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 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (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 3.22 2 1.5 1 0.5

Sample 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;}