1009. Product of Polynomials (25)

题目链接：http://www.patest.cn/contests/pat-a-practise/1009

题目：

  
  
  
  

   
   
   
   
 
    

      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
  
  
  
  

分析：

多项式相乘，没有什么难点的哦

AC代码：

#include<iostream>
using namespace std;
double x1[1001];//存放第一个多项式
double x2[1001];//存放第一个多项式
double ans[2002];//存放答案
int main(void){
 int i,j,k;
 while(scanf("%d",&k) != EOF){
  for(i = 0; i < 1001; i ++){
   x1[i] = 0;
   x2[i] = 0;
  }
  for(i = 0; i < 2002; i ++){
   ans[i] = 0;
  }        //init初始化
  for(i = 0; i < k; i ++){
   int tmp;
   scanf("%d",&tmp);
   scanf("%lf",&x1[tmp]);
  }
  scanf("%d", &k);
  for(i = 0; i < k; i ++){
   int tmp;
   scanf("%d",&tmp);
   scanf("%lf",&x2[tmp]);
  }
  for(i = 0; i < 1001; i ++){
   for(j = 0; j < 1001; j ++){
    ans[i + j] += x1[i] * x2[j];
   }
  }   //相乘
  int count = 0;
  for(i = 2002; i >= 0; i --){
   if(ans[i] != 0){
    count ++;
   }
  }
  printf("%d",count);
  for(i = 2002; i >= 0; i --){
   if(ans[i] != 0){
    printf(" %d %.1f",i,ans[i]);
   }
  }
 }
 return 0;
}

——Apie陈小旭