PAT 1002. A+B for Polynomials (25) c++版

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 aN1 N2 aN2 ... NK 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 <= 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

题目意思：分两行给出两个多项式，输入的格式为：多项式的项数，指数，系数，指数，系数...求出这两个多项式相加后的项数，指数，系数

思路：用一个数组来存放多项式，数组下标为指数，数组中存放系数，将数组初始化为0，在接收输入时，直接将系数加到对应位置的数上.

注意：	1、输出的格式中，系数取到小数点后一位

2、如果多项数相加后，所有项相加的和都为0，应输出一个0

#include <bits/stdc++.h>
using namespace std;

double temp[1005] = {0};

int main() {
int k;
for (int j = 0; j < 2; j++) {	//循环2次，读取两行的数据
scanf("%d", &k);			//读取该行的项数
for (int i = 0; i < k; i++) {
int num;
double t;
scanf("%d %lf", &num, &t);		//读取每一项的系数和指数
temp[num] += t;				//将指数作为数组的下标，对应的数组中存放系数的和
}
}
int count = 0;
//遍历查找多项式的项数
for (int i = 0; i < 1001; i++)
if (temp[i] != 0) count++;
printf("%d", count);
//按照指数从大到小的顺序输出指数和系数
for (int i = 1000; i >= 0; i--)
if (temp[i] != 0) printf(" %d %.1f", i, temp[i]);
return 0;
}