A1009. Product of 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 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

#include<cstdio>
#include<iostream>
using namespace std;
int main(){
    int K, n, count = 0;
    double a, poly1[1001] = {0}, re[2002] = {0};
    scanf("%d", &K);
    for(int i = 0; i < K; i++){
        scanf("%d%lf", &n, &a);
        poly1[n] = a;
    }
    scanf("%d", &K);
    for(int i = 0; i < K; i++){
        scanf("%d%lf", &n, &a);
        for(int j = 0; j < 1001; j++)
            re[n + j] = re[n + j] + poly1[j] * a;
    }
    for(int i = 2001; i >= 0; i--){
        if(re[i] != 0)
            count++;
    }
    printf("%d", count);
    for(int i = 2001; i >= 0; i--){
        if(re[i] != 0)
            printf(" %d %.1lf", i, re[i]);
    }
    cin >> K;
    return 0;
}

总结：

1、指数为1000的多项式乘法，结果最高为2000次幂。

2、为减少时间复杂度，可以将第一个多项式存储，第二个不存。第二个多项式边读入边直接遍历poly1并做乘法。还可以将两个多项式的系数与指数分别开数组存下来以减小复杂度。