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 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 #define N 100010
 6 #define lson l,mid,p<<1
 7 #define rson mid+1,r,p<<1|1
 8 #define demid int mid=(l+r)>>1
 9 
10 int segt[N<<2][3],col[N<<2];
11 
12 void cal(int p){
13     int tmp=segt[p][0];
14     segt[p][0]=segt[p][2];
15     segt[p][2]=segt[p][1];
16     segt[p][1]=tmp;
17 }
18 
19 void pushup(int p){
20     for(int i=0;i<3;i++) segt[p][i]=segt[p<<1][i]+segt[p<<1|1][i];
21 }
22 
23 void pushdown(int p){
24     if(col[p]){
25         col[p]%=3;
26         col[p<<1]+=col[p];
27         col[p<<1|1]+=col[p];
28         for(int i=0;i<col[p];i++) cal(p<<1),cal(p<<1|1);
29         col[p]=0;
30     }
31 }
32 
33 void build(int l,int r,int p){
34     if(l==r){
35         segt[p][0]=1;
36         return;
37     }
38     demid;
39     build(lson);
40     build(rson);
41     pushup(p);
42 }
43 
44 void update(int L,int R,int l,int r,int p){
45     //printf("%d %d %d %d\n",l,r,p,segt[p][0]);
46     if(L<=l && r<=R){
47         col[p]++;
48         cal(p);
49         return;
50     }
51     demid;
52     pushdown(p);
53     if(L<=mid) update(L,R,lson);
54     if(mid<R) update(L,R,rson);
55     pushup(p);
56 }
57 
58 int query(int L,int R,int l,int r,int p){
59     if(L<=l && r<=R) return segt[p][0];
60     demid;
61     pushdown(p);
62     int res=0;
63     if(L<=mid) res+=query(L,R,lson);
64     if(mid<R) res+=query(L,R,rson);
65     return res;
66 }
67 
68 int main(){
69     int t,cas=1;
70     cin>>t;
71     while(t--){
72         memset(segt,0,sizeof(segt));
73         memset(col,0,sizeof(col));
74         int n,m;
75         cin>>n>>m;
76         build(1,n,1);
77         printf("Case %d:\n",cas++);
78         while(m--){
79             int key,i,j;
80             scanf("%d%d%d",&key,&i,&j);
81             i++,j++;
82             if(key) printf("%d\n",query(i,j,1,n,1));
83             else update(i,j,1,n,1);
84         }
85     }
86     return 0;
87 }
88 /*
89 555
90 4 3
91 0 0 2
92 0 1 3
93 1 0 3
94 */

题目如下：

You have an array with n elements which is indexed from 0 to n - 1. Initially all elements are zero. Now you have to deal with two types of operations

1. Increase the numbers between indices i and j (inclusive) by 1. This is represented by the command '0 i j'.

2. Answer how many numbers between indices i and j (inclusive) are divisible by 3. This is represented by the command '1 i j'.

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 10⁵) and q (1 ≤ q ≤ 50000) denoting the number of queries. Each query will be either in the form '0 i j' or '1 i j' where i, j are integers and 0 ≤ i ≤ j < n.

Output

For each case, print the case number first. Then for each query in the form '1 i j', print the desired result.

Sample Input

Output for Sample Input

10 9

0 0 9

0 3 7

0 1 4

1 1 7

0 2 2

1 2 4

1 8 8

0 5 8

1 6 9

Case 1:

题意：两个操作，0 i j，代表i到j的值全部+1，1 i j输出i到j之间有多少个数%3==0；

线段树，区间更新，区间查询，维护3个值，%3==0，%3==1，%3==2的数的个数。

简单题，记得区域赛挺常出的

posted on 2013-03-28 21:43 SCAU_Xxm 阅读(461) 评论(0) 收藏举报