PAT 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 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 3.2
2 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <string>
#include <vector>
#include <strstream>
#include <map>

using namespace std;
double b[1005];
int tag[1005];
int a[45];

int k;
int main()
{ 
  scanf("%d",&k);
  int x;double y;
  int cnt=0;
  memset(tag,0,sizeof(tag));
  for(int i=1;i<=k;i++)
  {
    scanf("%d",&x);
    scanf("%lf",&y);
    if(!tag[x])
    {
            b[x]=y;
      a[cnt++]=x;
      tag[x]=1;
    }
    else
            b[x]+=y;
    
  }
  scanf("%d",&k);
  for(int i=1;i<=k;i++)
  {
    scanf("%d%lf",&x,&y);
    if(!tag[x])
    {
      b[x]=y;
        a[cnt++]=x;
      tag[x]=1;
    }
    else
      b[x]+=y;
  }
  
  sort(a,a+cnt);
  int num=0,num2;
  for(int i=cnt-1;i>=0;i--)
    if(b[a[i]]!=0) {num++;num2=i;}
  if(num==0)
  printf("%d\n",num);
  else
    printf("%d ",num);
  for(int i=cnt-1;i>=0;i--)
  {
    if(b[a[i]]==0)
      continue;
    if(i==num2)
      printf("%d %.1f\n",a[i],b[a[i]]);
    else
        printf("%d %.1f ",a[i],b[a[i]]);
  }

  return 0;
}