Lightoj 1135-count The multiples of 3 segment tree

http://www.lightoj.com/volume_showproblem.php?problem=1135

Test Instructions: given two operations, one for all elements of the interval plus 1, a query interval can be divisible by 3 of the number of how many.

idea: The requirement is divisible by 3, we can record 3 states, the current interval modulus 3 more than 1 of the remaining 2 of 0, then on a number increase, the direct exchange of the number of different remainder can be.

/** @Date    : 2016-12-06-20.00
  * @Author  : Lweleth ([email protected])
  * @Link    : https://github.com/
  * @Version :
  */

#include<bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

struct yuu
{
    int l, r;
    int add;
    int m0, m1, m2;
}tt[N << 2];

void pushup(int p)
{
    tt[p].m0 = tt[p << 1].m0 + tt[p << 1 | 1].m0;
    tt[p].m1 = tt[p << 1].m1 + tt[p << 1 | 1].m1;
    tt[p].m2 = tt[p << 1].m2 + tt[p << 1 | 1].m2;
}

void pushdown(int p)
{
    if(tt[p].add != 0)
    {
        tt[p].add %= 3;
        ///
        tt[p << 1].add += tt[p].add;
        if(tt[p].add == 2)
        {
            swap(tt[p << 1].m0 , tt[p << 1].m1);
            swap(tt[p << 1].m0 , tt[p << 1].m2);
        }
        else if(tt[p].add == 1)
        {
            swap(tt[p << 1].m0 , tt[p << 1].m2);
            swap(tt[p << 1].m1 , tt[p << 1].m0);
        }
        ///
        tt[p << 1 | 1].add += tt[p].add;
        if(tt[p].add == 2)
        {
            swap(tt[p << 1 | 1].m0 , tt[p << 1 | 1].m1);
            swap(tt[p << 1 | 1].m0 , tt[p << 1 | 1].m2);
        }
        else if(tt[p].add == 1)
        {
            swap(tt[p << 1 | 1].m0 , tt[p << 1 | 1].m2);
            swap(tt[p << 1 | 1].m1 , tt[p << 1 | 1].m0);
        }
        tt[p].add = 0;
    }
}

void build(int l, int r, int p)
{
    tt[p].l = l;
    tt[p].r = r;
    tt[p].add = tt[p].m0 = tt[p].m2 = tt[p].m1 = 0;
    if(l == r)
    {
        tt[p].m0 = 1;
        return ;
    }
    int mid = (l + r) >> 1;
    build(l , mid, p << 1);
    build(mid + 1, r, p << 1 | 1);
    pushup(p);
}

void updata(int l, int r, int v, int p)
{
    if(l <= tt[p].l && r >= tt[p].r)
    {
        tt[p].add += v;
        swap(tt[p].m0 , tt[p].m2);
        swap(tt[p].m1 , tt[p].m0);
        return ;
    }
    pushdown(p);
    int mid = (tt[p].l + tt[p].r) >> 1;
    if(l <= mid)
        updata(l, r, v, p << 1);
    if(r > mid)
        updata(l, r, v, p << 1 | 1);
    pushup(p);
}

int query(int l, int r, int p)
{
    if(l <= tt[p].l && r >= tt[p].r)
    {
        return tt[p].m0;
    }
    pushdown(p);
    int mid = (tt[p].l + tt[p].r) >> 1;
    int ans = 0;
    if(l <= mid)
        ans += query(l, r, p << 1);
    if(r > mid)
        ans += query(l, r, p << 1 | 1);
    return ans;
}
int main()
{
    int T;
    int cnt = 0;
    cin >> T;
    while(T--)
    {
        int n, q;
        scanf("%d%d", &n, &q);
        build(1, n, 1);
        printf("Case %d:\n", ++cnt);
        while(q--)
        {
            int t, x, y;
            scanf("%d%d%d", &t ,&x ,&y);
            if(t)
                printf("%d\n", query(x+1, y+1, 1));
            else
                updata(x+1, y+1, 1, 1);
        }
    }
    return 0;
}

Lightoj 1135-count The multiples of 3 segment tree