【PAT】1002. A+B for Polynomials (25)

题目描述：

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

翻译：这一次，你应当去完成两个多项式A和B的相加了。

INPUT FORMAT

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.

翻译：每一个输入文件包含一组测试数据。每一组测试数据包含两行，并且每一行包括一个多项式的信息：K N1 aN1 N2 aN2 … NK aNK，K是这个多项式非0的部分，Ni 和 aNi (i=1, 2, …, K)是各自的指数和系数。 1 <= K <= 10，0 <= NK < … < N2 < N1 <=1000。

OUTPUT FORMAT

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.

翻译：对于每组测试数据你需要在一行内输出A和B的和，并且和输入格式一样。注意输出末尾不能有空格，请（系数）精确到1位小数。

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

解题思路：

注意：如果两个多项式其中一项相加为0是不能输出的，其他就没什么了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#define INF 99999999
using namespace std;
int K,Max=0,ccount=0;
double a[1010];
int N[25];
int main(){
for(int i=0;i<2;i++){
scanf("%d",&K);
int n;
double an;
for(int j=0;j<K;j++){
scanf("%d%lf",&n,&an);
a
+=an;
Max=max(Max,n);
}
}
for(int i=Max;i>=0;i--){
if(a[i]){
N[ccount++]=i;
}
}
printf("%d",ccount);
for(int i=0;i<ccount;i++){
printf(" %d %.1lf",N[i],a[N[i]]);
}
printf("\n");
return 0;
}