[JSOI2008]最大数maxnumber

嘟嘟嘟

 

就是线段树板子题，还是单点修改区间查询。

用一个指针cnt记录当前序列里有几个数，然后操作1就是把++cnt的位置的数改为(n + t) % d；操作2就是查询cnt - L + 1到cnt的区间最大值。

我用的是先把线段树的节点开好的方法，所以这题按区间长度等于m开就行。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<cctype>
 8 #include<vector>
 9 #include<stack>
10 #include<queue>
11 using namespace std;
12 #define enter printf("\n")
13 #define space printf(" ")
14 #define Mem(a) memset(a, 0, sizeof(a))
15 typedef long long ll;
16 typedef double db;
17 const int INF = 0x3f3f3f3f;
18 const int eps = 1e-8;
19 const int maxn = 2e5 + 5;
20 inline ll read()
21 {
22     ll ans = 0;
23     char ch = getchar(), last = ' ';
24     while(!isdigit(ch)) {last = ch; ch = getchar();}
25     while(isdigit(ch))
26     {
27         ans = ans * 10 + ch - '0'; ch = getchar();
28     }
29     if(last == '-') ans = -ans;
30     return ans;
31 }
32 inline void write(ll x)
33 {
34     if(x < 0) x = -x, putchar('-');
35     if(x >= 10) write(x / 10);
36     putchar(x % 10 + '0');
37 }
38 
39 int m, cnt = 0;
40 ll mod;
41 ll t = 0;
42 
43 int l[maxn << 2], r[maxn << 2];
44 ll Max[maxn << 2];
45 void build(int L, int R, int now)
46 {
47     l[now] = L; r[now] = R;
48     if(L == R) return;
49     int mid = (L + R) >> 1;
50     build(L, mid, now << 1);
51     build(mid + 1, R, now << 1 | 1);
52 }
53 void update(int id, ll d, int now)
54 {
55     if(l[now] == r[now]) {Max[now] = d; return;}
56     int mid = (l[now] + r[now]) >> 1;
57     if(id <= mid) update(id, d, now << 1);
58     else update(id, d, now << 1 | 1);
59     Max[now] = max(Max[now << 1], Max[now << 1 | 1]);
60 }
61 ll query(int L, int R, int now)
62 {
63     if(l[now] == L && r[now] == R) return Max[now];
64     int mid = (l[now] + r[now]) >> 1;
65     if(R <= mid) return query(L, R, now << 1);
66     else if(L > mid) return query(L, R, now << 1 | 1);
67     else return max(query(L, mid, now << 1), query(mid + 1, R, now << 1 | 1));
68 }
69 
70 int main()
71 {
72     m = read(); mod = read();
73     build(1, m, 1);
74     for(int i = 1; i <= m; ++i)
75     {
76         char c; cin >> c;
77         if(c == 'A')
78         {
79             ll n = read(); 
80             update(++cnt, (n + t) % mod, 1);
81         }
82         else
83         {
84             int L = read();
85             t = query(cnt - L + 1, cnt, 1);
86             write(t); enter;
87         }
88     }
89     return 0;
90 }

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