PAT A1009

2021-10-05 22:07:18 阅读：125 来源： 互联网

标签：f1 f2 PAT int scanf each A1009 line

1009 Product of Polynomials (25 分)
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
结尾无空行
Sample Output:
3 3 3.6 2 6.0 1 1.6
结尾无空行

#include<cstdio>
int main()
{
	int k,n;
	float f1[2001]={0},f2[2001]={0},a;
	scanf("%d",&k);
	for(int i=0;i<k;i++)
	{
		scanf("%d%f",&n,&a);
		f1[n]=a;
	}

	scanf("%d",&k);
	int x;
	for(int i=0;i<k;i++)
	{
		scanf("%d%f",&n,&a);
		for(int j=0;j<1001;j++)
		{
			if(f1[j]!=0)
			{
				x=j+n;
				f2[x]+=f1[j]*a;
			}
		}
	}
	int count=0;
	for(int i=2000;i>=0;i--)
	{
		if(f2[i]!=0)
		{
			count++;
			
		}
	}
	printf("%d",count);
	for(int i=2000;i>=0;i--)
	{
		if(f2[i]!=0)
		{
			printf(" %d %.1f",i,f2[i]);
			
		}
	}
	return 0;
 }

标签：f1,f2,PAT,int,scanf,each,A1009,line
来源： https://blog.csdn.net/m0_51380817/article/details/120618951

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PAT A1009