1、【问题描述】 设计一个一元稀疏多项式简单计算器 【基本要求】 一元多项式简单计算器的基本功能是： 1，输入并建立多项式 ; 2，输出多项式，输出形式为整数序列:n,c1,e1,c2,c2,.,cn,en,其中 n 是多项式的项数，ci 和 ei 分别是第 i 项的系数和指数，序列按指数降序排列; 3,多项式 a 和 b 相加，建立多项式 a+b; 4,多项式 a 和 b 相减，建立多项式 a-b. 【测试数据】 1，(2x+5x8-3.1x11)+(7-5x8+11x9)=(-3.1x11+11x9+2x+7) 【实现提示】 用带表头结点的单链表存储多项式。 #include #include
2、 typedef struct nodefloat coef;int expn;struct node *next;Lnode, *polynmial;void create(polynmial /输入并建立多项式 Lvoid display(polynmial L); /显示，输出多项式 Lvoid sort(polynmial /多项式 L 按指数排序void reverse(polynmial /逆置void select(); /用户选择加减操作void add(polynmial La, polynmial Lb, polynmial /多项式 La,Lb 相加void subtra
3、ct(polynmial La, polynmial Lb, polynmial /多项式 La 减去Lb，结果给 Ldvoid create(polynmial static struct node *p;scanf(“%d“, L = (struct node *)malloc (sizeof(struct node);L-next = NULL;for(i = 0; i coef, p-next = L-next;L-next = p;void display(polynmial L)/显示，输出多项式 Lstruct node *p, *q;int flag = 0;int k = 0
4、;q = L-next;while(q)if(q-coef != 0)k+;q = q-next;printf(“%d, “, k);p = L-next;if(p-coef != 0)printf(“%.1f,%d, “, p-coef, p-expn);flag+;for(p = p-next; p; p = p-next)if(p-coef != 0)printf(“%.1f,%d, “, p-coef, p-expn);flag+;if(flag = 0)printf(“%dn“, flag);elseprintf(“n“);void sort(polynmial p = L-next
5、;L-next = NULL;while(p != NULL)r = L;q = L-next;while(q != NULL) q = q-next;u = p-next;r-next = p;p-next = q;p = u;void reverse(polynmial static struct node *p, *q, *s;H = (struct node*)malloc(sizeof(struct node);H-next = NULL;p = (struct node*)malloc(sizeof(struct node);s = L-next;p-coef = s-coef;p
6、-expn = s-expn;p-next = s-next;while(s)p-coef = s-coef;p-expn = s-expn;p-next = s-next;q = H-next;H-next = p;p-next = q;p = (struct node*)malloc(sizeof(struct node);s = s-next;p = H-next;q = L-next;while(p)q-coef = p-coef;q-expn = p-expn;q = q-next;p = p-next;void select() /用户选择加减操作printf(“请选择加减操作n“
7、);printf(“1.两个一元多项式相加 n“);printf(“2.两个一元多项式相减 n“);void add(polynmial La, polynmial Lb, polynmial static struct node *pc;Lc = (struct node*)malloc(sizeof(struct node);pa = La-next;pb = Lb-next;Lc-next = NULL;while(pa if(pa-expn expn)pc-next = Lc-next;Lc-next = pc;pc-coef = pa-coef;pc-expn = pa-expn;p
8、a = pa-next;elseif(pa-expn = pb-expn)pc-next = Lc-next;Lc-next = pc;pc-expn = pa-expn;pc-coef = pa-coef + pb-coef;pa = pa-next;pb = pb-next;elsepc-next = Lc-next;Lc-next = pc;pc-coef = pb-coef;pc-expn = pb-expn;pb = pb-next;while(pa)pc = (struct node*)malloc(sizeof(struct node);pc-next = Lc-next;Lc-
9、next = pc;pc-coef = pa-coef;pc-expn = pa-expn;pa = pa-next;while(pb)pc = (struct node*)malloc(sizeof(struct node);pc-next = Lc-next;Lc-next = pc;pc-coef = pb-coef;pc-expn = pb-expn;pb = pb-next;void subtract(polynmial La, polynmial Lb, polynmial static struct node *pd;Ld = (struct node*)malloc(sizeo
10、f(struct node);pa = La-next;pb = Lb-next;Ld-next = NULL;while(pa if(pa-expn expn)pd-next = Ld-next;Ld-next = pd;pd-coef = pa-coef;pd-expn = pa-expn;pa = pa-next;elseif(pa-expn = pb-expn)pd-next = Ld-next;Ld-next = pd;pd-expn = pa-expn;pd-coef = pa-coef - pb-coef;pa = pa-next;pb = pb-next;elsepd-next
11、 = Ld-next;Ld-next = pd;pd-coef = pb-coef;pd-expn = pb-expn;pb = pb-next;while(pa)pd = (struct node*)malloc(sizeof(struct node);pd-next = Ld-next;Ld-next = pd;pd-coef = pa-coef;pd-expn = pa-expn;pa = pa-next;while(pb)pd = (struct node*)malloc(sizeof(struct node);pd-next = Ld-next;Ld-next = pd;pd-coe
12、f = -pb-coef;pd-expn = pb-expn;pb = pb-next;int main()int sign;polynmial La, Lb, Lc, Ld;printf(“请输入第一个多项式： n“);create(La);sort(La);printf(“请输入第二个多项式： n“);create(Lb);sort(Lb);select();scanf(“%d“, switch(sign)case 1:printf(“多项式之和为 :n“);add(La, Lb, Lc);sort(Lc);reverse(Lc);display(Lc);break;default:printf(“多项式之差为 :n“);subtract(La, Lb, Ld);sort(Ld);reverse(Ld);display(Ld);break;return 0;
