1009 Product of Polynomials (25分)（一个字短！）

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本文介绍了一种计算两个多项式乘积的方法，通过使用double数组保存系数及其乘积后的系数，实现了多项式相乘的算法。输入包含两个多项式的系数和指数，输出为乘积多项式的系数和指数。

1009 Product of Polynomials (25分)

题目

This time, you are supposed to find A×BA\times BA×B where AAA and BBB 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 a_{N_K} where K is the number of nonzero terms in the polynomial, N_i and a_{N_i} (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤N_k<⋯<N₂<N₁≤10000.
Output Specification:
For each test case you should output the product of AAA and BBB 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

题意

求两个多项式的乘积

分析

用double数组保存系数以及相乘之后的系数

代码

#include<iostream>
#define maxn 1001
using namespace std;
int k1,k2;
double co1[maxn],co2[maxn],co[2*maxn];
int t1,t2,t;
int main(){
 	fill(co1,co1+maxn,0.0);
 	fill(co2,co2+maxn,0.0);
 	fill(co,co+2*maxn,0.0);
 	t1=t2=t=-1;
 	int e;
 	scanf("%d",&k1);
 	for(int j=0;j<k1;j++){
  		scanf("%d",&e);
  		if(t1<e)     
  		t1=e;
  		scanf("%lf",&co1[e]);
 	}
 	scanf("%d",&k2);
 	for(int j=0;j<k2;j++){
  		scanf("%d",&e);
  		if(t2<e)     
   			t2=e;
  		scanf("%lf",&co2[e]);
 	}
 	t=t1+t2;
 	for(int i=0;i<=t;i++){
  		for(int j=0;j<=i;j++){
   			if(j<=t1&&i-j<=t2)
   			co[i]+=co1[j]*co2[i-j];
  		}
 	}
 	int cnt=0;
 	for(int i=0;i<=t;i++){
  		if(co[i]!=0){
  			cnt++;
   			t1=i;
  		}
 	}
 	printf("%d",cnt);
 	for(int i=t1;i>=0;i--){
  		if(co[i]!=0)
  		printf(" %d %.1lf",i,co[i]);
 	}
 	return 0;
}