PAT 一元多项式的乘法与加法运算(链表 c++版)

题目按照指数递减的顺序给出两个一元多项式，输出两个多项式的乘积，还有 和 ，按照指数递减的顺序。

用链表实现一元多项式的乘法与加法运算。

首先来看加法运算

多项式 poly1   x^4 + 3x^3 + 6x

多项式 poly2   6x^5 + 4x^4 + 6

既然用到链表，自然少不了结构体

struct node{int coef; //系数int expn; //指数node* next;};

实现这加法，只要新创建一个空链表，用两个“游标” cur1，cur2 分别指向两个多项式的头节点，每次找到指数较大的项加入新链表，相应的cur = cur->next,

如果当前的两个对应项的指数相同，就把他们的系数相加， 如果和不为零，就把它加到新链表中。

乘法，模拟乘法即可，每次用poly1的每一项依次乘以 poly2的每一项，再调用求和函数即可。（因为每次求和都保证了当前序列是递减的）

具体来看代码吧。

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;struct node{int coef; //系数int expn; //指数node* next;};node* creat_list(int n){ //读入链表node  *head, *r;head = new node;r = head; int coef , expn;while(n--){scanf("%d%d",&coef,&expn);node* tmp = new node; //创建临时结点tmp->coef = coef;tmp->expn = expn;r->next = tmp; //临时结点接到链表中r = tmp;}r->next = NULL; //结尾设为 NULLreturn head;}node* add_list(node* a,node* b){node *r,*fans, *ans;node *ha,*hb; //为了防止修改指针本身的值，使用代理指针来完成操作，也就是游标。fans = new node;ans = fans; //ans 作为fans 的”游标“ha = a->next;hb = b->next;while(ha && hb){node* tmp = new node; //建立一次即可if(ha->expn > hb->expn){ //每次把指数（exponent）较大的加入链表fanstmp->coef = ha->coef;tmp->expn = ha->expn;ans->next = tmp;ans = tmp;ha = ha->next;}else if(ha->expn < hb->expn){tmp->coef = hb->coef;tmp->expn = hb->expn;ans->next = tmp;ans = tmp;hb = hb->next;}else{int mulOfcoef = (ha->coef)+(hb->coef); //如果指数相同， 就把系数求和。if(mulOfcoef!=0){tmp->coef = mulOfcoef;tmp->expn = ha->expn;ans->next = tmp;ans = tmp;}ha = ha->next; //注意这里 即使和为0 ，也要移动“游标”hb = hb->next;}}while(ha){node* tmp = new node;tmp->coef = ha->coef;tmp->expn = ha->expn;ans->next = tmp;ans = tmp;ha = ha->next;}while(hb){node* tmp = new node;tmp->coef = hb->coef;tmp->expn = hb->expn;ans->next = tmp;ans = tmp;hb = hb->next;}ans->next = NULL; //结尾设为 NULLreturn fans;}node* multi_list(node* a,node* b){node* ha, *hb;node* ans,*fans;ha = a->next;hb = b->next;fans = creat_list(0);if(ha == NULL || hb == NULL){return fans;}node* tmp;while(ha != NULL){tmp = new node;ans = tmp;hb = b->next; //每次都是从 b 的第一项开始乘；while(hb != NULL){node* ltmp = new node;ltmp->expn = ha->expn + hb->expn; //指数相加，系数相乘ltmp->coef = ha->coef * hb->coef;hb = hb->next;ans->next= ltmp;ans = ltmp;}ans->next = NULL;fans = add_list(fans,tmp); //将乘法 分解成一次次的加法ha = ha->next;}return fans;}void print_list(node* l){node *hc;int flag = 0;hc = l->next; //指针操作常用，用新创立的节点代替源节点操作if(hc == NULL){ //格式控制 。。 真坑！printf("0 0");}while(hc != NULL){if(flag)printf(" ");elseflag = 1;printf("%d %d",hc->coef,hc->expn);hc = hc->next;}}int main(){int n;scanf("%d",&n);node *a = creat_list(n);int m;scanf("%d",&m);node* b = creat_list(m);node* c = add_list(a,b);node* d = multi_list(a,b);print_list(d);printf("\n");print_list(c);printf("\n");return 0;}

C语言版本的戳笔者参考的这个网址C语言版本 ，

对于这个题，开始也用了模拟的方法做了，用到了map，写的较乱。。格式错误wa了无数次

给网上的测试数据吧

样例输入与输出：

序号输入输出1

4 3 4 -5 2  6 1  -2 03 5 20  -7 4  3 1

15 24 -25 22 30 21 -10 20 -21 8 35 6 -33 5 14 4 -15 3 18 2 -6 15 20 -4 4 -5 2 9 1 -2 0

2

2 1 2 1 02 1 2 -1 0

1 4 -1 02 2

3

2 -1000 1000 1000 02 1000 1000 -1000 0

-1000000 2000 2000000 1000 -1000000 00 0

4

01 999 1000

0 0999 1000

【结构体模拟版本】

#include <iostream>#include <cstdio>#include <algorithm>#include <map>using namespace std;const int maxn = 100000;struct node{int x, sub;}poly[maxn],poly2[maxn];int main(){int n;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d%d",&poly[i].x,&poly[i].sub);}int m;scanf("%d",&m);for(int i=0;i<m;i++){scanf("%d%d",&poly2[i].x,&poly2[i].sub);}map<int,int>mp;for(int i=0;i<n;i++){for(int j=0;j<m;j++){int a=poly[i].x*poly2[j].x;int sub = poly[i].sub+poly2[j].sub;mp[sub]+=a;}}int sign=0;if(!mp.empty()){map<int,int>::iterator it= mp.end();it--;for(;it!=mp.begin();it--){if(it->second!=0){if(!sign)sign=1;elsecout<<" ";cout<<it->second<<" "<<it->first;}}if(sign && it->second !=0){cout<<" "<<it->second<<" "<<it->first;}else if(!sign && it->second !=0){sign = 1;cout<<it->second<<" "<<it->first;}}if(!sign){cout<<"0 0";}cout<<endl;int beg=0;int beg2=0;sign=0;while(beg!=n && beg!=m){if(poly[beg].sub>poly2[beg2].sub){if(!sign)sign=1;elsecout<<" ";cout<<poly[beg].x<<" "<<poly[beg].sub;beg++;}else if(poly[beg].sub<poly2[beg2].sub){if(!sign)sign=1;elsecout<<" ";cout<<poly2[beg2].x<<" "<<poly2[beg2].sub;beg2++;}else{int tmpadd = poly[beg].x + poly2[beg2].x ;if( tmpadd == 0){}else{if(!sign)sign=1;elsecout<<" ";cout<<tmpadd<<" "<<poly[beg].sub;}beg++;beg2++;}}while(beg!=n){if(!sign)sign=1;elsecout<<" ";cout<<poly[beg].x<<" "<<poly[beg].sub;beg++;}while(beg2!=m){if(!sign)sign=1;elsecout<<" ";cout<<poly2[beg2].x<<" "<<poly2[beg2].sub;beg2++;}if(!sign)cout<<"0 0";cout<<endl;return 0;}