If you take 2 of 1, and 1 of 2.

Then you have weight
3, 3, 4 = 10, and value
5, 5, 7 = 17

#include

#define MAXWEIGHT 100

int n = 5;
int c[]={2,4,3,5,7}; //cost aka weight
int v[]={3,7,9,8,4}; //value aka benefit
int W=15; //maximum weight can take(bag size)
int main()
{
int table[W][n];
for(int i=0; i<=W; i++)
{
for(int j=0; j<n; j++)
{
table[i][j]=0;
}
}

int a[W]; //this hold the maximum value that can can be obtained using at most i weight

for(int i=0; i<=W; i++)
{
a[i]=0;
}

for(int i=1; i<=W; ++i)
{
for(int j=0; j<n; ++j)
{
if((c[j]<=i)&&(a[i]<=(a[i-c[j]]+v[j]))&&table[i-c[j]][j]!=1)
//c[j]<=i, to allow only if selected weight bucket is less than i
//(a[i]<=(a[i-c[j]]+v[j]), to check if the previous weight bucket has closer value.
//table[i-c[j]][j]!=1 is to prevent the items reuse.
{
for(int k=0; k=a[i])
{
table[i][j]=1;
}
a[i]=newTotalBenefit;
}
}
}

printf(“\t”);
for(int i=0; i<n; i++)
{
printf(“%d “,i);
}
printf(“\n”);
for(int i=0; i<=W; i++)
{
printf(“%d\t”, i);
for(int j=0; j<n; j++)
{
printf(“%d “,table[i][j]);
}
printf(“\n”);
}

printf(“\nmax benefit = %d”, a[W]);
}

//hope it helps.

To prevent it from reuse, I added a table to record down, if the item was used for a particular weight, put as 1, else 0.