LightOJ 1135 Count the Multiples of 3 （Segmengt + 懒惰标记）

Description

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

1

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

Sample Output

Case 1:

2

3

0

2

这不是一道裸线段树的题。。它涉及到区间更新，这里就需要用到懒惰标记了。。而且还要自己在考虑一下结构体中的数据元素都需要什么，这里需要左右区间指示标志，懒惰标记，记录区间中对3取余分别余1、余2、余0的元素个数，还有就是需要增加一个MOVE（）;它的功能是当访问到该区间时，，转换m0、m1、m2之间的值的操作。分析出来这些这道题就只剩下“ma”代码的工作了。

#include <iostream>#include <cstdio>#include <string>#include <cmath>#include <algorithm>#include <cstring>#include <map>#include <queue>#include <stack>#define INF 0x3f3f3f3f#define mem(a,b) memset(a,b,sizeof(a));#define For(a,b) for(int i = a;i<b;i++)#define LL long long#define MAX_N 100010using namespace std;int x[MAX_N];struct SegmentTree{    int l;    int r;    int m0;    int m1;    int m2;    int flag;}p[MAX_N<<2];void BuildTree(int root,int l,int r){    p[root].l = l;    p[root].r = r;    p[root].m1 = 0;    p[root].m2 = 0;    p[root].flag = 0;    if(l == r)    {        p[root].m0 = 1;        return ;    }    int mid = (l + r)/2;    BuildTree(root<<1,l,mid);    BuildTree(root<<1|1,mid+1,r);    p[root].m0 = p[root<<1].m0 + p[root<<1|1].m0;//    p[root].m1 = p[root<<1].m1 + p[root<<1|1].m1;//    p[root].m2 = p[root<<1].m2 + p[root<<1|1].m2;}void mov_e(SegmentTree &Tree){    int x = Tree.m0;    Tree.m0 = Tree.m2;    Tree.m2 = Tree.m1;    Tree.m1 = x;    return ;}void pushup(int now){    p[now].m0 = p[now<<1].m0 + p[now<<1|1].m0;    p[now].m1 = p[now<<1].m1 + p[now<<1|1].m1;    p[now].m2 = p[now<<1].m2 + p[now<<1|1].m2;    return ;}void pushdown(int now){    int tag = p[now].flag;    if(tag)    {        tag %= 3;        p[now<<1|1].flag += tag;        p[now<<1].flag += tag;        for(int i = 0; i<tag; i++)        {            mov_e(p[now<<1]);            mov_e(p[now<<1|1]);        }        p[now].flag = 0;    }    return ;}void Update(int root,int l,int r){    if(l<=p[root].l && p[root].r <=r)    {        p[root].flag += 1;        mov_e(p[root]);        return ;    }    pushdown(root);    int mid = (p[root].l+p[root].r)/2;    int ans = 0;    if(l>mid)        Update(root<<1|1,l,r);    else if(r<=mid)        Update(root<<1,l,r);    else    {        Update(root<<1|1,l,r);        Update(root<<1,l,r);    }    pushup(root);}int Query(int root,int l,int r){    if(l <= p[root].l && p[root].r <= r)    {        return p[root].m0;    }    pushdown(root);    int mid = (p[root].l + p[root].r)/2;    if(l>mid)        return Query(root<<1|1,l,r);    else if(r<=mid)        return Query(root<<1,l,r);    else    {        int ans = 0,ans1 = 0;        ans = Query(root<<1|1,l,r);        ans1 = Query(root<<1,l,r);        return (ans + ans1);    }}int main(){    int T;    while(~scanf("%d",&T))    {        for(int w = 1; w<=T; w++)        {            int n,q,x,y,ope;            scanf("%d%d",&n,&q);            BuildTree(1,1,n);            printf("Case %d:\n",w);            for(int i = 1; i<=q; i++)            {                scanf("%d",&ope);                if(ope == 0)                {                    scanf("%d %d",&x,&y);                    Update(1,x+1,y+1);                }                else if(ope == 1)                {                    int ans = 0;                    scanf("%d %d",&x,&y);                    printf("%d\n",Query(1,x+1,y+1));                }            }        }    }    return 0;}