PAT (Advanced Level) Practise 1002 解题报告

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• 问题描述
• 解题思路
• 代码
• 提交记录

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问题描述

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

大意是：

多项式加法

共两行输入数据，每行输入数据包括：K N1 aN1 N2 aN2 ... NK aNK，其中K是在多项式非零项的数量，Ni和aNi分别是指数和系数。

1<=K<=10，0<=NK<... < N2 < N1<=1000.

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解题思路

1. 数组下标作为次数，数值存储系数和；
2. 边读入边计算最终非零项个数。

注意点：

1. 系数和为0的项不计算在内；
2. 输出结果不能有多余空格。

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代码

错误版本

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
  int i,j,n,m,s,t;
  float a[1001],k;
  memset(a,0,sizeof(a));
  cin>>n;
  for (i=0;i<n;i++)
  {
    cin>>j>>k;
    a[j]=k;
  }
  t=n;
  cin>>n;
  for (i=0;i<n;i++)
  {
    cin>>j>>k;
    if (a[j]==0) t++; 
    a[j]+=k;
    if (a[j]==0) t--;
  }
  cout<<t<<' ';
  n=t; 
  t=0;
  for (i=1000;i>=0;i--)
  {
    if (a[i]!=0)
    {
        if (t<n)
        {
          printf("%d %.1f ",i,a[i]);
          t++;
        }
        else printf("%d %.1f",i,a[i]);
    }
  }
  return 0;
} 

评测结果为全部格式错误。经过检查，防止最后多出空格的做法错了 不应该偷懒。。t的初始值应该设为1，或改变if的判断条件。

正确版本（用了更简便的输出）

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
  int i,j,n,m,s,t;
  float a[1001],k;
  memset(a,0,sizeof(a));
  cin>>n;
  for (i=0;i<n;i++)
  {
    cin>>j>>k;
    a[j]=k;
  }
  t=n;
  cin>>n;
  for (i=0;i<n;i++)
  {
    cin>>j>>k;
    if (a[j]==0) t++; 
    a[j]+=k;
    if (a[j]==0) t--;
  }
  cout<<t;
  for (i=1000;i>=0;i--)
  if (a[i]!=0) printf(" %d %.1f",i,a[i]);
  return 0;
} 

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提交记录

错误版本

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