數(shù)據(jù)結(jié)構(gòu)-實(shí)驗(yàn)一-一元多項(xiàng)式相加

1、數(shù)據(jù)結(jié)構(gòu)實(shí)驗(yàn)報告實(shí)驗(yàn)一：一元多項(xiàng)式相加姓 名： 周 成 學(xué) 號： 13083511 專 業(yè)： 軟件工程 任 課 教 師： 馬慧珠 2013年12 月01 日1.實(shí)驗(yàn)名稱：一元多項(xiàng)式相加2.實(shí)驗(yàn)?zāi)康? 如何使用C語言實(shí)現(xiàn)鏈表的說明、創(chuàng)建以及結(jié)點(diǎn)的插入和刪除等操作。3.實(shí)驗(yàn)要求：對一元多項(xiàng)式能實(shí)現(xiàn)輸入、輸出，以及兩個一元多項(xiàng)式相加及結(jié)果顯示。4.實(shí)驗(yàn)內(nèi)容：一元多項(xiàng)式的表示在計算機(jī)內(nèi)用鏈表來實(shí)現(xiàn)，同時為了節(jié)省存儲空間，只存儲其中非零的項(xiàng)，鏈表中的每個節(jié)點(diǎn)存放多項(xiàng)式的系數(shù)非零項(xiàng)。它包含三個域，分別存放多項(xiàng)式的系數(shù)，指數(shù)，以及指向下一個項(xiàng)的指針。根據(jù)一元多項(xiàng)式相加的運(yùn)算規(guī)則：對于兩個一元多項(xiàng)式中所有指數(shù)

2、相同的項(xiàng)，對應(yīng)系數(shù)相加，若其和不為零，則構(gòu)成“和多項(xiàng)式”中的一項(xiàng)，對于兩個一元多項(xiàng)式中所有指數(shù)不相同的項(xiàng)，則分別復(fù)抄到“和多項(xiàng)式”中去。核心算法PolyAdd是把分別由pa和pb所指的兩個多項(xiàng)式相加，結(jié)果為pa所指的多項(xiàng)式。運(yùn)算規(guī)則如下：相加時，首先設(shè)兩個指針變量qa和qb分別從多項(xiàng)式的首項(xiàng)開始掃描，比較qa和qb所指結(jié)點(diǎn)指數(shù)域的值，可能出現(xiàn)下列三種情況之一：（1）qa->exp大于qb->exp,則qa繼續(xù)向后掃描。（2）qa->exp等于qb->exp,則將其系數(shù)相加。若相加結(jié)果不為零，將結(jié)果放入qa->coef中，并刪除qb所指結(jié)點(diǎn)，否則同時刪除qa和qb所

3、指結(jié)點(diǎn)。然后qa、qb繼續(xù)向后掃描。（3）qa->exp小于qb->exp,則將qb所指結(jié)點(diǎn)插入qa所指結(jié)點(diǎn)之前，然后qa、qb繼續(xù)向后掃描。掃描過程一直進(jìn)行到qa或qb有一個為空為止，然后將有剩余結(jié)點(diǎn)的鏈表接在結(jié)果表上。所得pa指向的鏈表即為兩個多項(xiàng)式之和。5.實(shí)驗(yàn)程序代碼及運(yùn)行結(jié)果：#include "stdafx.h"#include <stdio.h>#include <stdlib.h>#include <malloc.h>#include<stdio.h> #define NULL 0typedef s

4、truct NODE float coef; /系¦Ì數(shù)ºy int expn; /指?數(shù)ºy struct NODE *next; NODE;NODE *Creat(int n);void print(NODE *head);NODE *AddPolyn(NODE *head1, NODE *head2);NODE *Delfirst(NODE *head, NODE *q);void InsertBefore(NODE *p1, NODE *p2);int compare(int a, int b);/*創(chuàng)ä¡ä建

5、61;§鏈¢¡ä表À¨ª*/NODE *Creat(int n) NODE *current, *previous, *head; int i; head = (NODE *)malloc(sizeof(NODE); /*創(chuàng)ä¡ä建¡§頭ª¡¤結(jié)¨¢點(diǎn)Ì?*/ previous = head; for(i = 0; i < n; i+) current = (NODE *)malloc(sizeof(NODE)

6、; printf("請?輸º?入¨?系¦Ì數(shù)ºy和¨ª指?數(shù)ºy : "); scanf("%f%d", &current->coef, &current->expn); previous->next = current; previous = current; previous->next = NULL; return head;/*一°?元a多¨¤項(xiàng)?式º?的Ì?想?加¨&

7、#174;，ê?總Á¨¹體¬?考?慮?，ê?可¨¦分¤?qa的Ì?指?數(shù)ºy比À¨¨qb小?，ê?或¨°等̨¨于®¨²pb(如¨?果?系¦Ì數(shù)ºy相¨¤加¨®等̨¨于®¨²0和¨ª不?等Ì

8、8;¨于®¨²0),或¨°大䨮于®¨²pb里¤?面?由®¨¦InsertBefore和¨ªDelfirst兩¢?個?小?模¡ê塊¨¦組Á¨¦成¨¦一°?部?分¤?*/NODE *AddPolyn(NODE *head1, NODE *head2) NODE *ha, *hb, *qa, *qb;

9、 int a, b; float sum; ha = head1; /*ha和¨ªhb指?向¨°頭ª¡¤結(jié)¨¢點(diǎn)Ì?*/ hb = head2; qa = ha->next; /*qa和¨ªqb指?向¨°頭ª¡¤結(jié)¨¢點(diǎn)Ì?的Ì?下?一°?個?結(jié)¨¢點(diǎn)Ì?*/ qb = hb->next; while(qa && qb)

10、 /*qa和¨ªqb均¨´非¤?空?*/ a = qa->expn; b = qb->expn; switch(compare(a, b) case -1 : /*qa->expn < qb->expn*/ ha = qa; qa = qa->next; break; case 0 : sum = qa->coef + qb->coef; /*系¦Ì數(shù)ºy的Ì?和¨ª*/ if(sum != 0.0) /*如¨?果?不?是

11、86;?0.0*/ qa->coef = sum; /*改?變À?系¦Ì數(shù)ºy*/ ha = qa; else free(Delfirst(ha, qa); free(Delfirst(hb, qb); qa = ha->next; qb = hb->next; /*qb釋º¨ª放¤?后¨®要°a重?新?賦3值¦Ì*/ break; case 1 : /*如¨?果?qa-> expn > qb -> expn*/ Del

12、first(hb, qb); InsertBefore(ha, qb); /*把ã?qb插?入¨?到Ì?ha下?一°?個?結(jié)¨¢點(diǎn)Ì?之?前¡ã*/ qb = hb->next; ha = ha->next; break; if(qb) ha->next = qb; /*插?入¨?剩º¡ê余®¨¤的Ì?pb*/ free(head2); return head1; /*比À¨¨較

13、?*/int compare(int a, int b) if(a < b) return -1; else if(a > b) return 1; else return 0;/*刪¦?除y結(jié)¨¢點(diǎn)Ì?q*/NODE *Delfirst(NODE *p1, NODE *q) p1 -> next = q -> next; return (q); /*插?入¨?結(jié)¨¢點(diǎn)Ì?,引°y入¨?結(jié)¨¢點(diǎn)Ì?p,可¨¦以°

14、?讓¨?p插?入¨?到Ì?p2和¨ªp1之?間?*/void InsertBefore(NODE *p1, NODE *p2) NODE *p; p = p1->next; p1->next = p2; p2->next = p;/*打䨰印®?,為a了¢?美¨¤觀?程¨¬序¨°分¤?開a打䨰印®?*/void print(NODE *head) NODE *cu

15、rrent; current = head->next; while(current->next != NULL) printf("%0.f * x%d + ", current->coef, current->expn); current = current -> next; printf("%0.f * x%d", current->coef, current->expn); /system（ꡧ"pause"); int main() NODE *head

16、1, *head2, *head3; int n1, n2; printf("請?輸º?入¨?你?需¨¨要°a的Ì?多¨¤項(xiàng)?式º?的Ì?項(xiàng)?數(shù)ºy n1 : "); scanf("%d", &n1); head1 = Creat(n1); printf("第̨²一°?個?多¨¤項(xiàng)?式º?的Ì?顯?示º? : n"); prin

17、t(head1); printf("n請?輸º?入¨?你?需¨¨要°a的Ì?多¨¤項(xiàng)?式º?的Ì?項(xiàng)?數(shù)ºy n2 : "); scanf("%d", &n2); head2 = Creat(n2); printf("n第̨²二t個?多¨¤項(xiàng)?式º?的Ì?顯?示º? : n"); print(head2); head3 = AddPolyn(head1, head2); printf("n合?并¡é后¨®的Ì?多¨¤項(xiàng)?式º?的Ì?顯?示º? : n"); print(head3); printf("n");運(yùn)行結(jié)果：實(shí)驗(yàn)數(shù)據(jù)1如圖：輸入一個四次二項(xiàng)式X3+2X4，一個五次二項(xiàng)式X4+2X5，輸出如圖：實(shí)驗(yàn)數(shù)據(jù)2如圖：輸入一個五次四項(xiàng)式X2+X3+X4+X5，還有一個五次五項(xiàng)式1+X+X3+2X4+2X5輸出如圖所示實(shí)驗(yàn)數(shù)據(jù)3如圖：輸入一個七次三項(xiàng)式1+2x5+3X7，還有一個五次四