标签：删掉 highlight || 链表 rate float info stand while
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 Input2 1 2.4 0 3.2 2 2 1.5 1 0.5Sample 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。
标签：删掉 highlight || 链表 rate float info stand while
原文地址：https://www.cnblogs.com/LowBee/p/8976040.html
