PAT甲级 1002. A+B for Polynomials (25)

最新推荐文章于 2024-12-13 08:36:25 发布

Android_chunhui

最新推荐文章于 2024-12-13 08:36:25 发布

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文章标签： PAT甲级

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本文介绍了一种多项式求和的算法实现，该算法通过读取两个多项式的非零项并将其相加来计算多项式的和。输入包含两个多项式的系数和指数，输出则为求和后的多项式表达式。

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 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

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

注意：当求和后非零项数为0时后面不输出空格

#include <iostream>
#include <stdio.h>
#include <math.h>

using namespace std;

int main()
{
    /*
    *多项式求和，一般设置数组，指数项作为下标，系数项作为值 
    */
    float A[1001]={0};
    float B[1001]={0};
    int K;
    cin>>K;
    float c;
    int e;
    for(int j=0;j<K;j++)
    {
        cin>>e>>c;
        A[e]=c;
    }
    cin>>K;
    for(int j=0;j<K;j++)
    {
        cin>>e>>c;
        B[e]=c;
    }
    int nonzero_cnt=0;
    for(int i=0;i<1001;i++)
    {
        A[i]=A[i]+B[i];
        if(A[i])
            nonzero_cnt++;
    }
    printf("%d%s",nonzero_cnt,(nonzero_cnt?" ":""));
    int cnt=0;
    for(int i=1000;i>=0;i--)
    {
        if(A[i])
        {
            cnt++;
            printf("%d %.1f%s",i,A[i],(cnt<nonzero_cnt?" ":""));
        }
    }

    return 0;
}