This is the simplest way of finding repeated elements i.e. sorting the array. Here, I use a slower sorting algorithm (Selection Sort) that you might've learnt already. You can improve that. There are other ways, both horrible and better, like mentioned by the others, but there's no point getting into the details as you had to ask for pseudo-code for a simple problem. If you want to answer without sorting and also have your life made easier, switch to C++, and use STL containers like maps and sets.

Code:

void swap (int& L, int& R) {
int temp = L;
L = R;
R = temp;
}
void sort (int* Array, int Size) {
for (int i = 0; i < Size - 1; ++i) {
int idx = i;
for (int j = i + 1; j < Size; ++j)
if (Array [j] < Array [idx])
idx = j;
swap (Array [idx], Array [i]);
}
}
void printArray (int* Array, int Size) {
for (int i = 0; i < Size; ++i)
printf("%d ", Array [i]);
}
int main() {
// ios::sync_with_stdio(false);
// cin.tie(nullptr); cout.tie(nullptr);
// cout.precision(10);
// cout << fixed;// << boolalpha;
// Sufficiently large array
int Numbers [1000];
int NumberOfElements;
scanf("%d", &NumberOfElements);
assert(NumberOfElements <= 1000);
for (int i = 0; i < NumberOfElements; ++i)
scanf("%d", Numbers + i);
printf("\nThe Array you entered: ");
printArray(Numbers, NumberOfElements);
// The algorithm jimblumberg is talking about that sorts elements
sort(Numbers, NumberOfElements);
printf("\n\nThe sorted Array: ");
printArray(Numbers, NumberOfElements);
printf("\n\nStatistics:\n\n");
// The algorithm that even my 10 year old cousin can come up with and code without pseudo-code
for (int i = 0, j = 0; i < NumberOfElements; ++i) {
int CurrentElementBeingCheckedIfRepeated = Numbers [i];
// The variable Salem is talking about that counts number of duplicates
int RunLength = 1;
for (j = i + 1; j < NumberOfElements; ++j) {
if (Numbers [j] == CurrentElementBeingCheckedIfRepeated)
++RunLength;
else break;
}
i = j - 1;
if (RunLength != 0)
printf("Element %d has %d duplicates\n", CurrentElementBeingCheckedIfRepeated, RunLength);
}
return 0;
}