This time, you are supposed to find?A+B?where?A?and?B?are two polynomials.

Input Specification:

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?????(i=1,2,?,K) are the exponents and coefficients, respectively. It is given that?1≤K≤10，0≤N?K??<?<N?2??<N?1??≤1000.

Output Specification:

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

計(jì)算多項(xiàng)式A+B

輸入格式：

要求包含一個(gè)測試數(shù)據(jù)，每個(gè)測試數(shù)據(jù)要包含兩行，每行包括一個(gè)多項(xiàng)式：

K N1 an1 N2 an2....NK ank

K是多項(xiàng)式中非零項(xiàng)的個(gè)數(shù)，Ni和ani是分別指數(shù)和系數(shù)。k在【1，10】，NK逐個(gè)遞減【0，1000】

輸出格式：

每一個(gè)測試數(shù)據(jù)，你應(yīng)該在一行中輸出A和B的和，和輸入的格式一樣，要注意在每一行的最后不要包含空格，請(qǐng)精確到小數(shù)點(diǎn)后1位。

?

Sample Input:

2 1 2.4 0 3.2 2 2 1.5 1 0.5

## 多項(xiàng)式A:? 2.4*X^1 +3.2*X^0

## 多項(xiàng)式B:? 1.5*X^2 + 0.5 *X^1

A+B = 1.5*X^2 +2.9*X^1 + 3.2

轉(zhuǎn)換為輸出為： 3? 2 1.5 1 2.9 0 3.2??

?

Sample Output:

3 2 1.5 1 2.9 0 3.2

計(jì)算：

#include<stdio>
#include<map>
using namespace::std;
map<int,double> s;
int main()
{?? ?
?? ?int i1,t1;?? ?
?? ?double t2;?? ?
?? ?for(int z=0;z<2;z++)?? ?
?? ?{?? ??? ?
?? ??? ?scanf("%d",&i1);?? ??? ?
?? ??? ?for(int i=0;i<i1;i++)?? ??? ?
?? ??? ?{?? ??? ??? ?
?? ??? ??? ?scanf("%d%lf",&t1,&t2);?? ??? ??? ?
?? ??? ??? ?if(s.count(t1)==0)?? ??? ??? ??? ?
?? ??? ??? ??? ?s[t1]=t2;?? ??? ??? ?
?? ??? ??? ?else?? ??? ??? ??? ?
?? ??? ??? ??? ?s[t1]+=t2;?? ??? ?
?? ??? ?}?? ?
?? ?}?? ?
?? ?i1=0;?? ?
?? ?for(map<int,double>::const_iterator?
?? ??? ?m_it=s.begin();
?? ??? ?m_it!=s.end();m_it++)?? ?
?? ?{?? ??? ?
?? ??? ?if(m_it->second!=0.0&&m_it->second!=-0.0)?? ??? ??? ?
?? ??? ??? ?i1++;?? ?
?? ?}?? ?
?? ?printf("%d",i1);?? ?
?? ?for(map<int,double>::reverse_iterator m_it=s.rbegin();m_it!=s.rend();m_it++)?? ?
?? ?{?? ??? ?
?? ??? ?if(m_it->second!=0.0&&m_it->second!=-0.0)?? ??? ??? ?
?? ??? ??? ?printf(" %d %.1lf",m_it->first,m_it->second);?? ?
?? ?}?? ?
?? ?printf("\n");?? ?
?? ?return 0;
}

?

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