求解方程-源程序.doc

#include iostream.h #include stdio.h #include conio.h /基本输入输出文件#include string.h #include stdlib.h /定义杂项函数及内存分配函数#include ctype.h /字符处理#include math.h char terms9920;class Equation /定义类Equation public:Equation() void jecfc(int,int,int); /解二次方程void getterm(char *,char *,int &); /存放方程各项void getyz(int *,int &,int); /得到因子void jc(char *,char *,float); /降次void gettermf(char *,char *,int); /得到指定幂的项int getxs(char *); /得到系数（整数）int getqium(char *); /求幂int yang(char *,float); /检验根是否正确int exp(int,int); /求某数的次方（整数）float knyz(int *,int *,int,int,char *); /求所有可能的根float exp(float,int); /求某数的次方（实数）/返回equ的系数乘以val的equ的次数的次方的值float calcequval(char *,float); ;void main()cout使用说明（允许 & 不允许）：endl; coutendl; cout1. 在输入的方程后面不能输入 = ；endl; cout2. 所输入的方程必须是从高次到低次依次输入； endl; cout3. 若某项系数为0,可以不输入或输入0作为系数；endl; cout4. 不要用分数作为系数且第一项系数绝对值不小于一；endl; cout5. 要用xn表示指数,x n一可以,但不能用x n表示。endl; coutendl;Equation E;char *equ,*equ2;int loc=0,a=0,b=0,c=0,numterms;int deg,f_a1000,f_k1000,num_a,num_k;int i;float result;equ=new char5000; equ2=new char5000; cout请输入方程: endl;gets(equ);if(strlen(equ)=0)cout没有输入方程!endl;for(i=0;i99&equloc!=&equloc!=0&equloc!=n;i+)E.getterm(equ,termsi,loc); numterms=i;if(deg=E.getqium(terms0)2)cout无效输入,输入的方程最高次应大于等于2次endl;return;cout2)E.getyz(f_a,num_a,E.getqium(terms0);for(i=1;inumterms;i+)if( E.getqium(termsi)=0)c= E.getxs(termsi);E.getyz(f_k,num_k,c);if(result= E.knyz(f_a,f_k,num_a,num_k,equ)=0)if(result-0.05f)cout错误！不能解得根。endl;return;E.jc(equ,equ2,result);strcpy(equ,equ2);loc=0;for(i=0;i99&equloc!=&equloc!=0&equloc!=n;i+)E.getterm(equ,termsi,loc);numterms=i;deg-;for(i=1;inumterms;i+)if( E.getqium(termsi)=1)b= E.getxs(termsi);for(i=1;inumterms;i+)if( E.getqium(termsi)=0)c= E.getxs(termsi);E.jecfc( E.getxs(terms0),b,c);void Equation:jecfc(int a,int b,int c) /解二次方程int r1,r2,r3;float r11,r12;if(b*b-4*a*c0)r1=b*-1;r2=(b*b-4*a*c)*-1;r3=2*a;cout(r1+i*sqrt(r2)/r3;cout , ;cout(r1-i*sqrt(r2)/r3;elseif(sqrt(b*b-4*a*c)-(int)sqrt(b*b-4*a*c)=0)r11=(float)(-b+sqrt(b*b-4*a*c)/(2*a);r12=(float)(-b-sqrt(b*b-4*a*c)/(2*a);coutr11 , r12;elser1=b*-1;r2=b*b-4*a*c;r3=2*a;cout(r1+sqrt(r2)/r3;cout , ;cout(r1-sqrt(r2)/r3;coutendl;void Equation:getterm(char *src,char *dest,int &loc) /存放方程各项 int loc2=0;if(srcloc=-)destloc2=-;loc2+;for(;srcloc+loc2!=+&srcloc+loc2!=-&srcloc+loc2!=n&srcloc+loc2!=0&srcloc+loc2!=;loc2+)destloc2=srcloc+loc2; destloc2=0;loc+=loc2;if(srcloc!=-&srcloc!=0)loc+; void Equation:getyz(int *yz_list,int &num,int val) /得到因子int i,fcount=0;num=0;val=abs(val);for(i=1;i=val;i+)if(float)(float)(float)val/(float)i)-(int)(float)val/(float)i)=0.0f)num+;yz_listfcount=i;fcount+;int Equation:getxs(char *term) /得到系数char *temp;int i=0,mult=1;temp=new charstrlen(term);if(term0=-)mult=-1;i+;for(;!isdigit(termi)&!isalpha(termi);i+);if(termi=x)return mult;for(int j=0;isdigit(termi+j);j+)tempj=termi+j;return atoi(temp)*mult;int Equation:getqium(char *term) /求幂char temp10;if(strchr(term,x)!=NULL)if(strchr(term,)!=NULL)for(int i=0;termi!=;i+);i+;for(int j=0;isdigit(termi);i+,j+)tempj=termi;tempj=0;return atoi(temp);elsereturn 1;elsereturn 0;int Equation:yang(char *equ,float root) /检验根是否正确char terms9940;float val99,totalval=0;int loc=0;for(int i=0;equloc!=&equloc!=0&equloc!=n;i+)getterm(equ,termsi,loc);int numterms=i;for(i=0;i-0.002f&totalval*0.99f0.002f)return 0;elsereturn 1;int Equation: exp(int val,int e) /求某数的次方（整数）int v2=val;if(e=0) return 1;for(int i=1;ie;i+) v2*=val;return v2;float Equation:exp(float val,int e) /求某数的次方（实数）float v2=val;if(e=0) return 1.0f;for(int i=1;ie;i+) v2*=val; return v2;/返回equ的系数乘以val的equ的次数的次方的值float Equation:calcequval(char *equ,float val) int coeff= getxs(equ);int power= getqium(equ);return coeff* exp(val,power);void Equation:jc(char *base,char *newequ,float yz) /降次int synthterms99;int newterms99;int j=0,olddegree;char temp50=,exp10=,exp110=;char newequ120,newequ220; if( yang(base,yz)return;olddegree=getqium(terms0);for(int i=0;i=1;i-) newtermsi-1=(int)(float)(newtermsi*yz+synthtermsi);for(i=olddegree-1;newtermsi=0;i-);strcpy(newequ,);for(;i=0;i-)if(newtermsi!=0)switch(i)case 0:strcpy(exp,);break;case 1:strcpy(exp,x);break;default:itoa(i,exp1,10);strcpy(exp,x);strcat(exp,exp1);break;if(newtermsi!=1)if(newtermsi0)itoa(newtermsi,newequ1,10);strcat(newequ1,exp);elseif(iolddegree-1)strcpy(newequ1,+);itoa(newtermsi,newequ2,10);strcat(newequ1,newequ2);strcat(newequ1,exp);elseitoa(newtermsi,newequ1,10);strcat(newequ1,exp);elseif(newtermsi0)strcpy(newequ1,exp);elseif(i0)strcpy(newequ1,+);itoa(newtermsi,newequ2,10);strcat(newequ1,newequ2);elseitoa(newtermsi,newequ1,10);strcat(newequ,newequ1);/得到指定幂的项void Equation:gettermf(char *equ,char *dest,int power) int t=getqium(terms0);for(int i=0;it+1;i+)if( getqium(termsi)=power)strcpy(dest,termsi);return;strcpy(dest,0);/求所有可能的根float Equation:knyz(int *a,int *k,int n_a,int n_k,char *equ) for(