PAT 1009 Product of Polynomials
PAT 1009 Product of Polynomials
1009. Product of Polynomials (25)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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 0.5Sample Output
3 3 3.6 2 6.0 1 1.6
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <string>
#include <vector>
#include <strstream>
#include <map>
using namespace std;
struct Node
{
int x;
double y;
}a[15],b[15];
double tag[2005];
int n1,n2;
int main()
{
scanf("%d",&n1);
for(int i=1;i<=n1;i++)
scanf("%d%lf",&a[i].x,&a[i].y);
scanf("%d",&n2);
for(int i=1;i<=n2;i++)
scanf("%d%lf",&b[i].x,&b[i].y);
memset(tag,0,sizeof(tag));
for(int i=1;i<=n1;i++)
{
for(int j=1;j<=n2;j++)
{
tag[a[i].x+b[j].x]+=a[i].y*b[j].y;
}
}
int num=0;
for(int i=2000;i>=0;i--)
{
if(tag[i]!=0)
num++;
}
if(num==0)
{
printf("%d\n",num);
return 0;
}
else
printf("%d ",num);
int cnt=0;
for(int i=2000;i>=0;i--)
{
if(tag[i]!=0)
{
cnt++;
if(cnt==num)
printf("%d %.1f\n",i,tag[i]);
else
printf("%d %.1f ",i,tag[i]);
}
}
return 0;
}