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 }

View Code

——written by Lyon

posted @ 2013-06-21 21:12 LyonLys 阅读(181) 评论(0) 收藏举报

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