PAT 1002. A+B for Polynomials (25)

最新推荐文章于 2020-04-28 20:02:50 发布

ccDLlyy

最新推荐文章于 2020-04-28 20:02:50 发布

阅读量684

点赞数 1

CC 4.0 BY-SA版权

分类专栏： PTA数据结构题+大二习题 PAT甲级+PAT相关天梯赛训练

本文链接：https://blog.youkuaiyun.com/ccDLlyy/article/details/52717202

PAT甲级+PAT相关天梯赛训练同时被 2 个专栏收录

39 篇文章

订阅专栏

PTA数据结构题+大二习题

24 篇文章

订阅专栏

本文介绍了一种解决多项式相加问题的算法实现方法，输入包含两个多项式的系数及指数，输出为这两个多项式的和。文章通过示例详细解释了如何通过C语言程序来实现这一过程，并确保结果的准确性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 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

//浮点型和浮点型的比较肯定不可以通过==和!=比较，整型和浮点型的比较要根据情况而定

//对于判断p[i]为不为零，两个方法均过了

//开始：if(fabs(p[i])>0.0000000001)

//后来:if(p[i]!=0)

#include <stdio.h>
#include <stdlib.h>
int main()
{
    double p[1001];//存原始数据
    int count=0;//结果有多少项
    for(int i=0;i<=1000;i++)//初始化
        p[i]=0;
    for(int i=1;i<=2;i++){
        int k;
        int exponent;
        double coefficient;
        scanf("%d",&k);
        for(int j=1;j<=k;j++){
            scanf("%d%lf",&exponent,&coefficient);
            p[exponent]+=coefficient;
        }
    }
    for(int i=0;i<=1000;i++)
        if(fabs(p[i])>0.0000000001)
          count++;
    printf("%d",count);
    for(int t=1000;t>=0;t--){
       if(fabs(p[t])>0.0000000001)
          printf(" %d %.1f",t,p[t]);
    }
    return 0;
}