PAT 1009 Product of Polynomials-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_43130569/article/details/112615437

本文介绍如何使用C++实现两个多项式的乘法，处理大整数指数，保证输出精度，并提供样例输入输出。重点在于算法设计和结果格式的规范输出。

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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

#include <iostream>
#include <iomanip>
using namespace std;
const int maxn = 2001;
double ans[maxn] = {0};//存放结果 因为相乘后指数可能最大为2000，所以ans数组最大要开到2001
//当一个数组搞不定的时候 不要去死磕 就直接再开一个数组或者结构体数组就行
struct Poly{
    double c;//系数
    int e;//指数
}pol[1001]; //第一个多项式

int main(){
    int k1, k2, e;
    double c;
    int count = 0;   
    
    cin >> k1;
    for(int j = 0; j < k1; j++){ //记录第一个多项式的指数 与 系数
         cin >> pol[j].e >> pol[j].c;
    }
    
    cin >> k2;
    for(int i = 0; i < k2; i++){
        cin >> e >> c; //输入第二个多项式的 指数 和 系数
        for(int j = 0; j < k1; j++){ //第二个多项式的 指数和系数 直接与 第一个多项式相乘 累加到 对应指数的系数上去
            ans[e + pol[j].e] += (c * pol[j].c);
        }
    }
    
    for(int i=0; i<=2000; i++){
        if(ans[i] != 0) count++; //累计非零项数的项数
    }
    cout << count;
    for(int i=2000; i >= 0; i--){
        if(ans[i] != 0){
            cout << fixed << setprecision(1) << " " << i << " " << ans[i];
        }
    }
    return 0; 
}