标签：作者 highlight ati tmp back 倒序输出 line ext ice
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
思路
简单题，求两个多项式的乘积。
1.用一个vector poly存放系数乘积的结果，它的下标就是x的幂次方。两个多项式相乘后的最大次幂不会超过2000（因为Ni <= 1000）。
2.用一个countx统计不同次幂的个数，如果是第一次相加不为0，那countx加1。如果系数相加为0.那么countx就得减1，
3.输出countx，然后倒序输出poly中输出系数不为0的x幂次方就行。
代码
#include<iostream>
#include<vector>
#include<iomanip>
using namespace std;
vector<double> poly(2001,0);
class x
{
public:
int exp;
double co;
};
int main()
{
vector<x> fst,scd;
int k1;
cin >> k1;
for(int i = 0;i < k1;i++)
{
x tmp;
cin >> tmp.exp >> tmp.co;
fst.push_back(tmp);
}
int k2;
cin >> k2;
for(int i = 0;i < k2;i++)
{
x tmp;
cin >> tmp.exp >> tmp.co;
scd.push_back(tmp);
}
int countx = 0;
for(int i = 0;i < fst.size();i++)
{
for(int j = 0;j < scd.size();j++)
{
int index = fst[i].exp + scd[j].exp;
double res = fst[i].co * scd[j].co;
if(poly[index] == 0)
countx++;
poly[index] += res;
if(poly[index] == 0)
countx--;
}
}
cout << countx;
for(int i = poly.size() - 1;i >= 0;i--)
{
if(poly[i] != 0)
cout << " " << i << " " << fixed << setprecision(1) << poly[i];
}
cout << endl;
}
PAT1009:Product of Polynomials
标签：作者 highlight ati tmp back 倒序输出 line ext ice
原文地址：http://www.cnblogs.com/0kk470/p/7746441.html
