技術標籤：PAT甲級PATPAT甲級

立志用最少的程式碼做最高效的表達

PAT甲級最優題解——>傳送門

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​N​1​​​​ N​2​​ a​N​2​​​​ … N​K​​ a​N​K​​​​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​​<⋯<N​2​​<N​1​​≤1000.

Output Specification:
For each test case you should output the product 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 up to 1 decimal place.

Sample Input:
2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output:
3 3 3.6 2 6.0 1 1.6

測試點0：多項式相乘，可能出現係數為0的情況，因此，最後輸出的項數可能計算錯誤。

解法：陣列硬解， 其實換成結構體儲存效率會高一點，但考慮到最大資料量只有1k，因此也無所謂了。

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

gg a[1010] = {0}, b[1010] = {0}, c[2020] = {0};  

int main() {
	gg k,x; cin >> k; while(k--) {
		cin >> x; double d; cin >> d;
 

		a[x] = (gg)(d*100);
	}
	cin >> k; while(k--) {
		cin >> x; double d; cin >> d;
		b[x] = (gg)(d*100);
	}
	gg num = 0;
	for(gg i = 0; i <= 1000; i++) 
		for(gg j = 0; j <= 1000; j++) {
			if(a[i] == 0) continue;
			if(b[j] != 0) c[i+j] += a[i]*b[j]; 
		}
	
	for(int i = 0; i <= 2000; i++) if(c[i] != 0) num++;
	
	printf("%d", num);
	for(int i = 2000; i >= 0; i--) 
		if(c[i] != 0) printf(" %d %.1lf", i, c[i]/10000.0);
	
	
	return 0;
}