A1002 A+B for Polynomials (25 point(s))

多项式相加

1. 原文

原题

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 $N_1$ $a_{N1}$ $N_2$ $a_{N2}$ … $N_K$ $a_{NK}$

where K is the number of nonzero terms in the polynomial, $N_i$ and $a_{Ni}$ (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 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.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample output:

3 2 1.5 1 2.9 0 3.2

2. 解析

给定两行，每行表示一个多项式：第一个数表示该多项式中系数非零项的项数，后面每两个数表示一个项，分别表示项的幂次和系数。求两个多项式的和，并按指数从大到小输出。

Solution:

定义一个double数组num

指数相同时，系数相加 –> 使num[指数]+=系数

遍历数组，不为0的元素值即为多项式个数

从后往前遍历数组，输出不为0的元素

3. AC代码

#include<cstdio>
const int maxn=1010;
double num[maxn]={};
int main()
{
	int k;
	scanf("%d",&k);
	int a;
	double b;
	for (int i = 0; i < k; ++i)
	{
		scanf("%d%lf",&a,&b);
		num[a]+=b;
	}
	scanf("%d",&k);
	for (int i = 0; i < k; ++i)
	{
		scanf("%d%lf",&a,&b);
		num[a]+=b;
	}
	int cnt=0;
	for (int i = 0; i < maxn; ++i)
	{
		if (num[i]!=0)
		{
			cnt++;
		}
	}
	printf("%d",cnt);
	for (int i = maxn-1; i >= 0; i--)
	{
		if (num[i]!=0)
		{
			printf(" %d %.1f",i,num[i]);
		}
	}
	printf("\n");
	return 0;
}