Light 1135 - Count the Multiples of 3 (线段树 + 区间修改)
题意
区间统一增加1和询问区间内能被3整除的个数。
思路
一开始想直接维护答案,但是不行。
要维护区间内余数是0、1、2的个数。
复习了一下区间修改的写法。
代码
#include <stack>
#include <cstdio>
#include <list>
#include <cassert>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 1e5 + 10;
const int MOD = 9901;
const int MOD2 = 1e9 + 9;
const int seed = 188147;
const int dir[][2] = { {-1, 0}, {0, -1}, { 1, 0 }, { 0, 1 } };
int cases = 0;
typedef pair<int, int> pii;
struct SEGTREE
{
int cnt[3], lazy;
}t[MAXN<<2];
void maintain_node(int rt)
{
int *arr = t[rt].cnt;
int v0 = arr[0], v1 = arr[1], v2 = arr[2];
arr[0] = v2, arr[1] = v0, arr[2] = v1;
}
void Maintain(int rt)
{
for (int i = 0; i < 3; i++) t[rt].cnt[i] = t[LRT].cnt[i] + t[RRT].cnt[i];
}
void PushDown(int rt)
{
t[rt].lazy %= 3;
t[LRT].lazy += t[rt].lazy, t[RRT].lazy += t[rt].lazy;
for (int i = 0; i < t[rt].lazy; i++) maintain_node(LRT), maintain_node(RRT);
t[rt].lazy = 0;
}
int Query(int rt, int l, int r, int L, int R)
{
if (L <= l && r <= R) return t[rt].cnt[0];
if (t[rt].lazy) PushDown(rt);
int mid = MID(l, r), ans = 0;
if (L <= mid) ans = Query(LC, L, R);
if (R > mid) ans += Query(RC, L, R);
return ans;
}
void Update(int rt, int l, int r, int L, int R)
{
if (L <= l && r <= R)
{
maintain_node(rt);
t[rt].lazy++;
return;
}
if (t[rt].lazy) PushDown(rt);
int mid = MID(l, r);
if (L <= mid) Update(LC, L, R);
if (R > mid) Update(RC, L, R);
Maintain(rt);
}
void Build(int rt, int l, int r)
{
if (l == r)
{
t[rt].cnt[0] = 1;
return;
}
int mid = MID(l, r);
Build(LC); Build(RC);
Maintain(rt);
}
int main()
{
//ROP;
int T;
scanf("%d", &T);
while (T--)
{
MS(t, 0);
printf("Case %d:\n", ++cases);
int n, q;
scanf("%d%d", &n, &q);
Build(1, 1, n);
while (q--)
{
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
b++, c++;
if (a == 0) //increase
Update(1, 1, n, b, c);
else
printf("%d\n", Query(1, 1, n, b, c));
}
}
return 0;
}