PAT甲级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

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提交代码

思路：从多项式的乘法定义出发，就是用写循环

注意事项：边界，还有输出格式

#include 
struct Poly{
    int exp;//指数
    double cof;//系数
}poly[1001];

double ans [2001];//0-2000的幂都有
int main(){
    int n, m, number = 0;
    scanf("%d", &n);//第一个多项式中非零项的个数
    for(int i = 0; i < n; i++){
        scanf("%d %lf", &poly[i].exp, &poly[i].cof);//指数和系数
    }
    scanf("%d", &m);//第二个多项式中非零项的个数
    for(int i = 0; i < m; i++){
        int exp;
        double cof;
        scanf("%d %lf", &exp, &cof);//指数和系数
        for(int j = 0; j < n; j++){
            ans[exp + poly[j].exp] += (cof * poly[j].cof);
        }
        }
        for(int i = 0; i <= 2000; i++){
            if(ans[i] != 0.0) number++;
            }
        printf("%d", number);
        for(int i = 2000; i >= 0; i--){
            if(ans[i] != 0){
                printf(" %d %.1f", i, ans[i]);
            }
        }

    return 0;
}