PAT 甲级 1002题

题目：1002 A+B for Polynomials

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 aN1 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 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:

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2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

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3 2 1.5 1 2.9 0 3.2

解题思路：

可以设置一个数组，下标为指数，存储的数据是系数，然后通过输入依次将值按照指数（下标）存入数组中，再遍历数组，输出值不为0的项。

注意：

①测试点1：当输出的时候需要将系数保留1位小数；

②测试点3/4/5/6：当A、B对应指数的系数相加正好为0的时候，其输出项数需要仔细的计算（即为当系数为0的时候，需要将该项去掉）；

代码：

cpp 复制代码

#include<iostream>
using namespace std;
#define MAX 1002
int main()
{
	float arr[MAX] = {0.0f};    //对应指数作为数组下标，其存储的内容为该指数对应的系数
	int zhishu = 0;   //指数
	float xishu = 0;    //系数
	int K = 0;   //表示未知数的个数
	int K_Sum = 0;   //表示最终表达式的未知数个数

	//输入
	int j = 2;
	while(j--)   //输入两轮：A和B
	{
		cin>>K;
		K_Sum += K;
		while(K--)
		{
			cin>>zhishu>>xishu;
			for(int i=0;;i++)
			{
				if(i == zhishu)
				{
					if(arr[i] != 0.0f)   //如果A和B有相同的指数
					{
						K_Sum--;
						arr[i] += xishu;
						if(arr[i] == 0.0f)    //系数相加结果为0
						{
							K_Sum--;
						}
						break;
					}
					arr[i] += xishu;
					break;
				}
			}
		}
	}

	//输出
	cout<<K_Sum;
	for(int i=1001;i>=0;i--)
	{
		if(arr[i] != 0.0f)
		{
			cout<<" "<<i<<" ";
			printf("%.1f",arr[i]);
		}
	}
	cout<<endl;
	system("pause");
	return 0;
}