C语言解线性方程的四种方法

1、c语言解线性方程的四种方法发了好几天编了个解线性方程组的小程序，可第一次实战就大败而归。经过半天的调试，仍 找不出纠正的方法。因为并不是算法的问题，而是因为自己对编译器处理浮点函数的方法 不是很理解。明明d=0的方阵解出来不等于0 了，跟踪调试发现，计算过程程序对数据进 行了舍去处理，导致最终结果不对。不过如果没有浮点型的话，这个程序应该算不错了。 复制代码代码如下：# include<stdio.h>#in clude<math.h>#include<mem.h>#define num 100void print(void) /* 使用说明 */clrsc

2、rf);printf("nnnnntttt introduction n"); printf("t*n");printf("t*this program was design for compute linear equations.*n”)；printf("t*the way of use it is very simple.*n“);printf("t*first: input the number of the equation;(lnput 0 to exit) *n”)；prin tf("t*sec on

3、d: in put the coefficie nt of every equti on;*n“)；printf("t*third : input the constant of every equ廿on;*n");printf("t* last: chose the way you want use to solve the equtions; *n”)； printf("t* that's all, input any key to run it.*n");printf("t*by_tjx*nu);getchf);void

4、 chose(void)/*选择计算方法*/clrscrf); fflush(stdin);printf("nnnnntt*lntroduction* n");printf("tt* chose the way,please.* n")；printf("tt* a : gauss eliminant.* n");printf("tt* b : gauss_yd eliminant.* n");printf("tt* c : iterative way.* n");printf("tt*

5、 d : cramer way.* n");printf("tt* e : exit.* n");prin tf(mtt * * * by -pj* *、门11”printf(uttplease choose number:n");void inputfdouble *al,double bl,int num)/*数据输入*/int i jt;double *p;char del,de2;do printf("please input array a%d%d: n",num,num);prin tf("warn: the f

6、irst nu mber of the array must n't con tain zero! n”)； for(i=l;i< 二 nu m;i+)printf("please input array a%d: n"j);for(j=l;j<=nu m;j+)t=o；if(i=l&&j=l)doif(t=o) scanf(”lf”,&alij); t+;else printf("the input is invalid,please input again:n"); scanf(”f",&

7、alij); while(alij=o);else scanf(,%lf,/&alij);prin tf(" n please check the value of array a%d%d,press y to in put agai n.n”,num, num); dodel=getch();while(del!='y'&&del!='y'&&del!='n'&&del!='n");while(del='y,| |del='y');dop

8、rin tf("please in put array b%d: n",num);p=bl+l;for(i=l;i<=nu m;i+)scanf(”lf”,p+);prin tf("n pleasecheck the value of array b%d,press y to inputagaiwww.h un n”,n um);dode2=getch();while(de2!=y&&de2uy'&&de2un'&&de2!»n');while(de2='y'|

9、 |de2='y');int max(double *tl, double xl,int n) /*迭代子甫数*/int i,temp=0;for(i=l;i< 二 n ;i+)if(fabs(xli-tli)>le-2) temp=l;break;/* printf(" %d temp); */return temp;int ddcompute(double *al,double bl,double xl,int n) /*迭代法计算*/double *t;int i,j,k=o;double suml=0.0,sum2=0.0;t=(double*)m

10、alloc(n*sizeof(double);printf("nplease input the initial value of x:n");for(i=l;i<=n;i+)scanf("%lf"z&xli);do k+;for(i=l;i<=n;j+)ti=xli;for(i=l;i< 二 n ;i+)suml=0.0;sum2=0.0;for(j=l;j<=i-l;j+) suml=suml+alij*xlj; /*printf(" suml= %0.4f "suml);*/for(j=i+l;j

11、<=n;j 卄)sum2=sum2+alij*tj; /* printf(" sum2= %0.4fn,sum2);*/if(alii=o| |fabs(suml)>le+12| |fabs(sum2)>le+12)printf(" nwarning: these equtions can't be solve by this way!n press any key tocontinu e.")；getch();free(t);return 0;xli=(bli-suml-sum2)/alii;while(max(tzxl,n);/* f

12、or(i=l;i<=n;i+)if(i%3=0) printf("nk)；printf(n %.4f>11);*/free(t);return 1;int gscompute(double *al,double bl,double xl,int n) /*高斯消元法计算*/int ij,k;double m,sum;for(k=l;k<=n-l;k+)for(i=k+l;i<=n;i+) if(alkk=o) printf(” nthese equtions can't be solve is this n press any key to conti

13、nugetch();return 0;if(m=0-alik/alkk)=0) i+; continue;else for(j=k+l;j<=n;j+)alij=alij+alkj*m;bli=bli+blk*m;/* yi xia ji suan x zhi */xln=bln/alnn;for(i=n-l;i>=l;i-)sum=0.0;for(j=n;j>=i+l;j-)sum=sum+alij*xlj;xli=(bli-sum)/alii;return 1;int gs_ydcompute(double *al,double bl,double xl,int n)/*

14、高斯纟勺当法计算*/int ij,k；double m,sum;for(k=l;k<=n;k+)i=l；while(i< 二 n) if(alkk=o) printf nthese equtions can't be solve is this way.n press anykey to continue.n);getch();return 0;if(i!=k)if(m=0-alik/alkk)=0) i+; continue;else for(j=k+l;j<=n;j+)alij=alij+alkj*m;bli=bli+blk*m;i+；else i+;/* yi

15、xia ji suan x zhi */for(i=xli=bli/alii;return 1;double computed(double *a,int h,int i, int *cl,int n) /*计算系数行列式 d 值*/int i, j,p=l;double sum=0.0;讦(h二二n)sum=1.0;else i=+h;cll=0;for(j=l;j<=n;j+)if(clj)if(aij=o) p+;else sum=sum+aij*computed(aj,j,cl, n)*pow(l,l+p);p+;cll=l；return sum;void ncompute(do

16、uble *a,double b,double x,int n,int *c,double h)/*克莱姆法计算*/int ij;double tnum;for(j=l;j<=n;j+)for(i=l;i< 二 n ;i+)ti=aij;aij=bi;xj=computed(a,0,0,c, n)/h;for(i=l;i< 二 n ;i+)aij=ti;main()double xnum;double bnum;int i,j二2,n=0;int *c;double he;char m,decision;double *a;a=(double*)malloc(num*size

17、of(double*);for (i=0; i<num; i+)ai=(double*)malloc(num*sizeof(double);print();doclrscr();doif(n>=num) printfn is too large, please in put agai n:rt);elseprin tf(hplease in put the total nu mber of the equations n(n<n um): nh);scanf(" %d",&n);while(n>num);if(n=o) for(i=l; i&

18、lt;num; i+) free(ai);free(a); exit(l);input(a,b,n);c=(int *)malloc(n+l)*sizeof(int);memset(c,l,(n+l)*sizeof(int);he=computed(a,0,0,c, n);if(fabs(he)>le-4)other:chose();dom=getche();while(m!='a'&&m!二&&m!='a'&&m!='b'&&m!='c'&&m

19、!='c'&&m!='d'&&m!='d'&&m'e'&&m!二'e');switch(m)case 0:;case 'a1: j=gscompute(a,b,x,n); break;case 'b1:;case 'b': j=gs_ydcompute(a,b,x,n); break;case 'c1:;case c: j=ddcompute(a,b,x,n); break;case "d1:;case

20、 d: j=l; ncompute(abx,nche); break;case lel:;case e: j=2; break;default: j=2; break;if(j=d clrscr();printf(”nnnn“);printff' d=%.4f n",he);for(i=l;i<=n ;i+)if(i%5=0) printfcxn");printf(" %.4f“,xi);else if(j=o)printf(" nthese equtions can't be solve is this way.nplease chose the other way.")； goto other;el