前言

传送门

正文

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 aN​1 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 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

思路:

分别将两个多项式的系数记录在数组a和数组b中，再通过一个二重循环将结果存到数组c中，最后输出数组c中的非零元素即可

参考题解:

#include<cstdio>
const int max_len=1010; 
double a[max_len],b[max_len],c[2*max_len+1];
int main(){
	int k,m,count=0;
	double n;
	scanf("%d",&k);
	while(k--){
		scanf("%d%lf",&m,&n);
		a[m]+=n;
	}
	scanf("%d",&k);
	while(k--){
		scanf("%d%lf",&m,&n);
		b[m]+=n;
	}
	for(int i=0;i<max_len;i++){
		if(a[i]!=0){
			for(int j=0;j<max_len;j++){
				if(b[j]!=0){
					c[i+j]+=a[i]*b[j];
				}
			}
		}
	}
	for(int i=0;i<2*max_len+1;i++){
		if(c[i]!=0)count++;
	}
	printf("%d",count);
	for(int i=2*max_len;i>=0;i--){
		if(c[i]!=0)printf(" %d %.1f",i,c[i]);
	}
	return 0;
}

后记

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