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解题思路：PAT1002. 考察点是两个多项式的加法计算，本题考察的是两个多项式的乘法计算。
最高项是1000，所有两个多项式相乘得到的最高项是<=2000的。只需要输出非0项，即系数为0，不需要输出读入第2个多项式的时候，边读入边计算 代码如下： #include<stdio.h> // 定义结构体ploy数组保存第一组多项式的输入 struct Poly { int exp; double cof; }poly[1010]; // 保存运算的结果 double ans[2010]; int main() { int n, m, count = 0; // 读入第一个多项式 scanf("%d", &n); for (int i = 0; i < n; i++) scanf("%d %lf", &poly[i].exp, &poly[i].cof); // 读入第二个多项式 scanf("%d", &m); for (int i = 0; i < m; i++) { int exp; double cof; // 边读入边计算，每读入项都需要跟第一个多项式中每一项相乘 scanf("%d %lf", &exp, &cof); for (int j = 0; j < n; j++) ans[exp + poly[j].exp] += (cof*poly[j].cof); } // 统计非0项的个数 for (int i = 0; i <= 2000; i++) { if (ans[i] != 0.0) count++; } // 从高到低打印结果 printf("%d", count); for (int i = 2000; i >= 0; i--) { if (ans[i] != 0.0) printf(" %d %.1f", i, ans[i]); } return 0; }