Brute force is an option in your case: you can just use 2 nested loops and it takes less than 10000 tests only.
// pseudo code
for (i = 0; i <= 100; ++i)
for (j = 0; j <= 100; ++j) {
if ((i + j) > 100)
break;
k = 100 - i - j;
print(i, j, k);
}
If duplicates e.g. 0, 0, 100
and 0, 100, 0
should be excluded, you can use slightly modified code:
// pseudo code
for (i = 0; i <= 100; ++i)
for (j = i; j <= 100; ++j) {
if ((i + j) > 100)
break;
k = 100 - i - j;
if (j <= k)
print(i, j, k);
}