PAT-A 1002. A+B for Polynomials (25)-优快云博客

本文链接：https://blog.youkuaiyun.com/geekdoe/article/details/79495930

本文介绍了一种解决多项式加法问题的算法实现方法，输入两个多项式，输出它们相加的结果。程序使用C语言编写，通过扫描多项式的系数和指数来计算和，并输出相同形式的结果。

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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

#include <cstdio>  
#include <cstdlib>
#define Maxn 1001  
      
double result[Maxn];  
      
int main(void){  
    int K,exp,i;  
    double coef;  
    int count = 0;  
    scanf("%d",&K);  
    for(i=0;i<K;++i){  
        scanf("%d %lf",&exp,&coef);  
        result[exp] += coef;  
    }  
    scanf("%d",&K);  
    for(i=0;i<K;++i){  
        scanf("%d %lf",&exp,&coef);  
        result[exp] += coef;  
    }  
    for(i=0;i<Maxn;++i){  
        if(result[i]!=0)  
            count++;  
    }  
    printf("%d",count);  
    for(i=Maxn;i>=0;--i){  
        if(result[i] != 0){  
            printf(" %d %.1f",i,result[i]);  
        }  
    }  
    system("pause");  
    return 0;  
}