初一的多项式怎么才能学好，我一直学不好，求简单的方法

初一的多项式怎么才能学好，我一直学不好，求简单的方法

简单的方法就是你自己认为学不好，我是过来人，你静下心来，一个人钻一个下午，就可以完成了。我之前的经验是，多项式不好，是单项式不好，单项式就像你班上的同学，刚开始名字怎么记忆都会没有印象，你要去想这个人是谁，干嘛的，为什么是这样的，有谁和他很像，当你心态平和了你就发现当时怎么那么傻不就是他吗?~是不是晕了，总结一下，单项式要认清，OK。单项式组合起来就是多项式了。
跟你交个底，心态，很重要，从错误中成长，无论怎样，找方法说服自己把一道多项式题目弄懂，不要求多，发现之前的想法有了冲突了，再改。过程，熟悉就在过程中。

//一元多项式计算器 。非原创，我测试了下，还行，你再试试吧 #include #include #define ok 1; #define read(i) while (((i)=getchar())=='\n'); //读入合法字符 typedef int status; typedef struct polynode { float coef; int expn; struct polynode *next; }poly,*p_poly; status creatpoly(p_poly*,int); //创建多项式链表 void outp_poly(p_poly); //输出多项式 void addpoly(p_poly*,p_poly,p_poly); //求和 void decpoly(p_poly*,p_poly,p_poly); //求差 void delpoly(p_poly*); //删除并释放多项式 status main() { p_poly polyhead[26]={0}; char menu,name; for (;;) { printf ("1.建立多项式\n"); printf ("2.输出多项式\n"); printf ("3.求两个多项式之和(必须先创建),建立和式并输出\n"); printf ("4.求两个多项式之差(必须先创建),建立差式并输出\n"); printf ("5.删除多项式\n"); printf ("6.退出\n"); printf ("请选择指令:"); read(menu) switch (menu) { case '1': { int n; printf ("请输入要创建的多项式的名称（单个小写字母表示）:"); read(name) if (null!=polyhead[name-'a']) { printf ("此多项式已经存在!\n"); break; } printf ("请输入项数:"); scanf ("%d",&n); if(creatpoly(&polyhead[name-'a'],n)) printf ("多项式%c创建完毕!\n",name); break; } case '2': printf ("请输入要输出的多项式的名称:"); read(name) if (null==polyhead[name-'a']) { printf ("此多项式不存在!\n"); break; } outp_poly(polyhead[name-'a']); break; case '3': { char name1,name2; printf ("请输入加式一的名称:"); read(name1) if (null==polyhead[name1-'a']) { printf ("此多项式不存在!请先创建\n"); break; } printf ("请输入加式二的名称:"); read(name2) if (null==polyhead[name2-'a']) { printf ("此多项式不存在!请先创建\n"); break; } printf ("请输入要创建的和式的名称:"); read(name) if (null!=polyhead[name-'a']) { printf ("此多项式已经存在!\n"); break; } addpoly(&polyhead[name-'a'],polyhead[name1-'a'],polyhead[name2-'a']); outp_poly(polyhead[name-'a']); break; } case '4': { char name1,name2; printf ("请输入被减式的名称:"); read(name1) if (null==polyhead[name1-'a']) { printf ("此多项式不存在!请先创建\n"); break; } printf ("请输入减式的名称:"); read(name2) if (null==polyhead[name2-'a']) { printf ("此多项式不存在!请先创建\n"); break; } printf ("请输入要创建的差式的名称:"); read(name) if (null!=polyhead[name-'a']) { printf ("此多项式已经存在!\n"); break; } decpoly(&polyhead[name-'a'],polyhead[name1-'a'],polyhead[name2-'a']); outp_poly(polyhead[name-'a']); break; } case '5': printf ("请输入要删除的多项式名称:"); read(name) if (null==polyhead[name-'a']) { printf ("此多项式不存在!\n"); break; } delpoly(&polyhead[name-'a']); printf ("多项式%c已删除!\n",name); break; case '6': return ok; } } } status creatpoly(p_poly* p_polyhead,int n) { //为确保其他子函数正常运行，入口参数要求不含同类项、指数递减 int i=0; p_poly p,tail; *p_polyhead=(poly*)malloc(sizeof(poly)); (*p_polyhead)->expn=n; tail=*p_polyhead; printf ("请按照指数递减序输入最简形式的多项式\n"); for (;icoef),&(p->expn)); tail->next=p; tail=p; } tail->next=null; return ok; } void outp_poly(p_poly polyhead) { int i=0; p_poly p; printf ("%d ",polyhead->expn); p=polyhead->next; for (;iexpn;i++) { printf ("%f,%d ",p->coef,p->expn); p=p->next; } printf ("\n"); } void addpoly(p_poly* p_polyhead,p_poly polyhead1,p_poly polyhead2) { int i=0; p_poly p1,p2,p,tail; p1=polyhead1->next; p2=polyhead2->next; *p_polyhead=(p_poly)malloc(sizeof(poly)); p=tail=*p_polyhead; while(p1&&p2) { int e1,e2; e1=p1->expn; e2=p2->expn; if (e1>e2) { p=(p_poly)malloc(sizeof(poly)); tail->next=p; tail=p; *p=*p1; p1=p1->next; i++; } else if(e1==e2) { float sum_c=p1->coef+p2->coef; if (sum_c!=0.0) { p=(p_poly)malloc(sizeof(poly)); tail->next=p; tail=p; p->coef=sum_c; p->expn=e1; i++; } p1=p1->next; p2=p2->next; } else if (e1next=p; tail=p; *p=*p2; p2=p2->next; i++; } } while (p1) { p=(poly*)malloc(sizeof(poly)); *p=*p1; tail->next=p; tail=p; p1=p1->next; i++; } while (p2) { p=(poly*)malloc(sizeof(poly)); *p=*p2; tail->next=p; tail=p; p2=p2->next; i++; } (*p_polyhead)->expn=i; tail->next=null; } void decpoly(p_poly* p_polyhead,p_poly polyhead1,p_poly polyhead2) { int i=0; p_poly p1,p2,p,tail; p1=polyhead1->next; p2=polyhead2->next; *p_polyhead=(p_poly)malloc(sizeof(poly)); p=tail=*p_polyhead; while(p1&&p2) { int e1,e2; e1=p1->expn; e2=p2->expn; if (e1>e2) { p=(p_poly)malloc(sizeof(poly)); tail->next=p; tail=p; *p=*p1; p1=p1->next; i++; } else if(e1==e2) { float dif_c=p1->coef-p2->coef; if (dif_c!=0.0) { p=(p_poly)malloc(sizeof(poly)); tail->next=p; tail=p; p->coef=dif_c; p->expn=e1; i++; } p1=p1->next; p2=p2->next; } else if (e1next=p; tail=p; p->coef=-p2->coef; p->expn=p2->expn; p2=p2->next; i++; } } while (p1) { p=(poly*)malloc(sizeof(poly)); *p=*p1; tail->next=p; tail=p; p1=p1->next; i++; } while (p2) { p=(poly*)malloc(sizeof(poly)); *p=*p2; tail->next=p; tail=p; p2=p2->next; i++; } (*p_polyhead)->expn=i; tail->next=null; } void delpoly(p_poly* p_polyhead) { p_poly p,prior; p=prior=*p_polyhead; while (p->next) { p=p->next; free(prior); prior=p; } *p_polyhead=null; }