PAT.Product of polynomials(map)
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1009 Product of 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?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.
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
#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(" "); return 0; }
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1009 Product of Polynomials (25)(25 分)