PAT 甲级 1002. A+B for Polynomials

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 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

多项式相加，注意一下精度问题

#include<iostream>
#include<cstdio>
using namespace std;
double num[1010]={0},num1[1010]={0};
const double eps=1e-8;
int main(){
int n;
cin>>n;
while(n--){
int i;
scanf("%d",&i);
scanf("%lf",num+i);
}
cin>>n;
while(n--){
int i;
scanf("%d",&i);
scanf("%lf",num1+i);
}
int sum=0;
for(int i=1000;i>=0;i--){
if(!(num1[i]>=-eps&&num1[i]<=eps)) num[i]+=num1[i];
if(!(num[i]>=-eps&&num[i]<=eps)) sum++;
}
printf("%d",sum);
for(int i=1000;i>=0;i--){
if(!(num[i]>=-eps&&num[i]<=eps)) printf(" %d %.1f",i,num[i]);
}
}