1002 A+B for Polynomials

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博客围绕多项式A+B求和问题展开，介绍了输入规格，每个测试用例占2行，每行含多项式信息；输出规格要求按输入格式输出和，且行末无多余空格、精确到1位小数。还提醒对于系数为0的项要忽略。

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 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 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:

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

这个题很简单，但是要理解。对于系数为0的项，直接忽略，不输出，不计数。

#include <iostream>
#include <cstdio>
#include <string.h>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
// sn[10000][100];
struct stu {
    string name;
    int gao;
};

int cmp(int a,int b) {

    return a>b;

}
int main() {
    float N[1001]= {(float)0};

    int number=0;
    int num;
    cin>>num;

    for(int i=0; i<num; i++) {
        int ni;
        cin>>ni;
        
        float sum;
        cin>>sum;
        N[ni]+=sum;
    }
    cin>>num;

    for(int i=0; i<num; i++) {
        int ni;
        cin>>ni;
        
        float sum;
        cin>>sum;
        N[ni]+=sum;
    }
     number=0;
     for(int i=1000; i>=0; i--)
        if(N[i]!=(float)0)
        number++;
    cout<<number;
    for(int i=1000; i>=0; i--) {
        if(N[i]!=(float)0) {
                printf(" %d %.1f",i,N[i]);
            //cout<<' '<<i<<' '<<N[i];
        }
    }
    return 0;
}