PAT - 甲级 - 1002. A+B for Polynomials (25)

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 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 <= 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

题目大意：每组输入有两行，每行第一个数字表示多项式的个数，接着先输入一项的指数然后输入对应的系数，

要求输出两行的多项式相加的结果，以同样的格式输出，指数大小一次递减；

思路：指数的范围是0~1000，可以定义一个数组，数组小标表示指数，数组元素表示系数;两者之和系数相加，

例题是这样的   2.4*x^1 +3.2*x^0       和       1.5*x^2+0.5*x^1     相加就是   1.5*x^2 + 1.2*x^1 + 3.2*x^0

第一次提交wa    问题在于，如果多项式相加后为0 那么只输出0然后换行，而不是输出0 0 0;

#include
#include
using namespace std;
int main(){
	double ans[1001];
	int n,k;
	double N;
	for(int i=0 ;i<1001 ;i++){
		ans[i]=0;
	} 
	
	cin>>n;
	
	while(n--){
		cin>>k>>N;
		ans[k] += N;
	}
	cin>>n;
	while(n--){
		cin>>k>>N;
		ans[k] += N;
	}
	int flag = 1;
	int count = 0;
	for(int i=1000 ;i>=0 ;i--){
		if(ans[i]!=0){
			count ++;
		}
	}
	printf("%d",count);//注意这里  如果输出0  这样的话后面就没空格 
	
	for(int i=1000 ;i>=0 ;i--){
		if(ans[i]!=0){
			printf(" %d %.1f",i,ans[i]);
		}
	}
	
	printf("\n");
	
	return 0;
}