1009 Product of Polynomials (25 point(s))
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 0.5
Sample Output:
3 3 3.6 2 6.0 1 1.6
多项式 f = ax^n + bx^n-1 +...+ cx^0
1.注意点,不能想当然的把系数当成正的来看,所以还是要小心
exponents
指数
coefficients
系数
Polynomials
多项式
#include<iostream>
#include<vector>
#include<stdio.h>
using namespace std;
int main(){
const int MAXN = 2100;
float n1[1100];
float n2[1100];
float out[MAXN];
fill(n1,n1+MAXN,0.0);
fill(n2,n2+MAXN,0.0);
fill(out,out+MAXN,0.0);
int N,M;
cin>>N;
int max_n1 =0;
for(int i=0;i<N;i++){
int temp;
float flo;
cin>>temp>>flo;
n1[temp]+=flo;
if(temp>max_n1){
max_n1 = temp;
}
}
cin>>M;
int max_n2=0;
for(int i=0;i<M;i++){
int temp;
float flo;
cin>>temp>>flo;
n2[temp]+=flo;
if(temp>max_n2){
max_n2 = temp;
}
}
for(int i=0;i<=max_n1+1;i++){
for(int j=0;j<=max_n2+1;j++){
float temp = n1[i]*n2[j];
int index = i+j;
out[index] += temp;
// cout<<index <<" "<<out[index]<<endl;
}
}
int MAX_index=0;
int number=0;
for(int i=0;i<MAXN;i++){
if(out[i]!=0.0){
number++;
MAX_index = i;
}
}
cout<<number;
for(int i=MAX_index;i>=0;i--){
if(out[i]!=0.0){
printf(" %d %.1f",i,out[i]);
}
}
return 0;
}