1.直接建立1001长度的数组,分别进行存储,最后累加输出。
This time, you are supposed to find A+B where A and B are two polynomials.
Input
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
For each test case you should output the sum 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 to 1 decimal place.
Sample Input
2 1 2.4 0 3.2 2 2 1.5 1 0.5
Sample Output
3 2 1.5 1 2.9 0 3.2
AC代码:
//#include<string> //#include<stack> //#include<unordered_set> //#include <sstream> //#include "func.h" //#include <list> #include <iomanip> #include<unordered_map> #include<set> #include<queue> #include<map> #include<vector> #include <algorithm> #include<stdio.h> #include<iostream> #include<string> #include<memory.h> #include<limits.h> #include<stack> using namespace std; int main(void) { vector<float> N1(1002, 0); vector<float> N2(1002, 0); int n, m; cin >> n; //输入第一个多项式 for (int i = 0; i < n; i++) { int idx = 0; cin >> idx; cin >> N1[idx]; } cin >> m; //输入第二个多项式 for (int i = 0; i < m; i++) { int idx = 0; cin >> idx; cin >> N2[idx]; } //同阶的系数相加 for (int i = 0; i < N1.size(); i++) { N1[i] = N1[i] + N2[i]; } int sum = 0; for (int i = 0; i < N1.size(); i++) { if (N1[i] != 0) sum++; } cout << sum; for (int i = N1.size() - 1; i >= 0; i--) {//高位先输出 if (N1[i] != 0)//不为0,则输出 printf(" %d %.1f", i, N1[i]); } return 0; }