A+B for Polynomials (25)

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本文介绍了一种简单的多项式加法算法实现方法，通过使用数组来存储多项式的各项系数，实现了两个多项式的相加，并给出了完整的C++代码示例。

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A+B for Polynomials (25)

时间限制 1000 ms 内存限制 32768 KB 代码长度限制 100 KB 判断程序 Standard (来自小小)
题目描述
This time, you are supposed to find A+B where A and B are two polynomials.

输入描述:
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.

输出描述:
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.

输入例子:
2 1 2.4 0 3.2
2 2 1.5 1 0.5

输出例子:
3 2 1.5 1 2.9 0 3.2

题解：属于简单模拟，用p[max_n]数组来表示多项式，其中p[n]表示幂次为n的项的系数。

#include<bits/stdc++.h>
#include<stdlib.h>
#include<iostream>
using namespace std;
#define vi vector<int>
#define pii pair<int,int>
#define x first
#define y second
#define all(x) x.begin(),x.end()
#define pb push_back
#define mp make_pair
#define SZ(x) x.size()
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define pi acos(-1)
#define mod 1000000007
#define inf 1000000007
#define ll long long
#define DBG(x) cerr<<(#x)<<"="<<x<<"\n";
#define N 200010
template <class U,class T> void Max(U &x, T y){if(x<y)x=y;}
template <class U,class T> void Min(U &x, T y){if(x>y)x=y;}
template <class T> void add(int &a,T b){a=(a+b)%mod;}
const int maxn=1111;
int main()
{
    double p[maxn]={};
    int k,n,count=0;
    double x;
    cin>>k;
    rep(i,0,k){
      cin>>n>>x;
      p[n]+=x;
    }
    cin>>k;
    rep(i,0,k){
     cin>>n>>x;
      p[n]+=x;
    }
    rep(i,0,maxn)
    {
        if(p[i]!=0) count++;
    }
    cout<<count;
    per(i,0,maxn-1)
    {
        if(p[i]!=0)
            printf(" %d %.1f",i,p[i]);
    }

    return 0;
}