pat 甲级 1009. Product of Polynomials (25)

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

思路：多项式相乘，模拟即可，要注意的是最终结果中系数为0的项不需要输出。
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<set>
#include<queue>
using namespace std;
#define INF 0x3f3f3f
#define N_MAX 30+5
#define M_MAX 2001
struct x {
int exp;
double coef = 0;
bool vis = 0;
};
x poly1[N_MAX],poly2[N_MAX];
x poly[M_MAX];
int n1, n2;
int main() {
cin >> n1;
for (int i = 0; i < n1; i++)cin >> poly1[i].exp >> poly1[i].coef;
cin >> n2;
for (int i = 0; i < n2; i++)cin >> poly2[i].exp >> poly2[i].coef;
for (int i = 0; i < n1;i++) {
for (int j = 0; j < n2;j++) {
int exp = poly1[i].exp + poly2[j].exp;
poly[exp].vis = 1;
poly[exp].exp= exp;
poly[exp].coef+= poly1[i].coef*poly2[j].coef;
}
}

int num = 0;
//系数为0的项不用输出！！！！！！！！！
for (int i = M_MAX-1; i >= 0; i--) if (poly[i].vis&&poly[i].coef!=0) num++;
cout << num << " ";
for (int i = M_MAX-1; i >=0;i--) {
if (poly[i].vis&&poly[i].coef != 0) {
num--;
printf("%d %.1f%c",poly[i].exp,poly[i].coef,num==0?'\n':' ');
}
}
return 0;
}