divisors==k)
                    result++;
/* else reset the value of no. of divisors to 0*/
    }
/* printing the value for no. of integers form 1 to 100000 with K no. of divisors*/
        printf("There are %d numbers between 1 and 100000 inclusive which have exactly  \n",result  );
        printf("%d divisors.",k);
    return 0;
    }

Answer 1

事情比你想象的更严重 - 你可能会得到奇怪的结果，不仅仅是增加 1。(参见:你通过添加 2 来计算除数的数量，因为所有数字都可以有只有偶数个除数。)

问题出在这个声明中:

        for(i=1; i<sqrt(N); i++ ){

特别是在表达式中

        i<sqrt(N)

其中排除 sqrt(N) 可能的除数。

但是对于整数的2次方的数字(如1、4、9、16, ...) sqrt(N) 是它们的除数。

所以在你的代码中插入一些东西

        if (i * i == N)
            n_divisors++;

代码中如此感动的部分将看起来像(我添加了正确的缩进)

// loop to count the no. of divisors in an integer from 1 to 100000
for(N=1; N<=100000; N++){
    n_divisors=0;
    // loop for incrementing the no. divisors in an integer by 2
    for(i=1; i<sqrt(N); i++ ){
        // Testing if i is a divisor of the integer.
        r=(N%i);
        if( r==0)
            n_divisors+=2;
    }
    if (i * i == N)
        n_divisors++;
    //  If the no. of divisors is equal to k, then increment result//
    if(n_divisors==k)
        result++;
    // else reset the value of no. of divisors to 0//
}