// 求一个正数N的开方, 并且可以指定精度, 要求不能用库函数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的开方的整数部分        if( i*i-n > 0 )            break;    double answer = i-1;    double incr = 0.1;    for( j=1; j<=point; j++) // 第j次循环，求得i的开方的小数部分的第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("请输入正整数N: ");    scanf("%lf",&n);    printf("请输入精确到小数点后的位数: ");    scanf("%d",&point);    my_sqrt1(n,point);    my_sqrt2(n,point);    return 0;}