PAT (Advanced Level)1002解

因为最近在刷题库,想想就把本人可以想到的解法写到博客里,作为整理归纳。未必是最优解,还请各位高手多多包涵,能够指点指点。

题目要求
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​​ <⋯<N​2​​ <N​1​​ ≤1000.
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

解题思路1
采用数组解决
注意事项1

代码部分1

#include
#include
main(){
  float A[1001]={0.0};
  int num,x;
  double y;
//第一个多项式读入 
  scanf("%d",&num); 
  for(int i =0;i=0;i--){
    if(A[i]!=0.0)
    num++;
  } //统计总数
  printf("%d",num );
  for(int i =1000;i>=0;i--){
    if(A[i]!=0){
     printf(" %d",i);
    printf(" %.1lf",A[i]);   	
	}
  } //打印数据  
}

运行结果1
PAT (Advanced Level)1002解_第1张图片
解题思路2
采用map解决
注意事项2
代码2

#include
#include
#include
using namespace std;
main(){
	map mp;
  int num,x;
  double y;
//第一个多项式读入 
  scanf("%d",&num); 
  for(int i =0;i::iterator it=mp.begin();
  for(;it!=mp.end();it++){
  	if(it->second==0.0&&it->second==-0.0) num--;
}
  printf("%d",num );
   it=(mp.end());
  it--;
  for(;it!=mp.begin();it--){
  	if(it->second!=0.0&&it->second!=-0.0)
  	printf(" %d %.1lf",it->first,it->second);
}
	 it=mp.begin();
	 	if(it->second!=0.0&&it->second!=-0.0)
  	printf(" %d %.1lf",it->first,it->second);	
   //打印数据  
}

运行结果2
PAT (Advanced Level)1002解_第2张图片
从运行结果上来看,map的运行速度确实是比数组更慢,原则上应该是优先使用数组,map使用容易超时。

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