思路：
模拟。
实现：
1 #include <iostream> 2 using namespace std; 3 const int MAXN = 1000005; 4 int a[MAXN], nxt[MAXN]; 5 6 int main() 7 { 8 int l, r; 9 cin >> l >> r; 10 for (int i = 1; i <= r / 2; i++) { a[i] = 2 * i - 1; nxt[i] = i + 1; } 11 nxt[r / 2] = -1; 12 for (int k = 0; ; k++) 13 { 14 int i = 1, tmp = nxt[1], last = 1; 15 int tmp2 = tmp; 16 for (int j = 1; j < k + 1; j++) tmp2 = nxt[tmp2]; 17 int dx = a[tmp2]; 18 while (tmp != -1) 19 { 20 if ((i + 1) % dx == 0) nxt[last] = nxt[tmp]; 21 last = tmp; 22 tmp = nxt[tmp]; 23 i++; 24 } 25 if (dx > i) break; 26 } 27 int tmp = 1, cnt = 0; 28 while (tmp != -1) { if (a[tmp] > l && a[tmp] < r) cnt++; tmp = nxt[tmp]; } 29 cout << cnt << endl; 30 return 0; 31 }