标签：pac ann stream specific where pie pat col using
This time, you are supposed to find A×B where A and B are two polynomials.
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???? (,) are the exponents and coefficients, respectively. It is given that 1, 0.
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.
2 1 2.4 0 3.2
2 2 1.5 1 0.5
3 3 3.6 2 6.0 1 1.6
#include <iostream> #include <map> using namespace std; map <int, double> mp; const int maxn = 1000 + 5; int a1[maxn], a2[maxn]; double b1[maxn], b2[maxn]; int main() { int k1, k2; cin >> k1; for(int i = 0; i < k1; i ++) { cin >> a1[i] >> b1[i]; } cin >> k2; for(int i = 0; i < k2; i ++) { cin >> a2[i] >> b2[i]; } int a3; double b3; for(int i = 0; i < k1; i ++) { for(int j = 0; j < k2; j ++) { a3 = a1[i] + a2[j]; b3 = b1[i] * b2[j]; if(b3 != 0.0) mp[a3] += b3; } } map<int, double> :: reverse_iterator i; // 反向迭代器 for(i = mp.rbegin(); i!= mp.rend(); i ++) { if(i -> second == 0) mp.erase(i -> first); //删除值为零的元素，通过key删除 } printf("%d", mp.size()); for(i = mp.rbegin(); i != mp.rend(); i ++) { printf(" %d %.1f", i -> first, i -> second); } printf("\n"); return 0; }
PAT.Product of polynomials(map)
标签：pac ann stream specific where pie pat col using
原文地址：https://www.cnblogs.com/bianjunting/p/12492552.html
