PAT甲级-1002

最新推荐文章于 2025-01-22 23:38:46 发布

看身后

最新推荐文章于 2025-01-22 23:38:46 发布

阅读量141

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT甲级

本文链接：https://blog.youkuaiyun.com/weixin_44941429/article/details/90673888

PAT甲级专栏收录该内容

14 篇文章

订阅专栏

博客围绕多项式A+B求和问题展开，给出输入输出要求，输入包含多项式非零项数量、次幂和系数等信息，输出为A与B的和，格式同输入，需精确到小数点后一位。题解是先求多项式个数和，再从大到小输出次幂和系数。

1002. A+B for Polynomials (25)

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

题解：这题就是一元多次函数相加，输入一个k然后输入次幂输入系数，先输出多项式的个数和，然后从大到小输出次幂，系数。

Code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<stack>
int main()
{
    int n;
    float x[1010]={0};
    int a;
    float y;
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        scanf("%d%f",&a,&y);
        x[a]+=y;
    }
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        scanf("%d%f",&a,&y);
        x[a]+=y;
    }
    n=0;
    for(int i=0; i<1010; i++)
    {
        if(x[i]!=0)
        {
            n++;
        }
    }
    printf("%d",n);
    for(int i=1009; i>=0; i--)
    {
        if(x[i]!=0)
        {
            printf(" %d %.1f",i,x[i]);
        }
    }

    return 0;
}