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1009. Product of Polynomials (25)
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 Input2 1 2.4 0 3.2 2 2 1.5 1 0.5Sample Output
3 3 3.6 2 6.0 1 1.6
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;
}
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