傳送門

https://www.patest.cn/contests/pat-a-practise/1009

題目

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

分析

1.因為只有兩個變量，所以沒用結(jié)構(gòu)體數(shù)組戈咳，用的pair代替变隔，以first為指數(shù)孤荣，second為系數(shù)。
2.遍歷兩個多項式赞警，每次獲得臨時的指數(shù)和系數(shù)逸吵，根據(jù)指數(shù)磷账，將系數(shù)加到對應(yīng)下標的數(shù)組里，每次記錄一下最大指數(shù)蒸辆，便于后面統(tǒng)計多項式的項數(shù)征炼。
3.最后格式化輸出結(jié)果即可。

源代碼

//C/C++實現(xiàn)
#include <iostream>
#include <vector>

using namespace std;

double res[2001]; //result polynomial

int main(){
    int n;
    scanf("%d", &n);
    vector<pair<int, double>> v1(n); //the first polynomial
    for(int i = 0; i < n; ++i){
        scanf("%d %lf", &v1[i].first, &v1[i].second);
    }
    //input the second polynomial
    scanf("%d", &n);
    vector<pair<int, double>> v2(n); //the second polynomial
    for(int i = 0; i < n; ++i){
        scanf("%d %lf", &v2[i].first, &v2[i].second);
    }
    //compute
    int tmp_exp;
    double tmp_coef;
    int max_exp = 0;
    for(int i = 0; i < v1.size(); ++i){
        for(int j = 0; j < v2.size(); ++j){
            tmp_exp = v1[i].first + v2[j].first;
            tmp_coef = v1[i].second * v2[j].second;
            max_exp = (tmp_exp > max_exp) ? tmp_exp : max_exp;
            res[tmp_exp] += tmp_coef;
        }
    }
    //output
    int count = 0;
    for(int i = max_exp; i >= 0; --i){
        if(res[i] != (double)0){
            ++count;
        }
    }
    printf("%d", count);
    for(int i = max_exp; i >= 0; --i){
        if(res[i] != (double)0){
            printf(" %d %.1f", i, res[i]);
        }
    }
}