1009. 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 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

程序代码：

#include<stdio.h> #include<math.h> #define EPSION 0.1 double c[11]={0}; int e[11]={0}; double num[2001]={0}; int main() { int l1,l2; scanf("%d",&l1); int i; double c_0; int e_0; for(i=0;i<l1;i++) { scanf("%d%lf",&e[i],&c[i]); } scanf("%d",&l2); int j; double cTmp=0; int eTmp=0; for(i=0;i<l2;i++) { scanf("%d%lf",&e_0,&c_0); for(j=0;j<l1;j++) { cTmp=c[j]*c_0; eTmp=e[j]+e_0; num[eTmp]+=cTmp; } } int count = 0; for(i=2000;i>=0;i--) { if(fabs(num[i])>=EPSION) count++; } if(count == 0) { printf("0"); return 0; } printf("%d ",count); for(i=2000;i>=0;i--) { if(fabs(num[i])>=EPSION) { printf("%d %.1f",i,num[i]); count--; if(count>0) putchar(' '); } } return 0; }