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C++ code to find three numbers whose sum is n
Suppose we have a number n. We are going to find three numbers a, b and c, such that a + b + c = n and none of these three numbers are multiple of 3.
So, if the input is like n = 233, then the output will be [77, 77, 79]
Steps
To solve this, we will follow these steps −
if (n - 2) mod 3 is same as 0, then: return 1, 2, and n - 3 Otherwise return 1, 1, and n - 2
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
void solve(int n){
if ((n - 2) % 3 == 0)
cout << 1 << ", " << 2 << ", " << n - 3;
else
cout << 1 << ", " << 1 << ", " << n - 2;
}
int main(){
int n = 233;
solve(n);
}Input
233
Output
1, 2, 230
Updated on: 15-Mar-2022
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