1009. Product of Polynomials (25)

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本文介绍了一种计算两个多项式相乘的算法，并通过示例展示了输入输出格式及程序实现过程。要求输入两个多项式的系数和指数，输出其乘积结果。

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1009. Product of Polynomials (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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 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

#include<stdio.h>
int main(){
	double a[1001]={0};
	double b[1001]={0};
	double cc[2001]={0};
	int n1,n2,i,j,e;
	double c;
	scanf("%d",&n1);
	for(i=0;i<n1;i++){
		scanf("%d %lf",&e,&c);
		a[e]=c;
	}
	scanf("%d",&n2);
	for(i=0;i<n2;i++){		 
		scanf("%d %lf",&e,&c);
		b[e]=c;
	}	
	for(i=0;i<1001;i++){
		if(a[i]!=0){
			for(j=0;j<1001;j++){
				if(b[j]!=0){
					cc[i+j]=cc[i+j]+a[i]*b[j];
				}
			}
		}
	}
	int cou=0;
	for(i=2000;i>=0;i--){
		if(cc[i]!=0){
			cou++;
		}
	}
	printf("%d",cou);
	for(i=2000;i>=0;i--){
		if(cc[i]!=0){
			printf(" %d %.1lf",i,cc[i]);
		}
	}
}