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


int main(void)

{
    double x, average;
    double y =  1.0f;
    double judge = 0.00001f;


    printf("Enter a positive number: ");
    scanf("%lf", &x);


    do {
        average = (y + (x / y))/2;
        y = average;
    } while (fabs(y - average) > judge * y);


    printf("Square_root: %lf", y);

    return 0;
}

我觉得应该是循环能力算法出了问题，但是不知道该怎么改？求问各路大神....

judge 别乘以y

啊？那不然该怎么做......原题中要求说y的新旧值之差的绝对值小于0.00001和y的乘积时程序终止啊.....

你代码不是抄的吧
double y =  1.0f;
double judge = 0.00001f;

你的算法写完了，要回头看一眼
do {
    average = (y + x/y)/2;
    y = average;
} while (fabs(y - average) > judge * y);

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

int main( void )
{
    double x;
    double y =  1.0;
    double judge = 0.00001;

    printf( "Enter a positive number: " );
    scanf( "%lf", &x );

    while( fabs(y-(y+x/y)/2) > judge*y )
        y = (y+x/y)/2;

    printf( "Square_root: %lf", y );

    return 0;
}

是我自己写的。当时循环那里一直不通，后来想明白了。下面是我改了之后的。


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

int main(void)

{
    double x, average, temp;
    double y =  1.0f;
;


    printf("Enter a positive number: ");
    scanf("%lf", &x);


    do {
        average = (y + (x / y))/2;
        temp = fabs(y - average);
        y = average;
    } while (temp > 0.00001 * y);


    printf("Square_root: %lf", y);

    return 0;
}