[C/C++] 1002 A+B for Polynomials (25 分)

1002 A+B for Polynomials (25 分)

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 N​1 a​_N1 N2 a_N2… NK a_NK
where K is the number of nonzero terms in the polynomial, N​i and a_N​i (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N_​K<⋯2​​ 1 ≤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

---

#include

int main()
{
	const int NUM = 1010;
	double a[NUM] = {0};
	int k;
	int ept;
	double cft;
	int cnt = 0;
	for(int i=0; i<2; i++){
		scanf("%d",&k);
		for(int i=0; i<k; i++){
			scanf("%d%lf",&ept,&cft);
			a[ept] += cft;	
		}
	}
	for( int i=0; i<NUM; i++){
		if( a[i]!= 0) cnt++;
	}
	
	printf("%d",cnt);
	for(int i=NUM-1; i>=0; i--){
		if( a[i]!=0 ){
			printf(" %d %.1f",i,a[i]);
//			cnt--;
//			if( cnt>0 ) printf(" ");
		}
	}
	
	return 0;
} 

注意，如果输出格式依靠这种方法：

printf("%d ",cnt);
	for(int i=NUM-1; i>=0; i--){
		if( a[i]!=0 ){
			printf("%d %.1f",i,a[i]);
			cnt--;
			if( cnt>0 ) printf(" ");
		}
	}

会导致测试点6的格式错误，测试点6是非0项的个数为0时的特殊情况，只能输出0而无空格，应多加注意