题目:
1002. A+B for Polynomials (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
This time, you are supposed to find A+B where A and B are two polynomials.
Input
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10,0 <= NK < ... < N2 < N1 <=1000.
Output
For each test case you should output the sum 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 to 1 decimal place.
Sample Input
2 1 2.4 0 3.2 2 2 1.5 1 0.5
Sample Output
3 2 1.5 1 2.9 0 3.2 题目大意就是给定两个多项式 F1=aN1*x^N1+aN2*x^N2+aN3*x^N3+.....和F2=bN1*x^N1+bN2*x^N2+bN3*x^N3+..... 求两个多项式的和
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<iomanip> using namespace std; template <class T> class Link { public: T data1,data2; Link<T> *next; Link(const T d1,const T d2,Link<T> *nex=NULL) { data1=d1; data2=d2; next=nex; } Link() { data1=0; data2=0; next=NULL; } }; template <class T> class InkLink { private: Link<T> *head; int lon; public: InkLink() { head=new Link<T>; lon=0; } void display2() { Link<T> *p; p=head->next; cout << lon ; while(p!=NULL) { printf(" "); printf("%d %.1lf",(int)p->data1,p->data2); p=p->next; } cout <<endl; } bool insertDesc(const T value1,const T value2) { if(head->next==NULL) { head->next=new Link<T>(value1,value2); lon++; return true; } Link<T> *p,*q; p=head->next; q=head; while(p!=NULL&&((p->data1)>value1)) { q=p; p=p->next; } if(p==NULL||((p->data1)<value1)) { q->next=new Link<T>(value1,value2,p); lon++; } else if(p->data1==value1) { p->data2+=value2; if(p->data2==0) { if(p->next!=NULL)q->next=p->next; else q->next=NULL; delete p; lon--; } } return true; } void clearLink() { Link<T> *p,*q; p=head->next; q=p; while(p!=NULL) { p=p->next; delete q; q=p; } lon=0; head->next=NULL; } }; int main() { InkLink<double> l; int n; double a,b; while(scanf("%d",&n)!=EOF) { for(int i=0; i<n; i++) { scanf("%lf%lf",&a,&b); l.insertDesc(a,b); } scanf("%d",&n); for(int i=0; i<n; i++) { scanf("%lf%lf",&a,&b); l.insertDesc(a,b); } l.display2(); l.clearLink(); } return 0; } /* 2 1 2.4 0 3.2 2 2 1.5 0 -3.2 */
这道题交了无数回,踩过了所有的坑。。。首先多项式的系数可以为负数,当系数减到0的时候要删掉这个节点。还有就是精确到小数点后一位。最后要控制格式,答案末尾直接输出回车,不要有空格。最后一组样例的结果应该是0,同样不能输出空格,没过的可能需要注意一下这点。
原文地址:https://www.cnblogs.com/LowBee/p/8976040.html