PAT 甲级【1009 Product of Polynomials】
发布时间 2023-10-23 20:40:13作者: fishcanfly
/*
- 系数为0不输出
- 貌似runtime异常也显示答案不正确
*/
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
public class Main {
@SuppressWarnings("unchecked")
public static void main(String[] args) throws IOException {
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
in.nextToken();
int n = (int) in.nval;
int[] N = new int[n];
double[] a = new double[n];
for (int i = 0; i < n; i++) {
in.nextToken();
N[i] = (int) in.nval;
in.nextToken();
a[i] = in.nval;
}
in.nextToken();
int m = (int) in.nval;
int[] M = new int[m];
double[] b = new double[m];
for (int i = 0; i < m; i++) {
in.nextToken();
M[i] = (int) in.nval;
in.nextToken();
b[i] = in.nval;
}
double[] map = new double[N[0] + M[0]+1];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
map[N[i] + M[j]] += a[i] * b[j];
}
}
int k = 0;
for (int i = N[0] + M[0]; i >= N[n - 1] + M[m - 1]; i--) {
if (Math.abs(map[i]) > 0.0001) {
k++;
}
}
System.out.print(k);
for (int i = N[0] + M[0]; i >= N[n - 1] + M[m - 1]; i--) {
if (Math.abs(map[i]) > 0.0001) {
System.out.print(String.format(" %d %.1f", i, map[i]));
}
}
System.out.println();
}
}