uva 100 The 3n + 1 problem (RMQ)
uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=36
预处理，RMQ求区间最大值。
代码如下：
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <algorithm> 5 #include <cmath> 6 7 using namespace std; 8 9 typedef long long LL; 10 const int N = 4444444; 11 const int M = 22; 12 int pre[N], RMQ[M][N >> 2]; 13 int dfs(LL n) { 14 if (n <= 0) cout << n << endl; 15 if (n < N && pre[n]) return pre[n]; 16 int tmp; 17 if (n & 1) tmp = dfs(n * 3 + 1) + 1; 18 else tmp = dfs(n >> 1) + 1; 19 if (n < N) pre[n] = tmp; 20 return tmp; 21 } 22 23 void PRE() { 24 pre[1] = 1; 25 for (int i = 2, end = N >> 2; i < end; i++) if (pre[i] == 0) dfs(i); 26 // for (int i = 1; i < 20; i++) cout << i << ' ' << pre[i] << endl; 27 // prepare RMQ 28 for (int i = 0, end = N >> 2; i < end; i++) RMQ[0][i] = pre[i]; 29 for (int i = 1; i < M; i++) { 30 for (int j = 0, end = (N >> 2) - (1 << i); j <= end; j++) { 31 RMQ[i][j] = max(RMQ[i - 1][j], RMQ[i - 1][j + (1 << i - 1)]); 32 } 33 } 34 } 35 36 int query(int l, int r) { 37 if (l > r) swap(l, r); 38 int ep = (int) log2((double) r - l + 1); 39 return max(RMQ[ep][l], RMQ[ep][r - (1 << ep) + 1]); 40 } 41 42 int main() { 43 PRE(); 44 int l, r; 45 while (cin >> l >> r) cout << l << ' ' << r << ' ' << query(l, r) << endl; 46 return 0; 47 }
——written by Lyon
