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 N1 aN1 N2 aN2 ... NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=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 Input2 1 2.4 0 3.2 2 2 1.5 1 0.5Sample Output
3 3 3.6 2 6.0 1 1.6
有人用数组来存储多项式,用数组下标来表示指数,这对于大量的数据是有优势的,此题使用map来映射指数和系数
#include <iostream> #include <cstdio> #include <map> using namespace std; /* 两个多项式的乘积 * */ struct Poly { int ex; double co; }; Poly A[10]; Poly B[10]; map<int, double> C; int main(void) { int m, n; scanf("%d", &m); for (int i = 0; i < m; ++i) scanf("%d%lf", &A[i].ex, &A[i].co); scanf("%d", &n); for (int i = 0; i < n; ++i) scanf("%d%lf", &B[i].ex, &B[i].co); int exp; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { exp = A[i].ex + B[j].ex; C[exp] += A[i].co * B[j].co; if (C[exp] == 0) C.erase(exp); } } printf("%d", C.size()); for (map<int,double>::iterator iter = C.end() ; iter != C.begin(); ) { --iter; printf(" %d %.1lf", iter -> first, iter -> second); } printf("\n"); return 0;