find the count of all the stepping numbers in the range [n, m]. |
For each test case in a new line print an integer denoting the number of stepping numbers in the range between n and m.
Example:
Input:
3
0 21
10 15
0 1
Output:
13
2
2
Input:
3
0 21
10 15
0 1
Output:
13
2
2
#include <iostream> using namespace std; int test(int n) { int i = n % 10; n /= 10; while (n > 0) { if (abs(i - n % 10) != 1) return 0; i = n % 10; n /= 10; } return 1; } int main() { //code int t; cin >> t; while (t--) { int n, m, cnt = 0; cin >> n >> m; for (int i = n; i <= m; ++i) { if (test(i)) ++cnt; } cout << cnt << endl; } return 0; }