摘要：
/*O(lg(N）*/#include"iostream"using namespace std;long Countk2(long n,long k){ long count = 0; long i = 1; long current = 0,after = 0,before = 0; while((n / i) != 0) { current = (n / i) % 10; before = n / (i * 10); after = n - (n / i) * i; if (current > k) count = count + (before + 1) * 阅读全文
