Notes PAT

PAT-1002

Posted by MichaelChen on 2021-01-24
Estimated Reading Time 2 Minutes
Words 393 In Total
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A+B for Polynomials

Input Specification:

This time, you are supposed to find A+B where A and B are two polynomials.

Output Specification:

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 … 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≤N**K<⋯<N2<N1≤1000.

Sample Input:

1
2
2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:

1
3 2 1.5 1 2.9 0 3.2

Analysis

其K为多项式中非零项的个数，共输入两行。于是分别定义指数为正的数组polynomial，以及指数为负数的数组min。分别输入通过数组下标表示指数，相同则在输入的时候进行相加，不相同则添加到对应下标下。注意输出格式o(￣▽￣)o

Code

C

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#include <stdio.h>
#include "stdlib.h"

int main() {

    double polynomial[1005] = {0};
    double min[1005] = {0};
    int k,a,count = 0;
    double n = 0.0;

    scanf("%d",&k);
    for (int i = 0; i < k; ++i) {
        scanf("%d",&a);
        scanf("%lf",&n);
        if (a < 0){
            min[abs(a)] += n;
        } else{
            polynomial[a] += n;
        }
    }

    scanf("%d",&k);
    for (int i = 0; i < k; ++i) {
        scanf("%d",&a);
        scanf("%lf",&n);
        if (a < 0){
            min[abs(a)] += n;
        } else{
            polynomial[a] += n;
        }
    }

    for (int j = 0; j < 1005; ++j) {
        if (polynomial[j] != 0){
            count ++;
        }
        if (min[j] != 0){
            count ++;
        }
    }

    printf("%d",count);
    for (int l = 1004; l >= 0; --l) {
        if (polynomial[l] != 0){
            printf(" %d %.1lf",l,polynomial[l]);
        }
    }
    for (int m = 0; m <= 1004; ++m) {
        if (min[m] != 0){
            printf(" -%d %.1lf",m,min[m]);
        }
    }
    printf("\n");
    return 0;
}

Result

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