PATA 1009 Product of Polynomials (25分)

题目描述：This time, you are supposed to find A×B where A and B are two polynomials.

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输入格式：

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…

输出格式：

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.

输入样例：

2 1 2.4 0 3.2
2 2 1.5 1 0.5

输出样例：

3 3 3.6 2 6.0 1 1.6

题意：多项式对应的项相乘
思路：记录每个项再对应相加，简单题。
注意输入的&和输出对应的数据类型%lf！
优化：可以只用1个数组，可参考柳神的代码
注意判断double类型用0.0 虽然与0的效果一样，但0.0更准确

代码：

#include <iostream>
using namespace std;
int main(){
	int a,b;
	double pola[1001]={0},polb[1001]={0},pol[2001]={0};
	scanf("%d",&a);
	for(int i=0;i<a;i++){
		int num;
		double k;
		scanf("%d %lf",&num,&k);
		pola[num] = k; 
	}
	scanf("%d",&b);
	for(int i=0;i<b;i++){
		int num;
		double k;
		scanf("%d %lf",&num,&k); 
		polb[num] = k; 
	}
	for(int i=1000;i>=0;i--){
		for(int j=1000;j>=0;j--){
			pol[i+j] += pola[i]*polb[j];
		}
	}
	
	int n=0;//记录不为0 
	for(int i=0;i<2001;i++){		
		if(pol[i]!=0){
			n++;
		}
	}
	printf("%d",n);
	for(int i=2000;i>=0;i--){
		if(pol[i]!=0.0){
			printf(" %d %.1lf",i,pol[i]);
		}
	}	
	return 0;
} 
} 

GOGOGOGO！
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标签：lf,12,15,int,double,scanf,pol,num,2020
来源： https://blog.csdn.net/weixin_43743670/article/details/111195880