PAT(A) 1009. Product of Polynomials (25)

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>

const int MAXN=2005;
double poly[MAXN]={};   //多项式系数全都初始化为0
double poly2[MAXN]={};

int main()
{
    int k, exp, cnt=0, exp2;
    double coef, coef2;
    scanf("%d", &k);
    for(int i=0; i<k; i++){
        scanf("%d%lf", &exp, &coef);
        poly[exp] += coef;
    }
    //输入第二个多项式的相关数据
    scanf("%d", &k);
    for(int i=0; i<k; i++){
        scanf("%d%lf", &exp, &coef);
        for(int i=0; i<MAXN; i++){
            if(poly[i]!=0.0){
                coef2=coef*poly[i];
                exp2=exp+i;
                poly2[exp2] += coef2;
            }
        }
    }
    for(int i=0; i<MAXN; i++){
        if(poly2[i]!=0)
            cnt++;
    }
    printf("%d", cnt);
    for(int i=MAXN-1; i>=0; i--){
        if(poly2[i]!=0)
            printf(" %d %.1lf", i, poly2[i]);
    }
    return 0;
}