【测试点0分析】1009 Product of Polynomials (25 分)

立志用最少的代码做最高效的表达

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PAT甲级最优题解——>传送门

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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

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测试点0：多项式相乘，可能出现系数为0的情况，因此，最后输出的项数可能计算错误。

解法：数组硬解， 其实换成结构体存储效率会高一点，但考虑到最大数据量只有1k，因此也无所谓了。

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#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;
} 

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