PAT1002

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本文提供PAT1002多项式相加问题的详细解答及代码实现，强调输出精度控制，适合初学者参考。

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PAT1002是一个多项式相加的问题，其实很简单，但是多次提交总是有问题，后来发现原来是没有好好看题目，题目的最后有要求1decimal place，修改后就正确了，也正好忘记怎么使用cout来输出特定精度的双精度数，就在博客上写一下，不说别的了，上题目(原题：http://pat.zju.edu.cn/contests/pat-a-practise/1002)：

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

下面是鄙人的解法：

#include "iostream"
#include<string.h>
//#include<cstdio>
#include<iomanip>


using namespace std;










int main()
{
	int KA;
	int KB;
	int APolyExponent[10] = {0};
	double APolyCoeffcient[10] = {0};
	int BPolyExponent[10] = {0};
	double BPolyCoeffcient[10] = {0};
	int CPolyExponent[20] = {0};    // result;
	double CPolyCoeffcient[20] = {0};
	int KC = 0;
	//while(1)
	{
		int i;
		int Ap = 0;
		int Bp = 0;
		int Cp = 0;
		KC = 0;
		cin>>KA;
		for(i = 0;i < KA;i++)
		{
			cin>>APolyExponent[i];
			cin>>APolyCoeffcient[i];
		}
		cin>>KB;
		for(i = 0;i < KB;i++)
		{
			cin>>BPolyExponent[i];
			cin>>BPolyCoeffcient[i];
		}


		Ap = 0;
		Bp = 0;
		Cp = 0;
		while((Ap < KA) && (Bp <KB))
		{
			if(APolyExponent[Ap] == BPolyExponent[Bp])
			{
				CPolyExponent[Cp] = APolyExponent[Ap];
				CPolyCoeffcient[Cp] = APolyCoeffcient[Ap] + BPolyCoeffcient[Bp];
				Ap++;
				Bp++;
			}
			else if(APolyExponent[Ap] > BPolyExponent[Bp])
			{
				CPolyExponent[Cp] = APolyExponent[Ap];
				CPolyCoeffcient[Cp] = APolyCoeffcient[Ap];
				Ap++;
			}
			else
			{
				CPolyExponent[Cp] = BPolyExponent[Bp];
				CPolyCoeffcient[Cp] = BPolyCoeffcient[Bp];
				Bp++;
			}
			if(CPolyCoeffcient[Cp] != 0)
				Cp++;


		}
		while(Ap < KA)
		{
			CPolyExponent[Cp] = APolyExponent[Ap];
			CPolyCoeffcient[Cp] = APolyCoeffcient[Ap];
			Ap++;
			if(CPolyCoeffcient[Cp] != 0)
				Cp++;
		}
		while(Bp < KB)
		{
			CPolyExponent[Cp] = BPolyExponent[Bp];
			CPolyCoeffcient[Cp] = BPolyCoeffcient[Bp];
			Bp++;
			if(CPolyCoeffcient[Cp] != 0)
				Cp++;
		}
		cout << Cp ;
		for(i = 0;i < Cp;i++)
		{
			cout<<fixed<<setprecision(1);   // to cout double variable 
			cout<< " " << CPolyExponent[i]<< " "<<CPolyCoeffcient[i];
		}
		cout<<endl;
	}


	return 0;
}