【bzoj 1012】 Maximum number maxnumber 【JSOI2008】

Description

　　Now I ask you to maintain an array and provide the following two operations: 1. Query operation. Syntax: QL Function: Query
the largest number among the last L numbers in the current sequence, and output the value of this number. Restriction: L does not exceed the length of the current sequence. 2. Insert operation. Syntax: A n Function:
Add t, where t is the answer of the most recent query operation (if the query operation has not been performed, then t=0), and the result is
modulo a fixed constant D, the The resulting answer is inserted at the end of the sequence. Restrictions: n is a non-negative integer and is in the range of long integers. Note: Initially, the number column is empty, without a
number.

Input

　　The first line contains two integers, M and D, where M represents the number of operations (M <= 200,000), and D, as described above, satisfies D within longint. Next
M lines, query operation or insert operation.

Output

　　For each query operation, output one line. The row has only one number, the maximum number of the last L numbers in the sequence.

Sample Input

5 100
A 96
Q 1
A 97
Q 1
Q 2

Sample Output

96
93
96

This problem can use line segment tree or monotonic queue, the following is the program:

Segment tree:

[cpp]  view plain copy  

1. #include<stdio.h>  
2. #include<iostream>  
3. usingnamespace std;   
4. constint N=800005;   
5. struct tree{  
6.     int lc,rc,key;  
7. }t[N];  
8. int k;  
9. void build(int rt,int l,int r){  
10.     if(l==r){  
11.         t[rt].key=t[rt].lc=t[rt].rc=0;  
12.         return;  
13.     }  
14.     int m=(l+r)>>1;  
15.     t[rt].lc=++k;  
16.     t[rt].rc=++k;  
17.     build(t[rt].lc,l,m);  
18.     build(t[rt].rc,m+1,r);  
19.     t[rt].key=max(t[t[rt].lc].key,t[t[rt].rc].key);  
20. }  
21. void add(int rt,int l,int r,int k,int w){  
22.     if(l==r&&l==k){  
23.         t[rt].key=w;  
24.         return;  
25.     }  
26.     int m=(l+r)>>1;  
27.     if(k<=m){  
28.         add(t[rt].lc,l,m,k,w);  
29.     }  
30.     else{  
31.         add(t[rt].rc,m+1,r,k,w);  
32.     }  
33.     t[rt].key=max(t[t[rt].lc].key,t[t[rt].rc].key);  
34. }  
35. int ask(int rt,int l,int r,int a,int b){  
36.     if(l==a&&r==b){  
37.         return t[rt].key;  
38.     }  
39.     int m=(l+r)>>1;  
40.     if(b<=m){  
41.         return ask(t[rt].lc,l,m,a,b);  
42.     }  
43.     if(a>m){  
44.         return ask(t[rt].rc,m+1,r,a,b);  
45.     }  
46.     return max(ask(t[rt].lc,l,m,a,m),ask(t[rt].rc,m+1,r,m+1,b));  
47. }  
48. void read(int &s){  
49.     s=0;  
50.     int f=1;  
51.     char c=getchar();  
52.     while((c<'0'||c>'9')&&c!='-'){  
53.         c=getchar();  
54.     }  
55.     if(c=='-'){  
56.         f=-1;  
57.         c=getchar();  
58.     }  
59.     while(c>='0'&&c<='9'){  
60.         s*=10;  
61.         s+=c-'0';  
62.         c=getchar();  
63.     }  
64.     s*=f;  
65. }  
66. void ch(char &c){  
67.     c=getchar();  
68.     while(c!='A'&&c!='Q'){  
69.         c=getchar();  
70.     }  
71. }  
72. int main(){  
73.     int n,mod,q=0,x,l=0,i;  
74.     char c;  
75.     read(n);  
76.     read(mod);  
77.     build(0,1,n);  
78.     for(i=0;i<n;i++){  
79.         ch(c);  
80.         read(x);  
81.         if(c=='A'){  
82.             add(0,1,n,++l,(q+x)%mod);  
83.         }  
84.         else{  
85.             printf("%d\n",q=ask(0,1,n,l-x+1,l));  
86.         }  
87.     }  
88.     return 0;  
89. }  

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