1002. A+B for Polynomials (25)

1002. A+B for Polynomials (25)

时间限制

400 ms

内存限制

65536 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

写对这道题目真不容易，花了好几天换了好几个思路，附上代码和注意事项 

#include
#include
using namespace std;
#define MAX 1001
int main()
{
	double a[MAX];
	for(int i=0;i>k1;
	for(int i=0;i>m>>n;
		a[m]+=n;//把每一项的值放入对应的第几项中去 
	}
	cin>>k2;
	for(int i=0;i>m>>n;
		a[m]+=n;
	}
	int num=0;
	for(int i=0;i0;i--)//从大到小输出 
	{
		if(a[i]!=0)
		{
			cout<<" "<


首先，我做这道题，第一个思路就是弄一个结构体，比较方便的存储这里的项数和系数，但由于数字一直存不进去作罢。其实这个问题在我第二个思路里面也出现了，最后发现是我的项数和系数的类型都设成了int型，导致有小数点的项数根本存不进去，这是我后来发现的。
我的第二个思路是看了别人的博客才学会的，就是放进数组里面去，第几项放到数组里面的第几项里面去，然后系数就是它的值；
2.这里有很多格式要注意，比如，精确到1位小数，要用<是最后输出，注意格式


总之做了很久才做出来全对，开心