A1002. A+B for Polynomials

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

 

 1 #include<cstdio>
 2 #include<iostream>
 3 using namespace std;
 4 int main(){
 5     int K, n, count = 0;
 6     double poly1[1001] = {0}, poly2[1001] = {0}, a;
 7     scanf("%d", &K);
 8     for(int i = 0; i < K; i++){
 9         scanf("%d%lf", &n, &a);
10         poly1[n] = a;
11     }
12     scanf("%d", &K);
13     for(int i = 0; i < K; i++){
14         scanf("%d%lf", &n, &a);
15         poly2[n] = a;
16     }
17     for(int i = 1000; i >= 0; i--){
18         poly1[i] = poly1[i] + poly2[i];
19         if(poly1[i] != 0) 
20             count++;
21     }
22     printf("%d", count);
23     for(int i = 1000; i >= 0; i--){
24         if(poly1[i] != 0)
25             printf(" %d %.1lf", i, poly1[i]);
26     }
27     cin >> K;
28     return 0;
29 
30 }
View Code

 

总结：

1、多项式加法，在项数不多的情况下直接开数组。

2、注意不用输出系数为0的项

posted @ 2018-01-17 11:50  ZHUQW  阅读(98)  评论(0)  编辑  收藏  举报