C语言表编程：用二分法求一元三次方程的根 要求：又主函数调用求根子函数

C语言表编程：用二分法求一元三次方程的根 要求：又主函数调用求根子函数

二分法的基本思路是：任意两个点x1和x2，判断区间（x1,x2)内有无一个实根，如果f(x1)与f(x2)符号相反，则说明有一实根。接着取（x1,x2)的中点x，检查f(x)和f(x2)是否同号，如果不同号，说明实根在（x,x2)之间，如果同号，在比较（x1,x),这样就将范围缩小一半，然后按上述方法不断的递归调用，直到区间相当小（找出根为止）！

比如用二分法求f(x)=x^3-6x-1=0的实根。

代码如下（已调试）：

#include "math.h"
main()
{
float x,x1,x2;
float F(float x,float x1,float x2);
printf("请输入区间[x1,x2]\n");
scanf("%f%f",&x1,&x2);
printf("x=%f\n",F(x,x1,x2));

}
float F(float x,float x1,float x2)
{
float f,f1,f2;
do

{
f1=pow(x1,3)-6*x1-1.0;
f2=pow(x2,3)-6*x2-1.0;
}while(f1*f2>0); //确保输入的x1,x2使得f1,f2符号相反
do
{
x=(x1+x2)/2; //求x1,x2的中点
f=pow(x,3)-6*x-1.0;
if(f1*f>0) //当f与f1符号相同时
{x1=x;f1=f;}
else if(f2*f>0) //当f与f2符号相同时
{x2=x;f2=f;}
}while(fabs(f)>1e-6); //判断条件fabs(f)>1e-6的意思是f的值非常0
return x;
}

输入：1 5
则输出：x=2.528918
输入：-10 10
则输出：x=2.528918

代码如下，很完整 #include #include void main() { double x0,x1,xm,f0,f1,fm,x2,x3;//x2,x3是驻点，x0,x1,xm,f0,x1是二分法求根的工具。 double a[3],r[3]; int i,j=0; printf("input 3 coefficients:\n"); for(i=0;i<3;i++) { printf("a[%d]=",i); scanf("%lf",a+i); } printf("the function is:\n"); printf("x*x*x+%5.2f*x*x+%5.2f*x+%5.2f=0\n",a[0],a[1],a[2]); //y=3*x*x+2*a[0]*x+a[1] if(4*a[0]*a[0]-12*a[1]<0)//方程单调递增，与横轴只有一个交点。 { printf("input 2 numbers as your will:\n"); printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; if(f0*f1==0) { if(f0==0) { xm=x0; printf("the only root of the function is %.2f\n",xm); } else { xm=x1; printf("the only root of the function is %.2f\n",xm); } if(fabs(xm*xm*xm+a[0]*xm*xm+a[1]*xm+a[2])<1e-6) printf("comgratulations! the answer is right!\n"); else printf("sorry, you should try again.\n"); } else { while(f0*f1>=0) { printf("input 2 numbers again:\n"); printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; } do { xm=(x0+x1)/2; fm=xm*xm*xm+a[0]*xm*xm+a[1]*xm+a[2]; if(f0*fm>0) x0=xm; else x1=xm; }while(fabs(x0-x1)>1e-6); xm=(x0+x1)/2; printf("the only root of the function is %.2f\n",xm); if(fabs(xm*xm*xm+a[0]*xm*xm+a[1]*xm+a[2])<1e-6) printf("congratulations! the answer is right!\n"); else printf("sorry, you should try again.\n"); } } else//方程有增有减，但与横轴的交点不确定。 { x2=(-2*a[0]-sqrt(4*a[0]*a[0]-12*a[1]))/6; x3=(-2*a[0]+sqrt(4*a[0]*a[0]-12*a[1]))/6; printf("the stagnation of the function are:\n"); printf("x2=%.2f\nx3=%.2f\n",x2,x3); if((x2*x2*x2+a[0]*x2*x2+a[1]*x2+a[2])>0&&(x3*x3*x3+a[0]*x3*x3+a[1]*x3+a[2])>0) //方程左半单调递增支和横轴有交点。 { printf("input 2 numbers to start the procedure, and one of the number should be x2, and the other should be smaller than x2. you will get just one root.\n"); printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; while(f0*f1>=0) { printf("input 2 numbers again:\n"); printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; } do { xm=(x0+x1)/2; fm=xm*xm*xm+a[0]*xm*xm+a[1]*xm+a[2]; if(f0*fm>0) x0=xm; else x1=xm; }while(fabs(x0-x1)>1e-6); xm=(x0+x1)/2; printf("the only root of the function is %.2f\n",xm); if(fabs(xm*xm*xm+a[0]*xm*xm+a[1]*xm+a[2])<1e-6) printf("congratulations! the answer is right!\n"); else printf("sorry, you should try again.\n"); } else if((x2*x2*x2+a[0]*x2*x2+a[1]*x2+a[2])<0&&(x3*x3*x3+a[0]*x3*x3+a[1]*x3+a[2])<0) //方程右半单调递增支和横轴有交点 { printf("input 2 numbers to start the procedure, and one of the number should be x3, and the other should be bigger than x3.you will get just one root.\n"); printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; while(f0*f1>=0) { printf("input 2 numbers again:\n"); printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; } do { xm=(x0+x1)/2; fm=xm*xm*xm+a[0]*xm*xm+a[1]*xm+a[2]; if(f0*fm>0) x0=xm; else x1=xm; }while(fabs(x0-x1)>1e-6); xm=(x0+x1)/2; printf("the only root of the function is %.2f\n",xm); if(fabs(xm*xm*xm+a[0]*xm*xm+a[1]*xm+a[2])<1e-6) printf("congratulations! the answer is right!\n"); else printf("sorry, you should try again.\n"); } else if((x2*x2*x2+a[0]*x2*x2+a[1]*x2+a[2])*(x3*x3*x3+a[0]*x3*x3+a[1]*x3+a[2])<0)//一般方程有三个交点，分别位于增、减、增区间。 { printf("you will get 3 roots, type in 2 numbers 3 times, and in the first case, the bigger number you type in should be x2; and in the second case, the 2 numbers you type in should be x2 and x3, and in the third case, the smaller one should be x3.\n"); for(i=0;i<3;i++) { printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; while(f0*f1>=0) { printf("input 2 numbers again:\n"); printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; } do { xm=(x0+x1)/2; fm=xm*xm*xm+a[0]*xm*xm+a[1]*xm+a[2]; if(f0*fm>0) x0=xm; else x1=xm; }while(fabs(x0-x1)>1e-6); r[i]=(x0+x1)/2; printf("ok, next!\n"); }//三次循环找三个根。 printf("3 roots of the function are:\n"); for(i=0;i<3;i++) printf("r[%d]=%.2f\n",i,r[i]); for(i=0;i<3;i++) if(fabs(r[i]*r[i]*r[i]+a[0]*r[i]*r[i]+a[1]*r[i]+a[2])<1e-6) j++; if(j==3) printf("congratulations, the answer are all right!\n"); else printf("sorry, you should try again.\n"); } else { if(x2*x2*x2+a[0]*x2*x2+a[1]*x2+a[2]==0&&x3*x3*x3+a[0]*x3*x3+a[1]*x3+a[2]!=0)//x2是一个二重根 { r[0]=r[1]=x2; //补充剩下的 printf("input 2 numbers to start the procedure, and one of the number should be x3, and the other should be bigger than x3.you will get just one root.\n"); printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; while(f0*f1>=0) { printf("input 2 numbers again:\n"); printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; } do { xm=(x0+x1)/2; fm=xm*xm*xm+a[0]*xm*xm+a[1]*xm+a[2]; if(f0*fm>0) x0=xm; else x1=xm; }while(fabs(x0-x1)>1e-6); r[2]=(x0+x1)/2; printf("the 2 roots of the function are:\n"); for(i=0;i<3;i++) printf("r[%d]=%.2f\n",i,r[i]); for(i=0;i<3;i++) if(fabs(r[i]*r[i]*r[i]+a[0]*r[i]*r[i]+a[1]*r[i]+a[2])<1e-6) j++; if(j==3) printf("congratulations, the answer are all right!\n"); else printf("sorry, you should try again.\n"); } else if(x2*x2*x2+a[0]*x2*x2+a[1]*x2+a[2]!=0&&x3*x3*x3+a[0]*x3*x3+a[1]*x3+a[2]==0) { r[0]=r[1]=x3; //补充剩下的 printf("input 2 numbers to start the procedure, and one of the number should be x2, and the other should be smaller than x2.you will get just one root.\n"); printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; while(f0*f1>=0) { printf("input 2 numbers again:\n"); printf("x0="); scanf("%lf",&x0); printf("x1="); scanf("%lf",&x1); f0=x0*x0*x0+a[0]*x0*x0+a[1]*x0+a[2]; f1=x1*x1*x1+a[0]*x1*x1+a[1]*x1+a[2]; } do { xm=(x0+x1)/2; fm=xm*xm*xm+a[0]*xm*xm+a[1]*xm+a[2]; if(f0*fm>0) x0=xm; else x1=xm; }while(fabs(x0-x1)>1e-6); r[2]=(x0+x1)/2; printf("the 2 roots of the function are:\n"); for(i=0;i<3;i++) printf("r[%d]=%.2f\n",i,r[i]); for(i=0;i<3;i++) if(fabs(r[i]*r[i]*r[i]+a[0]*r[i]*r[i]+a[1]*r[i]+a[2])<1e-6) j++; if(j==3) printf("congratulations, the answer are all right!\n"); else printf("sorry, you should try again.\n"); } else { r[0]=r[1]=r[2]=x2; printf("3 roots are equal!\n"); printf("the 3 roots are:\n"); for(i=0;i<3;i++) printf("r[%d]=%.2f\n",i,r[i]); for(i=0;i<3;i++) if(fabs(r[i]*r[i]*r[i]+a[0]*r[i]*r[i]+a[1]*r[i]+a[2])<1e-6) j++; if(j==3) printf("congratulations, the answer are all right!\n"); else printf("sorry, you should try again.\n"); } } } }

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