# MyBatis-Source Code Analysis 5 Create SQLSession Dialogue

2023-01-06

### 1002 A+B for Polynomials 25 points

This time, you are supposed to find A+B where A and B are two polynomials.

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:
K   N 1   a N 1   N 2   a N 2   . . .   N K   a N K K \ N_1 \ a_{N_1} \ N_2 \ a_{N_2} \ … \ N_K \ a_{N_K}

where K K is the number of nonzero terms in the polynomial, N i N_i ​​ and a N ​ i ( i = 1 , 2 , ⋯ , K ) a_{N_​i}(i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1 ≤ K ≤ 10 , 0 ≤ N K < ⋯ < N 2 < N 1 ≤ 1000 1≤K≤10, 0≤N_K<⋯<N_2<N_1≤1000 .

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.

2 1 2.4 0 3.2
2 2 1.5 1 0.5

3 2 1.5 1 2.9 0 3.2

#include <iostream>
using namespace std;

void do_union(double exponents_all[]);

int main()
{

int n, exponents;
double coefficients;
double exponents_all = {
0 };
//int output_all;

for (int i = 0; i < 2; i++)
{

cin >> n;
while (n != 0)
{

cin >> exponents >> coefficients;
exponents_all[exponents] += coefficients;
n--;
}
}
do_union(exponents_all);
}

void do_union(double exponents_all[])
{

int count = 0;
for (int i = 1000; i >= 0; i--)
{

if (exponents_all[i] != 0)
count++;
}
cout << count;
for (int j = 1000; j >= 0; j--)
{

if (exponents_all[j] != 0)
printf(" %d %.1f", j, exponents_all[j]);
}
}

• After my code was submitted, there was a sample that did not pass it, and I was very puzzled after deducting two points.
• Since Chen Yueyu did not announce what the test was, according to my experience, there may be inputcoefficient 0The situation.
• According to the title, he has no restrictionscoefficient 0

The situation of, but for the sake of more concise the method, I amvoid do_union(double exponents_all[])Theforcycle jumps with a coefficient of 0.

• I didn’t expect a more concise method for the time being, I would be wrong when it was wrong. If a friend can see the problem, please be positive
• The above problem has been solved

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