164 lines
3.8 KiB
C
164 lines
3.8 KiB
C
// 1.问题描述
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// 设计一个一元稀疏多项式简单计算器。
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// 2.基本要求
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// 一元稀疏多项式简单计算器的基本功能是：
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// (1)输入并建立多项式；
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// (2)输出多项式，输出形式为整数序列：
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// n,c1,e1,c2,e2,…,cn,en,
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// 其中n是多项式的项数，ci,ei分别是第i项的系数和指数，序列按指数降序排列；
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// (3)多项式a和b相加，建立多项式a+b；
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// (4)多项式a和b相减，建立多项式a-b；
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// (5)计算多项式在x处的值；即给定x值，计算多项式值。
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// 3.实现提示
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// 用带表头结点的单链表存储多项式，多项式的项数存放在头结点中。
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#include <math.h>
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#include <stdio.h>
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#include <stdlib.h>
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typedef struct Node {
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int c, e;
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struct Node *next;
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} Node;
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typedef struct List {
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Node head;
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int len;
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} List;
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List *init_list() {
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List *l = (List *)malloc(sizeof(List));
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l->head.next = NULL;
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l->len = 0;
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return l;
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}
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// 降序插入
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void insert(List *l, int c, int e) {
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Node *p = &(l->head);
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while (p->next && p->next->e > e) {
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p = p->next;
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}
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if (p->next && p->next->e == e) {
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p->next->c += c;
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if (p->next->c == 0) {
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Node *q = p->next;
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p->next = q->next;
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free(q);
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l->len--;
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}
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} else {
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Node *q = (Node *)malloc(sizeof(Node));
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q->c = c, q->e = e;
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q->next = p->next;
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p->next = q;
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l->len++;
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}
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}
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// 只输出
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void output(List *l) {
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Node *p = l->head.next;
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printf("%d,", l->len);
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while (p) {
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printf("%d,%d", p->c, p->e);
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p = p->next;
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if (p)
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printf(",");
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}
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printf("\n");
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}
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// 只输出
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void plus(List *l1, List *l2) {
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Node *p1 = l1->head.next, *p2 = l2->head.next;
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while (p1 && p2) {
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if (p1->e > p2->e) {
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printf("%d,%d,", p1->c, p1->e);
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p1 = p1->next;
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} else if (p1->e < p2->e) {
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printf("%d,%d,", p2->c, p2->e);
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p2 = p2->next;
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} else {
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printf("%d,%d,", p1->c + p2->c, p1->e);
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p1 = p1->next;
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p2 = p2->next;
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}
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}
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while (p1) {
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printf("%d,%d,", p1->c, p1->e);
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p1 = p1->next;
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}
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while (p2) {
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printf("%d,%d,", p2->c, p2->e);
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p2 = p2->next;
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}
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printf("\n");
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}
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// 只输出
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void minus(List *l1, List *l2) {
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Node *p1 = l1->head.next, *p2 = l2->head.next;
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while (p1 && p2) {
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if (p1->e > p2->e) {
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printf("%d,%d,", p1->c, p1->e);
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p1 = p1->next;
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} else if (p1->e < p2->e) {
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printf("%d,%d,", -p2->c, p2->e);
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p2 = p2->next;
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} else {
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printf("%d,%d,", p1->c - p2->c, p1->e);
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p1 = p1->next;
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p2 = p2->next;
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}
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}
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while (p1) {
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printf("%d,%d,", p1->c, p1->e);
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p1 = p1->next;
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}
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while (p2) {
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printf("%d,%d,", -p2->c, p2->e);
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p2 = p2->next;
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}
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printf("\n");
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}
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int calc(List *l, int x) {
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int sum = 0;
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Node *p = l->head.next;
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while (p) {
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sum += p->c * pow(x, p->e);
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p = p->next;
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}
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return sum;
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}
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int main() {
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// freopen("data.in", "r", stdin);
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List *l = init_list();
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List *l2 = init_list();
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int n1, n2, c, e;
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scanf("%d", &n1);
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for (int i = 0; i < n1; i++) {
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scanf("%d%d", &c, &e);
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insert(l, c, e);
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}
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scanf("%d", &n2);
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for (int i = 0; i < n2; i++) {
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scanf("%d%d", &c, &e);
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insert(l2, c, e);
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}
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output(l);
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output(l2);
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plus(l, l2);
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minus(l, l2);
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while (1) {
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printf("请输入x的值：");
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scanf("%d", &c);
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if (c == -1)
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break;
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printf("%d\n", calc(l, c));
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}
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return 0;
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}
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