實現sqrt()函數

標籤：

 求一個正數N的開方, 並且可以指定精度, 要求不能用庫函數sqrt

方法一：如下所示，先求sqrt(N)的整數部分，再求小數點後1位，2位 ... ...

方法二：牛頓迭代法，根據公式 Ai+1 = (Ai+number/Ai)/2 ，其中Ai 的初始值，即A1任取，如1,2,3 ...

// 求一個正數N的開方, 並且可以指定精度, 要求不能用庫函數sqrt#include <stdio.h>#include <stdlib.h>double my_sqrt2(double number, int point){    double new_guess;    double last_guess;    if (number < 0) {        printf("Cannot compute the square root of a negative number!\n");        return -1;    }    printf("Method 2:\n");    new_guess = 1;    do {        last_guess = new_guess;        new_guess = (last_guess + number/last_guess) / 2;        printf("%.*lf\n", point, new_guess);    } while (new_guess != last_guess);    return new_guess;}double my_sqrt1(double n, int point){    if (n < 0) {        printf("Cannot compute the square root of a negative number!\n");        return -1;    }    int i,j;    for( i=1; i-n<0; i++ ) // 求得i的開方的整數部分        if( i*i-n > 0 )            break;    double answer = i-1;    double incr = 0.1;    for( j=1; j<=point; j++) // 第j次迴圈，求得i的開方的小數部分的第j位    {        for( i=1; i<10; i++ )        {            answer += incr;            if( answer*answer-n > 0 )                break;        }        answer -= incr;        incr /= 10;    }    incr *= 10;    printf("Method 1:\n");    printf("sqrt(%lf) is between %.*lf - %.*lf\n", n, point, answer, point, answer+incr);    return answer;}int main(void){    int point;    double n;    printf("請輸入正整數N: ");    scanf("%lf",&n);    printf("請輸入精確到小數點後的位元: ");    scanf("%d",&point);    my_sqrt1(n,point);    my_sqrt2(n,point);    return 0;}

 

實現sqrt()函數