# Maximize the value of x + y + z such that ax + by + cz = n in C++

We are given with integers a,b,c,n. The goal is to maximize the sum of x, y and z such that ax+by+cz=n.

From above formula,

cz=n-(ax+by)
z= (n- (ax+by))/c

By fixing x and y, calculate z using the above formula, for each x, y and z. Calculate sum and store the maximum such sum obtained.

## Input

n = 6, a = 3, b = 4, c = 5;

## Output

maximum x+y+z is 2.

Explanation − for x=2, y=0 and z=0 ax+by+cz=n.

3*2+0*4+0*5=6 = n

## Input

n = 4, a = 3, b = 1, c = 2;

## Output

maximum x+y+z=4

Explanation − for x=0, y=4 and z=4 ax+by+cz=n.

0*3+4*1+0*2=4 = n

## Approach used in the below program is as follows

• Integers a,b,c,and n are used for expression ax+by+cz=n.

• Function maximize(,int n,int a,int b,int c) takes a, b, c and n as input and returns the maximum possible sum of x, y and z such that ax+by+cz=n.

• Taking all possible ax values, for(i=0;i<=n;i+=a), also

• Taking all possible by values for(j=0;j<=n;j+=b),

• Calculate z= (n- (ax+by))/c.

• Now x=i/a and y=j/b. Calculate x+y+z and store in temp.

• If temp>=maxx so far, update maxx.

• Return maxx as desired sum.

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int maximize(int n, int a, int b, int c){
int maxx = 0;
// i for possible values of ax
for (int i = 0; i <= n; i += a)
// j for possible values of by
for (int j = 0; j <= n - i; j += b) {
float z = (n - (i + j)) / c;
// If z is an integer
if (floor(z) == ceil(z)) {
int x = i / a;
int y = j / b;
int temp=x+y+z;
if(temp>=maxx)
maxx=temp;
}
}
return maxx;
}
int main(){
int n = 6, a = 3, b = 4, c = 5;
cout <<"Maximized the value of x + y + z :"<<maximize(n, a, b, c);
return 0;
}

## Output

Maximized the value of x + y + z :2

Updated on: 28-Jul-2020

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