[JSOI2008]最大数

Posted 2020-09-15 蒟蒻zht的博客

 

题目传送门

 

1.线段树

线段树可以搞。

不过慢的要死1300+ms

 1 #include <cstdio>
 2 #include <iostream>
 3 
 4 using namespace std;
 5 
 6 int m, n, pos, ql, qr;
 7 long long c[2000001], x, d, t;
 8 char s;
 9 
10 void build(int o, int l, int r)
11 {
12     c[o] = -2147283647;
13     if(l == r) return;
14     int mid = (l + r) / 2;
15     build(o * 2, l, mid);
16     build(o * 2 + 1, mid + 1, r);
17 }
18 
19 void update(int o, int l, int r)
20 {
21     if(l == r)
22     {
23         c[o] = x;
24         return;
25     }
26     int mid = (l + r) / 2;
27     if(pos <= mid) update(o * 2, l, mid);
28     else update(o * 2 + 1, mid + 1, r);
29     c[o] = max(c[o * 2], c[o * 2 + 1]);
30 }
31 
32 int query(int o, int l, int r)
33 {
34     if(ql <= l && r <= qr) return c[o];
35     int mid = (l + r) / 2, ans = -2147283647;
36     if(ql <= mid) ans = max(ans, query(o * 2, l, mid));
37     if(qr > mid) ans = max(ans, query(o * 2 + 1, mid + 1, r));
38     return ans;
39 }
40 
41 int main()
42 {
43     int i, j;
44     scanf("%d%d", &m, &d);
45     build(1, 1, m);
46     for(i = 1; i <= m; i++)
47     {
48         scanf("%s%d", &s, &x);
49         if(s == ‘A‘)
50         {
51             x = (x + t) % d;
52             pos++;
53             update(1, 1, m);
54         }
55         else
56         {
57             ql = pos - x + 1;
58             qr = pos;
59             t = query(1, 1, m);
60             printf("%d\n", t);
61         }
62     }
63     return 0;
64 }

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2.单调栈

单调栈不仅快且代码量小。

300+ms

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 # include <string>
 5 # include <cmath>
 6 # include <vector>
 7 # include <map>
 8 # include <queue>
 9 # include <cstdlib>
10 # include <algorithm>
11 # define MAXN 200001
12 using namespace std;
13 
14 int m, d, t, len;
15 int top, s[MAXN], a[MAXN];
16 
17 int main()
18 {
19     int i, j, x, y;
20     char c[1];
21     scanf("%d %d", &m, &d);
22     while(m--)
23     {
24         scanf("%s %d", c, &x);
25         if(c[0] == ‘A‘)
26         {
27             x = (x + t) % d;
28             a[++len] = x;
29             while(top && a[s[top]] <= x) top--;
30             s[++top] = len;
31         }
32         else
33         {
34             y = lower_bound(s + 1, s + top + 1, len - x + 1) - s;
35             printf("%d\n", t = a[s[y]]);
36         }
37     }
38     return 0;
39 }

View Code