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PAT (Advanced Level) Practise 1009 Product of Polynomials (25)

前程有光 2022-11-09 阅读 71


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 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

和之前那个加法一样,注意0的问题

#include<cstdio>
#include<stack>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn = 1e5 + 10;
int n, x, cnt;
double a[maxn], b[maxn], c[maxn];

int main()
{
scanf("%d", &n);
while (n--)
{
scanf("%d", &x);
scanf("%lf", &a[x]);
}
scanf("%d", &n);
while (n--)
{
scanf("%d", &x);
scanf("%lf", &b[x]);
}
for (int i = 0; i < 1001; i++)
{
for (int j = 0; j < 1001; j++) c[i + j] += a[i] * b[j];
}
for (int i = 0; i < 2001; i++) if (c[i] != 0) cnt++;
printf("%d", cnt);
for (int i = 2000; i >= 0; i--) if (c[i] != 0) printf(" %d %.1lf", i, c[i]);
return 0;
}



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