Table of contents
topic
Decompose 2019 into the sum of 3 different positive integers, and require that each positive integer does not contain the numbers 2 and 4, how many different decomposition methods are there?
Note that swapping the order of 3 integers is considered the same way, e.g. 1000+1001+18 and 1001+1000+18 are considered the same.
Require
This is a question to fill in the blanks, you just need to calculate the result and submit it
train of thought
The idea is very simple - enumerate every number
Every positive integer does not contain the numbers 2 and 4, so the thousands, hundreds, and tens of each digit are extracted and judged separately
bool Panduan(int n)
{
int a = n / 1000;
int b = (n - a * 1000) / 100;
int c = (n - a * 1000 - b * 100) / 10;
int d = n % 10;
if (a == 2 || a == 4 || b == 2 || b == 4 || c == 2 || c == 4 || d == 2 || d == 4)
{
return false;
}
else
{
return true;
}
}Regarding the screening, only the numbers with different orders can be fixed in size, such as three numbers x, y, z fixed x>y>z, so that it will not cause more calculation results because of the different order
for (int i = 1; i<2019 ;i++)
{
for (int j = i+1; j < sum_1 - i; j++)
{
int z = sum_1 - i - j;
...
}
}final code
that is
#include <iostream>
using namespace std;
int sum_1 = 2019;
bool Panduan(int n)
{
int a = n / 1000;
int b = (n - a * 1000) / 100;
int c = (n - a * 1000 - b * 100) / 10;
int d = n % 10;
if (a == 2 || a == 4 || b == 2 || b == 4 || c == 2 || c == 4 || d == 2 || d == 4)
{
return false;
}
else
{
return true;
}
}
int main()
{
int sum_max = 0;
for (int i = 1; i<2019 ;i++)
{
for (int j = i+1; j < sum_1 - i; j++)
{
int z = sum_1 - i - j;
if (i == z || z == j || j >= z)
{
continue;
}
if (Panduan(i) && Panduan(j) && Panduan(z))
{
sum_max++;
}
else
{
continue;
}
}
}
cout << sum_max << endl;
return 0;
}result
40785