【PAT】【Advanced Level】1009. Product of Polynomials (25)

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

原题链接：

https://www.patest.cn/contests/pat-a-practise/1009

https://www.nowcoder.com/questionTerminal/9c06e801f8f54ebcb170c09117265132

思路：

由于精度问题，此题卡掉了map的做法，在牛客网上可过，在pat上不能通过所有测试点

就用朴素的的二维数组即可

仍然要注意系数为0的情况

code：

#include<iostream>#include<cstring>#include<map>#include<algorithm>#include<vector>#include<cstdio>#include<cmath>using namespace std;typedef struct S{  int z;  double x;};bool fl[2010];int  re[42];double jg[2010];int main(){    memset(fl,0,sizeof(fl));    memset(jg,0,sizeof(jg));    int n1,n2;    cin>>n1;    S a[n1];    for (int i=0;i<n1;i++)        cin>>a[i].z>>a[i].x;    cin>>n2;    S b[n2];    for (int i=0;i<n2;i++)    {        cin>>b[i].z>>b[i].x;        for (int j=0;j<n1;j++)        {            int s1=0;            double s2=0;            s1=b[i].z+a[j].z;            s2=b[i].x*a[j].x;            //cout<<s2<<endl;            jg[s1]+=s2;        }    }    int num=0;    for(int i=0;i<2010;i++)        if (jg[i]!=0) num++;    cout<<num;    for (int i=2010;i>=0;i--)        if (jg[i]!=0)            printf(" %d %.1lf",i,jg[i]);    printf("\n");    return 0;}