1009 Product of Polynomials (25)（25 point(s)）

problem

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 a~N1~ N2 a~N2~ ... NK a~NK~, where K is the number of nonzero terms in the polynomial, Ni and a~Ni~ (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

tip

求多项式的乘积。

answer

#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define Max 1000005

int N, M;
float num[Max];
pair<int ,float> n1[Max], n2[Max];

int main(){
//  freopen("test.txt", "r", stdin);
    memset(num, 0, sizeof(num));
    memset(n1, 0, sizeof(n1));
    memset(n2, 0, sizeof(n2));

    cin>>N;
    for(int i = 0; i < N; i++) {
        cin>>n1[i].first>>n1[i].second;
    }
    cin>>M;
    for(int i = 0; i < M; i++) {
        cin>>n2[i].first>>n2[i].second;
    }
    int number = 0;
    for(int i = 0; i < N; i++){
        for(int j = 0; j < M; j++){
            int ex = n1[i].first + n2[j].first;
            float co = n1[i].second * n2[j].second;
            num[ex] += co;
        }
    }
    for(int i = 0; i < Max; i++) if(num[i] != 0) number ++;
    cout<<number;
    for(int i = Max -1; i >= 0; i--){
        if(num[i] != 0){
            cout<<" "<<i<<" ";
            cout<<fixed<<setprecision(1)<<num[i];
        }
    }

    return 0;
}

experience

• 注意数组越界问题。
• 读清楚题意，多设计一组测试用例。

原文地址：https://www.cnblogs.com/yoyo-sincerely/p/9270837.html