PTA1002 A+B for Polynomials

问题描述：

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:

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

题目分析：

一开始读题没有读懂，看了别人的分析才看出来是指数排列。

输入时按照将相同底数的指数相加，遍历统计非0组的个数，再次遍历按格式输出即可，注意特殊的0。

代码如下：

#include
   
    using namespace std;const int N=1e3+1;double a[N];int main(){   	int n,x;	double y;	cin>>n;	for (int i = 0; i < n; i++)	{   		cin>>x>>y;		a[x]+=y;	}	cin>>n;	for (int i = 0; i < n; i++)	{   		cin>>x>>y;		a[x]+=y;	}	int len=0;	for (int i = 0; i <= 1000; i++)	{   		if (a[i]!=0.0)		{   			len++;		}	}	if(len)	{   		int i;		cout<
    
     <<' ';		for ( i = 1000; i >= 0; i--)		{   				if (a[i]!=0.0)			{   				printf("%d %.1f",i,a[i]);				len--;				if(len) cout<<' ';			}		}	}	else	{   		cout<<"0";	}	return 0;}

总结

这道题主要的难点还是读题，看懂题后很好做，通过率低的原因还是对于英文题面的理解。

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