PAT (Advanced Level) 1002. A+B for Polynomials (25) 解题报告_pat advanced level 1002. a+b for polynomials (25)(-优快云博客

本文介绍了一个简单的算法，用于计算两个多项式的和，并给出了相应的C++实现代码。输入为两个多项式，输出为相加后的多项式，保留一位小数。

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

分析：输入两行多项式，输出相加后的式子

代码：

#include <cstdio>
#include <iostream>
#include <map>
using namespace std;

int main()
{
    int n, k, sum = 0;
    double s;
    map<int, double> M;
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
    {
        scanf("%d %lf", &k, &s);
        M[k] += s;
    }
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
    {
        scanf("%d %lf", &k, &s);
        M[k] += s;
    }
    for(int i = 1000; i >= 0; i--)
    {
        if(M[i] == 0) continue;
        sum++;
    }
    printf("%d", sum);
    for(int i = 1000; i >= 0; i--)
    {
        if(M[i] == 0) continue;
        printf(" %d %.1lf", i, M[i]);
    }
    printf("\n");
    return 0;
}