一个雅克比迭代算法解线性方程组示例
不要喷我,我只是新手,多多指导啦。。。
方程组:
10x1-x2+2*x3=-11
8x2-x3+3*x4=-11
2*x1-x2+10x3=6
- x1+3*x2-x3+11x4=-25
精度1e-6
代码:
#include<stdio.h>
#include<math.h>
int main(){
double x1=0,x2=0,x3=0,x4=0,x1t,x2t,x3t,x4t;
do{
x1t=x1;
x2t=x2;
x3t=x3;
x4t=x4;
x1=(x2-2*x3-11)/10;
x2=(x3-3*x4-11)/8;
x3=(-2*x1t+x2t+6)/10;
x4=(x1-3*x2t+x3t+25)/11;
}while((fabs(x1-x1t)>1e-6)&&(fabs(x2-x2t)>1e-6)&&(fabs(x3-x3t)>1e-6)&&(fabs(x4-x4t)>1e-6));
printf("x1=%f\n",x1);
printf("x2=%f\n",x2);
printf("x3=%f\n",x3);
printf("x4=%f\n",x4);
return 0;
}
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