PAT 1002 A+B for Polynomials（25 分）

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最新推荐文章于 2024-03-12 00:56:10 发布

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本文介绍了一种解决多项式加法问题的算法实现方法。输入为两个多项式的非零项系数及对应指数，输出为这两个多项式的和。文章提供了一个C语言程序示例，能够处理最多10项的多项式，并能准确输出结果到一位小数。

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1002 A+B for Polynomials（25 分）

This time, you are supposed to find A+B where A and B are two polynomials.
Input Specification:
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:
K $N_1\ a_{N_1}\ N_2\ a_{N_2} ... N_K\ a_{N_K}$
where K is the number of nonzero terms in the polynomial, $N_i$ and $a_{N_i}$ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that $1 \leq K \leq 10$ , $0≤N_K<⋯<N_2<N_1 ≤1000.$
Output Specification:
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

解析

这题有两个要注意的点：
①题目有：with the same format as the input
你的系数是浮点，且输出时只能有一位小数
② 两个指数相同时，它们系数的和可能是0.要注意这一点

Code:

#include <stdio.h>
double polynomial[1001];
int main()
{
    int k, exp;
    double coe;
    scanf("%d", &k);
    while ( k-- ){
        scanf("%d %lf", &exp, &coe);
        polynomial[exp] = coe;
    }
    scanf("%d", &k);
    while ( k-- ){
        scanf("%d %lf", &exp, &coe);
        polynomial[exp] += coe;
    }
    int count = 0;
    for( int i=0; i<1001; i++)
        if ( polynomial[i])
            ++count;
    printf("%d", count);
    for( int i=1001; i>=0; i--)
        if ( polynomial[i])
            printf(" %d %.1f", i, polynomial[i]);
}