PAT A 1002. A+B for Polynomials (25)
原题：
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 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
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
求多项式和，无难度。
和的系数如果在 (-0.05,0.05) 则相应项应舍弃。
代码：
#include <iostream>
using namespace std;
int main()
{
int n1,n2,n,i,j;
int e1[11],e2[11],e[20]; //n1指数，n2指数，和指数
double c1[10],c2[10],c[20]; //n1系数，n2系数，和系数
cin>>n1; //输入
for(i=0;i<n1;i++)
{
cin>>e1[i];
cin>>c1[i];
}
e1[i]=-1;
cin>>n2;
for(i=0;i<n2;i++)
{
cin>>e2[i];
cin>>c2[i];
}
e2[i]=-1;
n=0;
i=0;
j=0;
while(i<n1||j<n2) //求和
{
if(e1[i]>e2[j]) //确切说，前两个判断也需要增加数的范围，因为没有限定输入符合0.1的要求
{
e[n]=e1[i];
c[n++]=c1[i++];
}
else if(e1[i]<e2[j])
{
e[n]=e2[j];
c[n++]=c2[j++];
}
else if(c1[i]+c2[j]>=0.05||c1[i]+c2[j]<=-0.05) //注意一位精度，导致和可能不到0.1，则不显示
{
e[n]=e1[i];
c[n++]=c1[i++]+c2[j++];
}
else //注意一位精度，导致和可能不到0.1，则不显示
{
i++;
j++;
}
}
cout<<fixed; //输出
cout.precision(1);
cout<<n;
for(i=0;i<n;i++)
cout<<" "<<e[i]<<" "<<c[i];
return 0;
}