/*
*C語言實(shí)現(xiàn)Gauss-Seidel迭代法解方程組;
*實(shí)現(xiàn)特定方程組的求解并比較他們的收斂的快慢，E=10e-6;
*author:黃翔 date:2008-6-5
*/

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

#define eps 0.000001
#define true 1

int main()
{	
	double A[10][10]={{4,-1,0,0,0,0,0,0,0,0},  //系數(shù)矩陣
	                  {-1,4,-1,0,0,0,0,0,0,0},
					  {0,-1,4,-1,0,0,0,0,0,0},
                      {0,0,-1,4,-1,0,0,0,0,0},
					  {0,0,0,-1,4,-1,0,0,0,0},
					  {0,0,0,0,-1,4,-1,0,0,0},
					  {0,0,0,0,0,-1,4,-1,0,0},
					  {0,0,0,0,0,0,-1,4,-1,0},
					  {0,0,0,0,0,0,0,-1,4,-1},
					  {0,0,0,0,0,0,0,0,-1,4,}};
	double B[10]={4.0,11.0,15.0,18.0,21.0,24.0,27.0,30.0,37.0,45.5};//常數(shù)項(xiàng)
	
	double x[10]={0}; //初始解
	int i,j;          //用于循環(huán)條件數(shù)組下標(biāo)
	int count=0;      //迭代次數(shù)計(jì)數(shù)器
	double M,N,G;     //迭代系數(shù)
	double t;
	double E[10], e;  //范數(shù)數(shù)組,用作停機(jī)條件判斷;

//Gauss-Seidel迭代	

	while(true){
		for(i=0;i<10;i++){
			for(j=0,M=0;j<=i-1;j++){
				M-=A[i][j]*x[j];
			}
			for(j=i+1,N=0;j<10;j++){
				N-=A[i][j]*x[j];
			}
			G=1/A[i][i];
			t=x[i];
			x[i]=G*(M+N+B[i]);		
			E[i]=fabs(t-x[i]);
		}
	
		for(i=0,e=E[0];i<10;i++){//
			if(e<E[i]){
				e=E[i];
			}
		}
	    count++;
		if(e<eps) break;  //
	}

	printf("迭代次數(shù)\t%d\n",count);
	printf("求得方程組的解為:\n");
	for(i=0;i<10;i++)
		printf("%.10f\n",x[i]);	
}