标签:count Product PAT int 0.0 result 1009 exponents coefficients
题目
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 a**N1 N2 a**N2 ... N**K aNK
where K is the number of nonzero terms in the polynomial, N**i and aNi (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤N**K<⋯<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
理解与算法
多项式乘法,思路也不难,遍历两个多项式,相乘即可,M*N
的复杂度,需要考虑的是存储方式。
题目里给出的范围是1-1000
,所以数组大小设置为2001就足够了,但是,如果指数范围非常大,大到几万,几十万,就不太好用数组来实现,可以考虑用map来实现。
代码与实现
//
// Created by tanknee on 2021/1/24.
//
#include <iostream>
#include <map>
using namespace std;
int main() {
int count, exponents;
double coefficients, m1[1001] = {0.0}, result[2001] = {0.0};
cin >> count;
for (int i = 0; i < count; ++i) {
cin >> exponents >> coefficients;
m1[exponents] = coefficients;
}
cin >> count;
for (int i = 0; i < count; ++i) {
cin >> exponents >> coefficients;
for (int j = 0; j < 1001; ++j) {
if (m1[j] != 0.0)
result[j + exponents] += m1[j] * coefficients;
}
}
count = 0;
for (int i = 2000; i >= 0; i--) {
if (result[i] != 0.0)
count++;
}
printf("%d", count);
for (int i = 2000; i >= 0; i--) {
if (result[i] != 0.0)
printf(" %d %.1f", i, result[i]);
}
return 0;
}
标签:count,Product,PAT,int,0.0,result,1009,exponents,coefficients 来源: https://www.cnblogs.com/tanknee/p/14320813.html
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