PAT甲级——A+B for Polynomials

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​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N​K​​<⋯<N​2​​<N​1​​≤1000.

Output Specification:

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

思路1：

     题目要求的是多项式的和，并输出所有系数非零的多项式数目，以及按照指数从大到小的顺序输出多项式每一项的指数和系数。首先可以考虑建立一个map映射，将多项式的指数和系数对应起来。输入第一组数据的时候可以直接建立映射，并且建立一个布尔型的hashTable[]，用来记录已经出现过的指数，再输入第二组数据的时候，如果输入的指数之前已经存在了，那么就需要将指数对应的系数在原来的基础上加上后面输入的系数，如果输入的指数之前没有出现过那么就建立新的指数和系数的映射。输出数据的时候由于map会按照键的值从小到达输出，那么我们需要利用反向迭代器map<int,float>::reverse_iterator it来从后往前输出，起点为mp.rbegin()(包含），终点为mp.rend()(不包含）。

代码：

#include <iostream> #include<map> using namespace std; map<int, float>mp; bool hashTable[1010] = { false }; int main() { int n1, n2, N; float a; scanf("%d", &n1); for (int i = 0; i < n1; i++) { scanf("%d %f", &N, &a); mp[N] = a; hashTable[N] = true; } scanf("%d", &n2); for (int i = 0; i < n2; i++) { scanf("%d %f", &N, &a); if (hashTable[N] == true) mp[N] += a; else mp[N] = a; } map<int, float>::iterator ip; map<int, float>::reverse_iterator it; int len = mp.size(); int count = 0; for (ip = mp.begin(); ip != mp.end(); ip++) { if ((int)ip->second != 0) count++; } printf("%d", count); for (it = mp.rbegin(); it != mp.rend(); it++) { if ((int)it->second != 0) printf(" %d %.1f", it->first, it->second); } return 0; }

思路2：

        设立a数组，⻓度为指数的最大值，a[i] = j表示指数i的系数为j，接收a和b输入的同时将对应指数 的系数加入到a中，累计a中所有非零系数的个数，然后从后往前输出所有系数不为0的指数和系数。

代码：  

#include <iostream> using namespace std; int main() { float a[1010] = { 0 }; int m, n, t; float num; scanf_s("%d", &n); for (int i = 0; i < n; i++) { scanf_s("%d%f", &t, &num); a[t] += num; } scanf_s("%d", &m); for (int i = 0; i < m; i++) { scanf_s("%d%f", &t, &num); a[t] += num; } int count = 0; for (int i = 0; i < 1010; i++) { if ((int)a[i] != 0) count++; } printf_s("%d", count); for (int i = 1009; i >= 0; i--) { if ((int)a[i] != 0) printf_s(" %d %.1f", i, a[i]); } return 0; }