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  • 1009. Product of Polynomials (25)

    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
    
     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<stdlib.h>
     4 
     5 double str1[1001],str2[1001],ans[2002];
     6 int main()
     7 {
     8     int n,m,zi,xi,cnt=0;
     9     int i,j;
    10     scanf("%d",&n);
    11     for( i=0; i<n; i++)
    12     {
    13         scanf("%d",&zi);
    14         scanf("%lf",&str1[zi]);
    15     }
    16     scanf("%d",&m);
    17     for( i=0; i<m; i++)
    18     {
    19         scanf("%d",&zi);
    20         scanf("%lf",&str2[zi]);
    21     }
    22     for( i=0; i<1001; i++)
    23     {
    24         if( str1[i])
    25         {
    26             for( j=0; j<1001; j++)
    27                 if( str2[j])
    28                     ans[i+j] += str1[i]*str2[j];
    29         }
    30     }
    31     for( i=0; i<2001; i++)
    32         if( ans[i])
    33             cnt++;
    34     printf("%d",cnt);
    35     for( i=2001; i>=0; i--)
    36     {
    37         if( ans[i])
    38             printf(" %d %.1f",i,ans[i]);
    39     }
    40     return 0;
    41 }
    在这个国度中,必须不停地奔跑,才能使你保持在原地。如果想要寻求突破,就要以两倍现在速度奔跑!
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  • 原文地址:https://www.cnblogs.com/yuxiaoba/p/8543394.html
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