1002 A+B for Polynomials （25 分)

2019-03-04

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 N1 a**N1 N2 a**N2 … N**K aNK

where K is the number of nonzero terms in the polynomial, N**i and aNi (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N**K<⋯<N2<N1≤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:

1 2	2 1 2.4 0 3.2 2 2 1.5 1 0.5

Sample Output:

1	3 2 1.5 1 2.9 0 3.2

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

给出多项式（指数，系数），求多项式的和

分析

利用结构体po存储一项，vector<po>存储一个多项式，相加后排序输出即可。系数为0的项不输出。

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class po{
public:
    int n;
    double an;

    po(double n, double an) : n(n), an(an) {}
};
int cmp(po a,po b){
    if(a.n>b.n)
        return  true;
    return false;
}
int main() {
    int k1,k2,n;
    double an;
    cin>>k1;
    vector<po> p;
    for(int i=0;i<k1;i++){
        cin>>n>>an;
        po newp(n,an);
        p.push_back(newp);
    }

    cin>>k2;
    for(int i=0;i<k2;i++){
        int f=0;
        cin>>n>>an;
        for(auto it=p.begin();it!=p.end();it++){
            if(it->n==n) {
                it->an += an;
                f=1;
                break;
            }
        }
        if(!f){
            po newp(n,an);
            p.push_back(newp);
        }
    }
    sort(p.begin(),p.end(),cmp);
    int l=0;
    for(auto po:p)
        if(po.an!=0)
            l++;
    cout<<l;
    for(po val:p){
        if(val.an!=0)

        printf(" %d %.1f",val.n,val.an);
    }
    return 0;
}