A-1002

题目内容：

1002. A+B for Polynomials (25)

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 3.22 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2                                                   

思路：

用枚举法后相加，但是要注意相加抵消的情况

代码：

#include<stdio.h>int main(){    float a[1002]={0.0},b[1002]={0.0},c[1002]={0.0};                //数组的范围要看准    int M,T,i=0,j=0,count=0;    float temp;    scanf("%d",&M);    while(M--)    {        scanf("%d%f",&i,&temp);        a[i]=temp;    }    scanf("%d",&T);    while(T--)    {        scanf("%d%f",&i,&temp);        b[i]=temp;    }    for(i=0;i<1002;i++)    {      c[i]=a[i]+b[i];    //正负抵消问题      if(c[i]!=0.0)      {        count++;    }    }    if(count==0)        printf("0");    else    printf("%d ",count);    for(i=1001;i>=0;i--)    {        if(c[i]!=0.0)        {            printf("%d %.1f",i,c[i]);            count--;            if(count!=0)                printf(" ");        }    }    return 0;}