PAT A1002 A+B for Polynomials (25 分)

一、题目

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 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 <= N_K < … < N₂< N₁ <=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

二、题意

单词	意思
polynomial	多项式
exponent	指数
coefficient	系数
respectively	分别地，各自地
1 decimal place	小数点第一位

给出两行，每行均为一个多项式。K为多项式中非零项的个数，Ni为多项式的指数，a_Ni为多项式的系数。求两个多项式的和，并以输入格式输出结果。非零项项数，以及高次到低次输出非零项的指数和系数

三、思路

定义一个数组，其中p[n]表示指数n的项的系数，初值为0。

四、注意

1、注意输出格式，从高次到低次
2、精确到一位小数
3、count计数非零项项数时，不要选择在读入时计数，存在正负相抵问题

五、代码

#include <cstdio>
using namespace std;
int main() {int m, n, e;float c;float a[1005] = { 0 };scanf_s("%d", &m);for (int i = 1; i <= m; i++) {scanf_s("%d %f", &e, &c);a[e] += c;}scanf_s("%d", &n);for (int i = 1; i <= n; i++) {scanf_s("%d %f", &e, &c);a[e] += c;}int count = 0;for (int i = 0; i <= 1000; i++) {if (a[i] != 0) count++;}printf("%d", count);for (int i = 1000; i >=0; i--) {//输出格式应该从高次到低次if (a[i] != 0.0)//判断非零项printf(" %d %.1f", i, a[i]);//%.1f精确到小数点后一位}return 0;
}