ZOJ 4009 And Another Data Structure Problem（ZOJ Monthly, March 2018 Problem F，发现循环节 + 线段树 + 永久标记）

题目链接  ZOJ Monthly, March 2018 Problem F

题意很明确

这个模数很奇妙，在$[0, mod)$的所有数满足任意一个数立方$48$次对$mod$取模之后会回到本身。

所以开$48$棵线段树，和一个永久标记。当对某个区间操作时对这个区间加一层永久标记。

即当前我要查找的第$x$层，实际找的是第$up[i] + x$层。

时间复杂度$O(48nlogn)$

 

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define	ls		i << 1
#define	rs		i << 1 | 1
#define	mid		((l + r) >> 1)
#define lson		ls, l, mid
#define	rson		rs, mid + 1, r

typedef long long LL;

const int N   = 1e5 + 10;
const int mod = 99971;

int T;
int a[N], nxt[N], cal[N][49];
int t[N << 2][49], up[N << 2];
int n, m;

void pushup(int i){
	rep(j, 0, 47){
		t[i][j] = t[ls][(j + up[ls] + 48) % 48] + t[rs][(j + up[rs] + 48) % 48];
		t[i][j] %= mod;
	}
}

void build(int i, int l, int r){
	up[i] = 0;
	if (l == r){
		rep(j, 0, 47) t[i][j] = cal[a[l]][j];
		return;
	}

	build(lson);
	build(rson);
	pushup(i);
}

void update(int i, int l, int r, int L, int R){
	if (L <= l && r <= R){
		++up[i];
		up[i] %= 48;
		return;
	}

	if (L <= mid) update(lson, L, R);
	if (R  > mid) update(rson, L, R);
	pushup(i);
}

int query(int i, int l, int r, int L, int R, int cnt){
	if (L <= l && r <= R){
		return t[i][(up[i] + cnt) % 48];
	}

	int ret = 0;
	int now = cnt + up[i];
	if (L <= mid) ret += query(lson, L, R, now);
	if (R  > mid) ret += query(rson, L, R, now);
	ret %= mod;
	return ret;
}	

int main(){

	scanf("%d", &T);
	rep(i, 1, mod - 1) nxt[i] = 1ll * i * i * i % mod;
	rep(i, 1, mod - 1){
		cal[i][0] = i;
		rep(j, 1, 47){
			cal[i][j] = nxt[cal[i][j - 1]];
		}
	}

	while (T--){
		scanf("%d%d", &n, &m);
		rep(i, 1, n){
			scanf("%d", a + i);
			a[i] %= mod;
		}

		build(1, 1, n);

		while (m--){
			int op, l, r;
			scanf("%d%d%d", &op, &l, &r);
			if (op == 1) update(1, 1, n, l, r);
			else printf("%d\n", query(1, 1, n, l, r, 0));
		}
	}

	return 0;
}