【PAT 甲級】1009 Product of Polynomials (25)

1009 Product of Polynomials (25)

 

題目描述

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

 

輸入描述:

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.

 

輸出描述:

For each test case you should output the product 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 up to 1 decimal place.

 

輸入例子:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

 

輸出例子:

3 3 3.6 2 6.0 1 1.6

我的思路

多項式相乘，用下標代替指數，數組內存的數據爲係數。

 

注意事項

一開始我用的float，小數點進位非常奇怪的報錯，改成double後就通過了。

 

我的代碼 

#include <cstdio>
int main()
{
	int K1, K2, t, max1=0, index, max2=0, c=0;
	double temp;
	double a[1010]={0.0}, b[2020]={0.0};
	
	scanf("%d", &K1);
	for (int i=1; i<=K1; i++)
	{
		scanf("%d %lf", &t, &temp);
		a[t] = temp;
		max1 = max1>t?max1:t;
	}
	
	scanf("%d", &K2);
	for (int j=1; j<=K2; j++)
	{
		scanf("%d %lf", &t, &temp);
		for (int i=max1; i>=0; i--)
		{
			if (a[i]!=0.0)
			{
				index = i+t;  //相乘後新的次數 
				max2 = max2>index?max2:index;
				b[index] += a[i]*temp;  //相乘後的係數 
			}
		}
	}
	
	//統計個數
	for (int i=max2; i>=0; i--)
	{
		if (b[i]!=0.0)
		{
			c++;
		}
	}
	
	//輸出 
	printf("%d", c); 
	for (int i=max2; i>=0; i--)
	{
		if (b[i]!=0.0)
		{
			printf(" %d %.1lf", i, b[i]);
		}
	}	
	
	return 0;
}

 

《算法筆記》上面的，用了結構體的正確代碼：

#include <cstdio>

struct Poly
{
	int exp;
	double cof;
}poly[1001];

double ans[2001]; 

int main()
{
	int n, m, number = 0;
	scanf("%d", &n);
	for (int i=0; i<n; i++)
	{
		scanf("%d %lf", &poly[i].exp, &poly[i].cof);
	}
	scanf("%d", &m);
	for (int i=0; i<m; i++)
	{
		int exp;
		double cof;
		scanf("%d %lf", &exp, &cof);
		for (int j=0; j<n; j++)
		{
			ans[exp + poly[j].exp] += (cof * poly[j].cof);
		}
	}
	
	for (int i=0; i<=2000; i++)
	{
		if (ans[i]!=0.0) number++;
	}
	
	printf("%d", number);
	for (int i=2000; i>=0; i--)
	{
		if (ans[i]!=0.0)
		{
			printf(" %d %.1lf", i, ans[i]);
		}
	}
	return 0;
}