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

/* 输入：两行数据表示两个多项式，K表示K个系数非零的项，N1表示指数，An1表示系数，以此类推 输出：A和B两个多项式的乘积，输出格式与输入同，保留一位小数。注意：数组要开的够大。 */#include<iostream>#include<string.h>#include<stdio.h>#include<iomanip>using namespace std;int main(){double a[1050],b[1050],sum[2050];int k;int max1=0,max2=0,nonzero=0;memset(a,0,sizeof(a));memset(b,0,sizeof(b));memset(sum,0,sizeof(sum));cin>>k;while(k--){int n;double an;cin>>n>>an;a[n]=an;if(n>max1)max1=n;} cin>>k;while(k--){int n;double an;cin>>n>>an;b[n]=an;if(n>max2)max2=n;} for(int i=0;i<=max1;i++){for(int j=0;j<=max2;j++){sum[i+j]+=(a[i]*b[j]);}}for(int i=0;i<=(max1+max2);i++){if(sum[i]!=0.0)nonzero++;}cout<<nonzero;for(int i=(max1+max2);i>=0;i--){if(sum[i]!=0.0){cout<<" "<<i;              cout<<fixed<<setprecision(1);              cout<<" "<<sum[i];  }}cout<<endl;return 0;}