（甲）pat-1002（多项式加法）

最新推荐文章于 2022-09-09 18:45:23 发布

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最新推荐文章于 2022-09-09 18:45:23 发布

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分类专栏： aa题目题解

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本文介绍了一种解决多项式相加问题的算法实现。通过输入两个多项式的系数和指数，该算法能准确地计算出两个多项式的和，并以相同的格式输出结果。文章提供了完整的C++代码实现，包括多项式的比较、排序以及求和过程。

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1002 A+B for Polynomials (25)（25 分）

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

心得：看清数据。

#include<bits/stdc++.h>
using namespace std;
struct N{
	int exp;
	double cof;
};

bool cmp(N a,N b)
{
	return a.exp>b.exp;
}
int main(void)
{
	int k1,k2,i,j,k;
	while(~scanf("%d",&k1))
	{
		N a[15],b[15],c[35];
		for(i=0;i<k1;i++) scanf(" %d %lf",&a[i].exp,&a[i].cof);
		scanf("%d",&k2);
		for(i=0;i<k2;i++) scanf(" %d %lf",&b[i].exp,&b[i].cof);
		sort(a,a+k1,cmp);
		sort(b,b+k2,cmp); 
		
		i=0,j=0,k=0;
		while(i<k1&&j<k2)
		{
			if(a[i].exp>b[j].exp)
			{
				c[k].exp=a[i].exp;
				c[k].cof=a[i].cof;
				i++;
				k++;
			}
			else if(a[i].exp<b[j].exp)
			{
				c[k].exp=b[j].exp;
				c[k].cof=b[j].cof;
				j++;
				k++;
			}
			else if(a[i].exp==b[j].exp)
			{
				double tp=a[i].cof+b[j].cof;
				if(tp==0)
				{
					i++;
					j++;
					continue;
				}
				else
				{
					c[k].cof=tp;
					c[k].exp=a[i].exp;
					i++;
					j++;
					k++;
				}
			}
		}
		
		while(i<k1)
		{
			c[k].cof=a[i].cof;
			c[k].exp=a[i].exp;
			k++;
			i++;
		}
		while(j<k2)
		{
			c[k].cof=b[j].cof;
			c[k].exp=b[j].exp;
			k++;
			j++;
		}
		
		printf("%d",k);
		for(i=0;i<k;i++)
		{
			printf(" %d %.1f",c[i].exp,c[i].cof);
		}
		printf("\n");
	}
	return 0;
}