# C++ Program to Find Minimum Value of any Algebraic Expression

C++Server Side ProgrammingProgramming

This is a C++ Program to Find Minimum Value of any Algebraic Expression.An algebraic expression of the form (x1 + x2 + x3 + . . . + xa) * (y1 + y2 + . . . + yb) and (a + b) integers is given.consider all possible combinations of a numbers and remaining b numbers and calculating their values, from which minimum value can be derived.

## Algorithm

Begin
function MaxValue() :
Arguments:
a[] = array which store the elements.
x,y = integers.
Body of the function:
1) Find the sum of array elements.
2) Initialize s=0.
3) Make for loop i = 0 to (x + y)-1 Shift the integers by 25 so that they become positive.
4) Declare a boolean array p[i][j] that represents true if sum j can be reachable by choosing i numbers.
5) Initialization of the array.
6) Make for loop i = 0 to (x + y)-1 to determine If p[i][j] is true, that means it is possible to select i numbers from (x + y) numbers to sum upto j.
7) Initialize min_value=INF.
8) Make for loop i = 0 to (MAX * MAX + 1)-1 to Check if a particular sum can be reachable by choosing x numbers.
if (p[x][i])
Get the actual sum as we shifted the numbers by 25 to avoid negative indexing in array .
9) Print the min_value.
End

## Example

#include <bits/stdc++.h>
using namespace std;
#define INF 1e9
#define MAX 25
int MinValue(int a[], int x, int y) {
int s= 0;
for (int i = 0; i < (x + y); i++) {
s+= a[i];
a[i] += 25;
}
bool p[MAX+1][MAX * MAX + 1];
//Initialize the array to 01.
memset(p, 0, sizeof(p));
p = 1;
for (int i = 0; i < (x + y); i++) {
// k can be at max x because the
// left expression has x numbers
for (int k = min(x, i + 1); k >= 1; k--) {
for (int j = 0; j < MAX * MAX + 1; j++) {
if (p[k - 1][j])
p[k][j + a[i]] = 1;
}
}
}
int min_value = INF;
for (int i = 0; i < MAX * MAX + 1; i++) {
if (p[x][i]) {
int tmp = i - 25 * x;
min_value = min(min_value, tmp * (s - tmp));
}
}
cout << "Minimum Value: " << min_value ;
}
int main() {
int x = 2, y = 2; //input is taken of x and y.
int ar[] = { 7,6,4,3 };
MinValue(ar, x, y);
return 0;
}

## Output

Minimum Value: 91