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 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 3.22 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;}