编程题目：PAT(Advanced Level) Practice 1002. A+B for Polynomials (25)

1002. A+B for Polynomials (25)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

        看起来是很简单的一个题目，但是对于各种情况的考量可能不到位，起初用的方法始终存在问题，代码如下，在测试点3不能通过，尚未找出原因，若哪位看出问题所在，请加以指点。

/*
http://pat.zju.edu.cn/contests/pat-a-practise/1002 1002. A+B for Polynomials (25)
*/

#include<iostream>
#include<string.h>
#include<iomanip>
#include<vector>
using namespace std;

struct term
{
	int exponent;//指数
	double coefficient;//系数
};

int main()
{
	int K1,K2;
	vector<term> v1,v2,v3;//分别存输入项，结果项
	

	cin >> K1;
	while(K1--)
	{
		term temp; //临时保存一项
		cin>>temp.exponent>>temp.coefficient;
		v1.push_back(temp);
		//K1--;
	}
	cin >> K2;
	while(K2--)
	{
		term temp; //临时保存一项
		cin>>temp.exponent>>temp.coefficient;
		v2.push_back(temp);
		//K2--;
	}
	int total1 = v1.size();
	int total2 = v2.size();
	int p1 = 0,p2 = 0;
	while(p1<v1.size() && p2<v2.size())
	{
		term temp1 = v1[p1];
		term temp2 = v2[p2];
		term temp3;
		if(temp1.exponent == temp2.exponent)//指数项相等
		{
			temp3.exponent = temp1.exponent;
			temp3.coefficient = temp1.coefficient + temp2.coefficient;
			if(temp3.coefficient!=0)
			{
				v3.push_back(temp3);
			}
			p1++;
			p2++;
		}
		else if(temp1.exponent> temp2.exponent)
		{
			if(temp1.coefficient!=0)
			{
				v3.push_back(temp1);
			}
			p1++;
		}
		else
		{
			if(temp2.coefficient!=0)
			{
				v3.push_back(temp2);
			}
			p2++;
		}
	}
	while(p1<v1.size())
	{
		if(v1[p1].coefficient!=0)
		{
			v3.push_back(v1[p1]);
		}
		p1++;
		
	}
	while(p2<v2.size())
	{
		if(v1[p2].coefficient!=0)
		{
			v3.push_back(v2[p2++]);
		}
		p2++;
	}
	//输出
	cout<<v3.size();
	
	for(int i =0;i<v3.size();i++)
	{
		cout<<fixed<<setprecision(1);
		cout<<" "<<v3[i].exponent<<" "<<v3[i].coefficient;
	}

	system("pause");
	return 0;
}

        在几番尝试未果之后，决定换一种方式，由于题目给出的指数范围至多为1000，座椅这里可以采用大数组的形式保存结果，数组下标即对应着指数值，操作起来方便很多，最后注意调整输出格式，设置setprecision(1)，就完美通过，而如果不设置这一句，就会出错，可能测试点上有输入不止一位小数的测试用例吧。另外还有一个要注意的点就是两项相加后系数为0，就不要计入结果中了。一下为参考代码，可通过。

http://pat.zju.edu.cn/contests/pat-a-practise/1002 1002. A+B for Polynomials (25)
*/

#include<iostream>
#include<string.h>
#include<iomanip>
#include<vector>
using namespace std;

int main()
{
	double coe[1001];
	int k;
	int exponent;
	double coefficient;
	
	memset(coe,0,sizeof(coe));
	cin>> k;
	while(k--)
	{
		cin>>exponent>> coefficient;
		coe[exponent] += coefficient; 
	}
	cin>>k;
	while(k--)
	{
		cin>>exponent>> coefficient;
		coe[exponent] += coefficient; 
	}
	int count = 0;
	for(int i = 0;i<1001;i++)
	{
		if(coe[i]!=0)
			count++;
	}
	cout<<count;
	for(int i =1000 ;i>=0;i--)
	{
		if(coe[i]!=0)
		{
			cout<<fixed<<setprecision(1);
			cout<<" "<<i<<" "<<coe[i];
		}
	}
	system("pause");
	return 0;
}