【PAT A1002】 1002 A+B for Polynomials (25)（25 分）

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

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一定要审好题，第一遍做的时候看错了，以为前面的是指数，后面的是系数。所以做成了这样：

/*用stl

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正确的答案是这样的：
想用hash，因为这种一一对应的问题应用hash可以避免很多查询（循环）的问题。
但是简单的用hash存储 项--->系数 的映射会出现最后无法排序输出的问题。
最后折中一下，开一个struct存储弄好的结构体，直接用sort排序结构体即可。
（什么鬼是系数从大到小的排序。。。怨念）

#include 
#include 
#include 
using namespace std;
#define maxn 1010
float h[maxn];
struct poly{
   int a;
   float b;
}po[maxn];
bool cmp(poly x,poly y){
   return x.a>y.a;
}
int main() {
   int m,n,a;
   float b;
   int sum=0;
   scanf("%d",&m);
   while(m--){
   scanf("%d%f",&a,&b);
       h[a]=b;
   }
   scanf("%d",&n);
   while(n--){
       scanf("%d%f",&a,&b);
       if(h[a]!=0){
           h[a]+=b;
       }
       else{
           h[a]=b;
       }
   }
   for(int i=0;i