A

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 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

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                                                   
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");elseprintf("%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;
}