PAT 1002. A+B for Polynomials (25)

This time, you are supposed to find A+B where A and B are two polynomials.

Input

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

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

题目分析：

A+B运算的变形，类似于MergeSort算法。

代码：

#include<stdafx.h>

#include <iostream>
using namespace std;

int main()
{
	int k1, k2;
	scanf("%d", &k1);

	int N1[1005];
	float NK1[1005];
	int i;
	for (i = 0;i<k1;i++)
	{
		scanf("%d %f", &N1[i], &NK1[i]);
	}

	scanf("%d", &k2);

	int N2[1005];
	float NK2[1005];

	for (i = 0;i<k2;i++)
	{
		scanf("%d %f", &N2[i], &NK2[i]);
	}


	int j;

	int k3, N3[2005];
	float NK3[2005];
	int k = 0;

	i = 0;
	j = 0;

	while (i < k1 && j < k2)
	{
		if (N1[i] < N2[j])
		{
			N3[k] = N2[j];
			NK3[k] = NK2[j];
			k++;
			j++;
		}
		else if (N1[i] == N2[j])
		{
			float tmp = NK1[i] + NK2[j];
			if (tmp != 0)
			{
				N3[k] = N1[i];
				NK3[k] = tmp;
				k++;
			}
			i++;
			j++;
		}
		else
		{
			N3[k] = N1[i];
			NK3[k] = NK1[i];
			k++;
			i++;
		}
	}

	if (i == k1)
	{
		while(j < k2)
		{
			N3[k] = N2[j];
			NK3[k] = NK2[j];
			k++;
			j++;
		}
	}

	if (j == k2)
	{
		while(i < k1)
		{
			N3[k] = N1[i];
			NK3[k] = NK1[i];
			k++;
			i++;
		}

	}

	cout << k;
	for (i = 0;i<k;i++)
	{
		printf(" %d %.1f", N3[i], NK3[i]);
		//cout << " " << N3[i] << " " << NK3[i]; 
	}
	return 0;
}