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.

intmain(int argc, char *argv[]){ int K1; scanf("%d", &K1); Poly *N1 = new Poly[K1]; for (int i = 0; i < K1; i++) scanf("%d %lf", &N1[i].zhishu, &N1[i].xishu);

int K2; scanf("%d", &K2); double *result = newdouble[2001]();//结果数组，下标为指数，指数最大值为2000，值为系数，带()及调用构造函数初始化为0.0 //memset为什么不行？

int zhishu; double xishu; int count = 0;

for (int i = 0; i < K2; i++) {//一边读一边算 scanf("%d %lf", &zhishu, &xishu); for (int j = 0; j < K1; j++) { if (fabs(result[zhishu + N1[j].zhishu]) <1e-6) count++; result[zhishu + N1[j].zhishu] += xishu * N1[j].xishu; if (fabs(result[zhishu + N1[j].zhishu]) < 1e-6) count--; } }

if (count != 0) { printf("%d", count); for (int i = 2000; i >= 0; i--) if (fabs(result[i]) > 1e-6) printf(" %d %.1lf", i, result[i]); } else printf("0");