What is the complexity of each of these loops? If an array is mentioned, they use arr of size N, and possibly arr2 of and M. Consider upper and lower bounds.

1. for(int i = 0; i < arr.size(); ++i) sum += arr[i];
2. for(int i = 0; i < arr.size(); i += 2) sum += arr[i];
3. for(int i = 1; i <= arr.size(); i *= 2) sum += arr[i];
4. int a = 0, b = arr.size()-1; while(a < b) { int p = (a + b)/2; if(x < arr[p]) b = p-1; else if(x > arr[p]) a = p+1; else break; }
5. for(int i = 0; i < arr.size(); i++) for(int j = 0; j < arr.size(); j++) sum += arr[i]*arr[j];
6. for(int i = 0; i < arr.size(); i++) for(int j = 1; j <= arr.size(); j *= 2) sum += arr[i]*arr[j];
7. for(int i = 1; i <= arr.size(); i++) for(int j = i; j < arr.size(); j++) sum += arr[i]*arr[j];
8. for(int i = 0; i < arr.size(); i++) for(int j = i; j < arr.size(); j += i+1) sum += arr[i]*arr[j];
9. for(int i = 0; i < arr.size(); ++i) arr[i] = 0; while(true) { for(int i = 0; i < arr.size(); ++i) print arr[i]; int j = arr.size() - 1 for( ; j >= 0 && arr[j] == 9; --j) arr[j] = 0; if(j < 0) break; else ++arr[j]; }
Think about what the output looks like. That should give you the number of iterations for the outer loop.
10. For the last two, b and e are the input values to the computation. Consider e to be the size of the problem, and find a complexity as a function of e.
p = 1; for(int i = 0; i < e; ++i) p *= b;
11. p = 1; while(e > 0) { if(e % 2 == 0) { p *= p; e /= 2; } else { p *= b; --e; } }
12. Zheng's Problem