【PAT - 甲级1009】Product of Polynomials (25分)（模拟，细节）

题干：

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

题目大意：

按照输入的格式给定两个多项式A和B，让你按照相同的格式输出两个多项式的乘积，也就是A×B。

解题报告：

就是个模拟题，但是注意细节不少。

1.首先观察输入可以发现当系数为0的时候这一项是需要省略的。

2.其次最后输出的第一个数不是最高次项，而是后面需要跟着的数的个数。

3.次数需要从高往低输出。

4.注意进行多项式乘积的时候，跟输入的两个K是没有关系的，而是跟两组N1...Nk是有关系的，又因为这里保证了N是由大到小的，也就是只和两个N1是有关系的。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
double A[MAX],B[MAX],C[MAX],b;
int n,m,mx,a;
int main()
{cin>>n;for(int i = 1; i<=n; i++) {scanf("%d%lf",&a,&b);A[a] = b;if(i == 1) mx = a;}cin>>m;for(int i = 1; i<=m; i++) {scanf("%d%lf",&a,&b);B[a] = b;if(i == 1) mx += a;}for(int i = 0; i<=mx; i++) {for(int j = 0; j<=mx; j++) {C[i+j] += A[i]*B[j];}}int ans = 0;for(int i = 0; i<=mx; i++) {if(C[i] != 0) ans++;}printf("%d",ans);for(int i = mx; i>=0; i--) {if(C[i] == 0) continue;printf(" %d %.1f",i,C[i]);}return 0 ;
}