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
#include<cstdio>
#include<iostream>
using namespace std;
int main(){
int K, n, count = ;
double poly1[] = {}, poly2[] = {}, a;
scanf("%d", &K);
for(int i = ; i < K; i++){
scanf("%d%lf", &n, &a);
poly1[n] = a;
}
scanf("%d", &K);
for(int i = ; i < K; i++){
scanf("%d%lf", &n, &a);
poly2[n] = a;
}
for(int i = ; i >= ; i--){
poly1[i] = poly1[i] + poly2[i];
if(poly1[i] != )
count++;
}
printf("%d", count);
for(int i = ; i >= ; i--){
if(poly1[i] != )
printf(" %d %.1lf", i, poly1[i]);
}
cin >> K;
return ; }
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
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