Question

用C语言编程二分法求解方程组

求解的方程组为Ax^m+Bx^n=1，其中A,B为自己在程序中输入的常数（整数或者小数、分数）。x的幂m,n为自己在程序中输入的常数（整数或者小数、分数）且m,n均小于2，求解x的值。用二分法编程，多谢高手帮忙啦！

Answer 1

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

#define EQUATION(A, B, m, n, x) (A*pow(x, m) + B*pow(x, n) - 1)/* 定义方程*/

double A, B, m, n;/*方程参数*/

void find_extent(double *x1, double *x2)/*找出根所在的区间*/
{
double i = 0, j = 0;
*x1 = 0;
if(EQUATION(A, B, m, n, 0) > 0)
{
while(EQUATION(A, B, m, n, ++i) > 0 && EQUATION(A, B, m, n, --j) > 0);
*x2 = EQUATION(A, B, m, n, i) <= 0 ? i : j;
}
else
{
while(EQUATION(A, B, m, n, ++i) < 0 && EQUATION(A, B, m, n, --j) < 0);
*x2 = EQUATION(A, B, m, n, i) >= 0 ? i : j;
}
}

/*二分法解方程*/
/*输入参数：根所在区间[x1, x2]， 和精度*/
double get_result(double x1, double x2, double accuracy)
{
double x = (x1 + x2) / 2;
while(fabs(EQUATION(A, B, m, n, x)) > accuracy)
{
if(EQUATION(A, B, m, n, x) * EQUATION(A, B, m, n, x1) > 0)
x1 = x;
else
x2= x;
x = (x1 + x2)/2;
}
return x;
}

int main()
{
double x;
double x1,x2;
scanf("%lf%lf%lf%lf", &A, &B, &m, &n);
find_extent(&x1, &x2);
x = get_result(x1, x2, 0.00001);
printf("%lf", x);
return 0;
}

Answer 2

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

#define FUNCTION(x, A, B, m, n) (A*pow(x, m) + B*pow(x, n) - 1)
#define MAXSIZE 100

double A[MAXSIZE], B[MAXSIZE];
double m[MAXSIZE], n[MAXSIZE];

double esp = 1e-4;//误差控制

void process(double x1, double x2, double A, double B, double m, double n)
{
int flag = 1;
double x0;

do
{
x0 = (x1 + x2)/2;

if ((0 < FUNCTION(x1, A, B, m, n)) && (0 < FUNCTION(x2, A, B, m, n)) && (0 < FUNCTION(x0, A, B, m, n)))
{
flag = 0;
break;
}
if((0 > FUNCTION(x1, A, B, m, n)) && (0 > FUNCTION(x2, A, B, m, n)) && (0 > FUNCTION(x0, A, B, m, n)))
{
flag = 0;
break;
}

if (0 > FUNCTION(x0, A, B, m, n)*FUNCTION(x1, A, B, m, n))
{
x2 = x0;
}
else
{
x1 = x0;
}

} while (esp <= fabs(FUNCTION(x0, A, B, m, n)));

if (1 == flag)
{
printf("The root is %.4lf\n",x0);
}
}

int main(void)
{
int i;
double j;

for (i=0; i<MAXSIZE; i++)
{
printf("input A and B:\n");
scanf("%lf%lf", &A[i], &B[i]);

printf("input m and n:\n");
scanf("%lf%lf", &m[i], &n[i]);

for (j=-10; j<10; j+=esp)
{
process(j, j+esp, A[i], B[i], m[i], n[i]);
}
printf("\n\n");
}

return 0;
}

Answer 3

#include

Answer 4

什么样的方程组啊。

追问

A,Bm,n都是需要在程序中输入的。我知道这个题目很难。。。多谢高手了！

追答

今天要下班了，明天来帮你解吧。。。最好举个方程的例子，我看看

追问

好的，非常感谢您！比如我举个例子，用C语言编程求解1.1x^1.5+1.4x^2=1，这个x如何求？因为我A，B有很多数据组合，最好A，B是能一次大量输入的。多谢多谢！