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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 }
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