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PAT (Advanced Level) Practice — 1009 Product of Polynomials (25 分)

題目連結:https://pintia.cn/problem-sets/994805342720868352/problems/994805509540921344

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 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.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5
           

Sample Output:

3 3 3.6 2 6.0 1 1.6
           

:題意  多項式相乘,即指數相加,系數相乘,數組a、b的下标分别存兩個多項式的指數,數組的值用來存儲系數

注意多項式乘積→指數相加之後範圍是2000,ans數組範圍要大于2000.

多項式相加  PAT甲級—1001 

#include<iostream>
#include<cstdio> 
#include<cstring>
using namespace std;
int main(){
	int n,m;
	double a[1011],b[1011];
	double ans[2022];
	memset(ans,0,sizeof(ans));
	int x;
	double xi;
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>x>>xi;
		a[x]=xi;
	}
	cin>>m;
	for(int i=0;i<m;i++){
		cin>>x>>xi;
		b[x]=xi;
	}
	for(int i=0;i<1001;i++){
		for(int j=0;j<1001;j++){
			if(a[i]!=0||b[j]!=0){
				ans[i+j]+=a[i]*b[j];
			}
		}
	}
	int cnt=0;
	for(int i=0;i<2001;i++){
		if(ans[i]!=0){
			cnt++;
		}
	}
	cout<<cnt;
	for(int i=2000;i>=0;i--){
		if(ans[i]!=0){
			printf(" %d %.1lf",i,ans[i]);
		}
	}
	cout<<endl;
	return 0;
}
           
PAT (Advanced Level) Practice — 1009 Product of Polynomials (25 分)