标签：ott number exponents product read ica his term input
This time, you are supposed to find A×B where A and B are two polynomials.
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:
K N?1?? a?N?1???? N?2?? a?N?2???? ... N?K?? a?N?K????
where K is the number of nonzero terms in the polynomial, N?i?? and a?N?i???? (,) are the exponents and coefficients, respectively. It is given that 1, 0.
For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.
2 1 2.4 0 3.2
2 2 1.5 1 0.53 3 3.6 2 6.0 1 1.6
方法一：链表法（不建议使用）
#include<iostream> #include<algorithm> #include<cstdio> #include<cstdlib> #include<cmath> #include<cstring> #include<string.h> using namespace std; //多项式相乘与相加 const int maxn=110010; typedef struct Node{ int expon; double coef; struct Node *next; }node,*pNode; pNode ReadPoly(pNode p){ int K; pNode q,r; scanf("%d",&K); p=new node; p->next=NULL; r=p; while(K--){ q=new node; scanf("%d %lf",&q->expon,&q->coef); r->next=q; r=r->next; } r->next=NULL; return p; } void printPoly(pNode p){ pNode q,r; int k=0; r=p->next; while(r) { if(r->coef!=0) { k++; } r=r->next; } printf("%d ",k); q=p->next; while(q){ if(q->coef==0){//系数为0，则不必输出 q=q->next; continue; } if(q->next!=NULL) { printf("%d %.1f ",q->expon,q->coef); } else{ printf("%d %.1f",q->expon,q->coef); } q=q->next; } printf("\n"); } pNode Mult(pNode p1,pNode p2){ pNode ps,psn,ppp; ps=new node;//新链表头部 ps->expon=0; ps->coef=0; ps->next=NULL; // psn=pp;//新链表尾部 pNode p11,p22; p11=p1->next; int flog=0; // p22=p2->next; while(p11){ p22=p2->next;//返回链表第一个结点 while(p22){ ppp=new node; ppp->expon=p11->expon+p22->expon; ppp->coef=p11->coef*p22->coef; psn=ps;//返回链表头节点 flog=0; while(psn->next) { if(psn->next->expon==ppp->expon){ flog=1; break; } else{ psn=psn->next; } } if(flog==1){ psn->next->expon=ppp->expon; psn->next->coef=psn->next->coef+ppp->coef;//注意不要写错哦 } else{ psn->next=ppp; psn=psn->next; psn->next=NULL; } p22=p22->next; } p11=p11->next; } return ps; } pNode Add(pNode p1,pNode p2){ pNode ps,pp1,ppp;//pp指向链表头部，pp1指向链表尾部，ppp指向新节点 pNode p11,p22;//p11，p22都指向链表尾部 p11=p1->next; p22=p2->next; ps=new node; pp1=ps; while(p11&&p22){ ppp=new node; if(p11->expon>p22->expon){ ppp->expon=p11->expon; ppp->coef=p11->coef; pp1->next=ppp; pp1=pp1->next; p11=p11->next; } else if(p11->expon<p22->expon){ ppp->expon=p22->expon; ppp->coef=p22->coef; pp1->next=ppp; pp1=pp1->next; pp1->next=NULL; p22=p22->next; } else if(p11->expon==p22->expon){ ppp->expon=p11->expon; ppp->coef=p11->coef+p22->coef; pp1->next=ppp; pp1=pp1->next; p11=p11->next; p22=p22->next; } } if(p11!=NULL){ pp1->next=p11; } if(p22!=NULL){ pp1->next=p22; } return ps; } pNode sort(pNode p){//冒泡排序 pNode r,t; int ti; double td; r=p->next; while(r){ t=r->next; while(t){ if(t->expon>r->expon){ ti=t->expon; td=t->coef; t->expon=r->expon; t->coef=r->coef; r->expon=ti; r->coef=td; } t=t->next; } r=r->next; } return p; } int main(){ pNode p1,p2,pp,ps; p1=ReadPoly(p1); p2=ReadPoly(p2); pp=Mult(p1,p2); pp=sort(pp); printPoly(pp); // ps=Add(p1,p2); // pp=sort(pp); // printPoly(ps); return 0; }
方法二：利用数组
#include<iostream> #include<algorithm> #include<cstdio> #include<cstdlib> #include<cmath> #include<cstring> #include<vector> #include<string.h> using namespace std; const int maxn=101000; struct node{ int expon; double coef; }; bool compare(node a,node b){ return a.expon>b.expon; } int main(){ vector<node> p1,p2,p3; int m,n,e; node nd; double c; scanf("%d",&m); while(m--){ scanf("%d %lf",&nd.expon,&nd.coef); p1.push_back(nd); } scanf("%d",&n); while(n--){ scanf("%d %lf",&nd.expon,&nd.coef); p2.push_back(nd); } node p; int flog=0,k; for(int i=0;i<p1.size();i++){ for(int j=0;j<p2.size();j++){ p.expon=p1[i].expon+p2[j].expon; p.coef=p1[i].coef*p2[j].coef; flog=0; for(k=0;k<p3.size();k++){ if(p3[k].expon==p.expon){ flog=1; p3[k].coef=p3[k].coef+p.coef; break; } } if(flog==0){ p3.push_back(p); } } } sort(p3.begin(),p3.end(),compare); int t=0; for(int i=0;i<p3.size();i++){ if(p3[i].coef!=0){ t++; } } printf("%d ",t); for(int i=0;i<p3.size();i++){ if(p3[i].coef==0){ continue; } if(i<p3.size()-1) printf("%d %.1f ",p3[i].expon,p3[i].coef); else{ printf("%d %.1f",p3[i].expon,p3[i].coef); } } printf("\n"); return 0; }
1009 Product of Polynomials (25分)
标签：ott number exponents product read ica his term input
原文地址：https://www.cnblogs.com/dre-52yao/p/14284553.html
