【PAT甲级】1002 A+B for Polynomials (25分)

1002 A+B for Polynomials (25分)

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 N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ ... N​K​​ a​N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N​K​​<⋯

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

三个个坑

一个是 系数为0 不能算进去

一个是 输出格式又要去保留一位小数点Please be accurate to 1 decimal place.

一个是 不能==0 判断 double （其实可以 哈哈）但是 我用 abs（num-0）<0.005 不对 还是精度太小 if( abs(num[i]) >= 1e-15 ) 

所以要判断一个单精度浮点数：则是if( abs(f) <= 1e-6)；
要判断一个双精度浮点数：则是if( abs(f) <= 1e-15 )；

 

自己的答案

#include
#include
using namespace std;
double num[10005];
int main(){
	int n;
	sets;
	cin>>n;
	for(int i=0;i>a>>b;
		num[a]+=b;
	}
	cin>>n;
	for(int i=0;i>a>>b;
		num[a]+=b;
	}
    int count=0;
    for(int i=0;i<10005;i++){
        if(num[i]!=0){
            count++;
        }
    }
	cout<=0;i--){
		if(num[i]!=0){
            printf(" %d %0.1f", i, num[i]);
			//cout<<" "<

网搜答案

#include
const int max_n = 1111;
double a[max_n] = {};
int main(){
  
  int k, n, count = 0;
  double e;
  scanf("%d", &k);
  for(int i = 0; i < k; i ++){
    scanf("%d%lf", &n, &e);
    a[n] = e;
  }
  
  scanf("%d", &k);//这里k只是为了存储，所以这儿k是可以重复存放值。
  for(int i = 0; i < k; i ++){
    scanf("%d%lf", &n, &e);
    a[n] += e;//这儿其实就是有相同的就相加，没有相同的就不想加
  }
  
  for(int i = 0; i < max_n; i ++){
    if(a[i] != 0){
      count ++;
    }
  }
  printf("%d", count);
  for(int i = max_n - 1; i >=0; i --){
    if(a[i] != 0) printf(" %d %.1f", i, a[i]);
  }
  return 0;
}