/* Driver program to test above functions */
int main()
{
int arr[] = {1, 5, 8, 9, 10, 17, 17, 20};
int size = sizeof(arr)/sizeof(arr[0]);
printf("Maximum Obtainable Value is %d\n", cutRod(arr, size));
getchar();
return 0;
}
int main()
{
int size = sizeof(arr)/sizeof(arr[0]);
int r[size];
memset(r, -1, sizeof(arr));
printf("Maximum Obtainable Value is %d\n", cutRod(size, r));
return 0;
}
int cutRod(int size){
if(size <= 0)
return 0;
int q = INT_MIN;
for(int i=0;i<size;i++){
q = max(q, arr[i] + cutRod(size-i-1));
}
return q;
}
int cutRod(int price[], int n)
{
int i;
if (n <= 0)
return 0;
int max_val = INT_MIN;
for (i = 0; i<n; i++)
max_val = max(max_val, price[i] + cutRod(price, n-i-1));
return max_val;
}
int cutRod(int size, int r[]){
int q = INT_MIN;
if(r[size]>=0)
return r[size];
if(size == 0)
q = 0;
for(int i=0;i<size;i++){
q = max(q, arr[i] + cutRod(size-i-1, r));
}
r[size] = q;
return q;
}
/* Returns the best obtainable price for a rod of length n and
price[] as prices of different pieces */
int cutRod(int price[], int n)
{
if (n <= 0)
return 0;
int max_val = INT_MIN;
// Recursively cut the rod in different pieces and compare different
// configurations
for (int i = 0; i<n; i++)
max_val = max(max_val, price[i] + cutRod(price, n-i-1));
return max_val;
}
int main()
{
int size, arr[50];
int i;
printf("Enter Size of rod:");
scanf("%d",&size);
printf("Enter the prices of the pieces:");
for(i=0;i<size;i++)
scanf("%d",&arr[i]);
printf("Maximum Obtainable Value is %d\n", cutRod(arr, size));
getchar();
return 0;
}
void cutRod(int *price, int n, int * res) {
if (n <= 0) {
*res = 0;
return;
}
int max_val = INT_MIN;
int aux[n];
memset(aux, INT_MIN, n);
int i;
for(i = 0; i<n; i++) {
cutRod(price, n-i-1, &aux[i]);
}
#pragma omp taskwait
for(i = 0; i<n; i++) {
max_val = max(max_val, price[i] + aux[i]);
}
*res = max_val;
}
int main()
{
int size = sizeof(arr)/sizeof(arr[0]);
printf("Maximum Obtainable Value is %d\n", cutRod(size));
return 0;
}
