PAT-A 1009. Product of Polynomials
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;
}