1002. A+B for Polynomials

#include<stdio.h>
int main ( )
{
    int i , j , K1 , K2 ;
    int a , B[1001] ;
    float A[1001] = {0} , b ;

    scanf("%d",&K1 );
    for ( i = 1 ; i <= K1 ; i++ )
    {
        scanf("%d %f",&a, &b );
        A[a] = b ;
    }
    scanf("%d",&K2 );
    for ( i = 1 ; i <= K2 ; i++ )
    {
        scanf("%d %f",&a, &b );
        A[a] += b ;
    }

    int count = 0 ;
    for ( i = 0 ; i <= 1000 ; i++ )
        if ( A[i] )
            B[count++] = i ;
    printf("%d",count );
    for ( i = count-1 ; i >= 0 ; i-- )
        printf(" %d %.1f",B[i] , A[B[i]] );
    printf("\n");
    return 0;
}

辣鸡无疑了，这么简单的一道题折腾了这么久，还想用循环存储相同和不相同下标的值。

注意变量的值。

原题描述如下：

This time, you are supposed to find A+B where A and B are two polynomials.

Input

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 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1, 2, ..., K) are the exponents and coefficients, respectively.  It is given that 1 <= K <= 10，0 <= NK < ... < N2 < N1 <=1000.  

Output

For each test case you should output the sum 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 to 1 decimal place.

Sample Input

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

 

就是多项式求和的问题。。