PAT (Advanced Level) Practice 1002 A+B for Polynomials

【Problem Description】:
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 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 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.
【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

【解题思路】
考虑用链表来做这道题，用STL中的list容器，构建项结构体包含指数和系数，依次构建链表A和B，按降幂顺序排好，按规则加入和多项式C，注意三点：
① 输出要按照幂次从大到小顺序，格式上需要保留一位小数。
② 题目的输入和输出都是系数非零项。
③ 注意正负抵消的问题，两个多项式的某一个相同幂次的项刚好为相反数，相加之后就为 0 了所以不用输出。
【C++代码】

#include <iostream>
#include <list>
#include <algorithm>
#include <iomanip>
using namespace std;
struct term {	//多项式的项
int exp;	//指数
double coe;	//系数
};
bool cmp(term a,term b) {	//按指数从大到小排序
return a.exp>b.exp;
}
int main() {
list<term> A,B,C;
int n,m;
term t;
cin>>n;
while(n--) {	//构建多项式A
cin>>t.exp>>t.coe;
A.push_back(t);
}
A.sort(cmp);
cin>>m;
while(m--) {	//构建多项式B
cin>>t.exp>>t.coe;
B.push_back(t);
}
B.sort(cmp);
while((!A.empty())&&(!B.empty())) {	//直到两个多项式链表中有一个为空
if(A.front().exp==B.front().exp) {
t.exp=A.front().exp;
t.coe=A.front().coe+B.front().coe;
C.push_back(t);
A.pop_front();
B.pop_front();
} else if(A.front().exp>B.front().exp) {
C.push_back(A.front());
A.pop_front();
} else {
C.push_back(B.front());
B.pop_front();
}
}
if(!A.empty()) {	//将不为空的链表的全部项加入和多项式
while(!A.empty()) {
C.push_back(A.front());
A.pop_front();
}
} else if(!B.empty()) {
while(!B.empty()) {
C.push_back(B.front());
B.pop_front();
}
}
int cnt=0;	//统计系数为0的项
for(list<term>::iterator it=C.begin(); it!=C.end(); it++) {
if((*it).coe==0) {
cnt++;
}
}
C.sort(cmp);
cout<<C.size()-cnt;	//项的个数
for(list<term>::iterator it=C.begin(); it!=C.end(); it++) {
if((*it).coe!=0) {
cout<<" "<<(*it).exp<<" "<<setiosflags(ios::fixed)<<setprecision(1)<<(*it).coe;//注意保留1位小数
}
}
cout<<endl;
return 0;
}

【Java代码】

import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
float polynomial[] = new float[1001];
int n = scan.nextInt();
for(int i = 0; i < n; i++) {
int exp = scan.nextInt();
polynomial[exp] = scan.nextFloat();
}
int m = scan.nextInt();
for(int i = 0; i < m; i++) {
int exp = scan.nextInt();
polynomial[exp] += scan.nextFloat();
}
scan.close();
int cnt = 0;
for(int i = 0; i < polynomial.length; i++) {
if(polynomial[i] != 0) {
cnt++;
}
}
System.out.print(cnt);
for(int i = polynomial.length - 1; i >= 0 ; i--) {
if(polynomial[i] != 0) {
System.out.printf(" %d %.1f",i,polynomial[i]);
}
}
}
}

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