1.题目要求两个多项式相乘的结果。
2.直接创建1000*1000+1的数组,然后遍历所有相乘的组合情况。
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 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
AC代码:
//#include<string> //#include <iomanip> #include<vector> #include <algorithm> //#include<stack> #include<set> #include<queue> #include<map> //#include<unordered_set> //#include<unordered_map> //#include <sstream> //#include "func.h" //#include <list> #include<stdio.h> #include<iostream> #include<string> #include<memory.h> using namespace std; int main(void) { int n, m; cin >> n; vector<double> NK1(1001, 0); vector<double> NK2(1001, 0); vector<double> NK(1000001, 0);//直接创建1000*1000+1的数组 //输入第一个多项式 for (int i = 0; i < n; i++) { int temp; cin >> temp; cin >> NK1[temp]; } cin >> m; //输入第二个多项式 for (int i = 0; i < m; i++) { int temp; cin >> temp; cin >> NK2[temp]; } //进行相乘 for (int i = 0; i < 1001; i++) { for (int j = 0; j < 1001; j++) { if (NK1[i] != 0 && NK2[j] != 0) { NK[i + j] += NK1[i] * NK2[j]; } } } int sum = 0; for (int i = 0; i < 1000001; i++) {//统计有多少项非零 if (NK[i] != 0) sum++; } cout << sum; if (sum != 0) cout << " "; for (int i = 1000000; i >= 0; i--) { if (NK[i] != 0) {//输出非零的多项式 printf("%d %.1lf", i, NK[i]); sum--; if (sum != 0) cout << " "; } } return 0; }