1009 Product of Polynomials (25 分)
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 aNKN_1\ a_{N_1}\ N_2\ a_{N2}\ ...\ N_K\ a_{N_K}N1 aN1 N2 aN2 ... NK aNK
where K is the number of nonzero terms in the polynomial, NiN_iNi and aNia_{N_i}aNi
(i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤101≤K≤101≤K≤10, 0≤NK<⋯<N2<N1≤10000≤N_K<⋯<N_2<N_1≤10000≤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
解析
这题要注意的是:多项式相乘后:有的项的系数可能为0.注意筛选。
多项式相乘:用双重循环相乘后,生成的多项式:可能指数不是按从大到小排序的。
比如说:
polynomial1 x3+x1x^3+x^1x3+x1
polynomial2 x100+x1x^{100}+x^{1}x100+x1
polynomial x103+x4+x101+x2x^{103}+x^{4}+x^{101}+x^{2}x103+x4+x101+x2
无论用什么数据结构:都要注意这上面的两点哟。
Code:
#include <stdio.h>
double polynomial[2020];
double temp[1010];
int main()
{
int k, exp;
double coe;
scanf("%d", &k);
while ( k-- ){
scanf("%d %lf", &exp, &coe);
temp[exp] = coe;
}
scanf("%d", &k);
while ( k-- ){
scanf("%d %lf", &exp, &coe);
for(int i=0; i<1010; i++)
if(temp[i] != 0.0)
polynomial[i+exp] += coe*temp[i];
}
int count = 0;
for( int i=0; i<2020; i++)
if ( polynomial[i] != 0.0)
++count;
printf("%d", count);
for( int i=2019; i>=0; i--)
if ( polynomial[i] != 0.0)
printf(" %d %.1f", i, polynomial[i]);
}