Codeforces Round #539 (Div. 1) E

如果mod是质数就好做了，但是做除法的时候对于合数mod可能没有逆元。所以就只有存一下mod的每个质因数(最多9个)的幂，和剩下一坨与mod互质的一部分。然后就能做了。有点恶心。

CODE

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 100005;
const int MAXP = 9;
int n, q, a[MAXN], p[MAXP], cnt, mod, phi;
inline int qpow(int a, int b) {
	int re=1;
	for(;b;a=1ll*a*a%mod,b>>=1)if(b&1)re=1ll*re*a%mod;
	return re;
}
struct node {
	int ind[MAXP];
	node () { memset(ind, 0, sizeof ind); }
	inline int init(int x) {
		memset(ind, 0, sizeof ind);
		for(int i = 0; i < cnt; ++i)
			while(x % p[i] == 0)
				x /= p[i], ind[i]++;
		return x;
	}
	inline void operator *=(const node &o) {
		for(int i = 0; i < cnt; ++i)
			ind[i] += o.ind[i];
	}
	inline void operator /=(const node &o) {
		for(int i = 0; i < cnt; ++i)
			ind[i] -= o.ind[i];
	}
	inline int calc() {
		int re = 1;
		for(int i = 0; i < cnt; ++i)
			re = 1ll * re * qpow(p[i], ind[i]) % mod;
		return re;
	}
}tag[MAXN<<2]; //tag存mod的质因数的幂
int res[MAXN<<2], rlz[MAXN<<2]; //res存与mod互质的部分的积
int sum[MAXN<<2], lz[MAXN<<2]; //sum存答案

inline void pre(int x) {
	phi = x;
	for(int i = 2; i*i <= x; ++i)
		if(x % i == 0) {
			p[cnt++] = i, phi = phi / i * (i-1);
			while(x % i == 0) x /= i;
		}
	if(x > 1) p[cnt++] = x, phi = phi / x * (x-1);
}

inline void upd(int i) { sum[i] = (sum[i<<1] + sum[i<<1|1]) % mod; }
inline void pd(int i) {
	tag[i<<1] *= tag[i], tag[i<<1|1] *= tag[i];
	memset(tag[i].ind, 0, sizeof tag[i].ind);
	if(lz[i] != 1) {
		sum[i<<1] = 1ll * sum[i<<1] * lz[i] % mod;
		lz[i<<1] = 1ll * lz[i<<1] * lz[i] % mod;
		sum[i<<1|1] = 1ll * sum[i<<1|1] * lz[i] % mod;
		lz[i<<1|1] = 1ll * lz[i<<1|1] * lz[i] % mod;
		lz[i] = 1;
	}
	if(rlz[i] != 1) {
		res[i<<1] = 1ll * res[i<<1] * rlz[i] % mod;
		rlz[i<<1] = 1ll * rlz[i<<1] * rlz[i] % mod;
		res[i<<1|1] = 1ll * res[i<<1|1] * rlz[i] % mod;
		rlz[i<<1|1] = 1ll * rlz[i<<1|1] * rlz[i] % mod;
		rlz[i] = 1;
	}
}

void build(int i, int l, int r) {
	//printf("SSS (%d,%d,%d)
", i, l, r);
	lz[i] = rlz[i] = 1;
	if(l == r) {
		int x; scanf("%d", &x);
		res[i] = tag[i].init(x) % mod;
		sum[i] = x % mod;
		return;
	}
	int mid = (l + r) >> 1;
	build(i<<1, l, mid);
	build(i<<1|1, mid+1, r);
	upd(i);
}

void multi(int i, int l, int r, int x, int y, int num, node v, int val) {
	if(x <= l && r <= y) {
		tag[i] *= v;
		lz[i] = 1ll * lz[i] * val % mod;
		sum[i] = 1ll * sum[i] * val % mod;
		rlz[i] = 1ll * rlz[i] * num % mod;
		res[i] = 1ll * res[i] * num % mod;
		return;
	}
	pd(i);
	int mid = (l + r) >> 1;
	if(x <= mid) multi(i<<1, l, mid, x, y, num, v, val);
	if(y > mid) multi(i<<1|1, mid+1, r, x, y, num, v, val);
	upd(i);
}

void divide(int i, int l, int r, int x, int num, node v) {
	if(l == r) {
		sum[i] = res[i] = 1ll * res[i] * qpow(num, phi-1) % mod;
		tag[i] /= v; sum[i] = 1ll * sum[i] * tag[i].calc() % mod;
		return;
	}
	pd(i);
	int mid = (l + r) >> 1;
	if(x <= mid) divide(i<<1, l, mid, x, num, v);
	else divide(i<<1|1, mid+1, r, x, num, v);
	upd(i);
}

inline int query(int i, int l, int r, int x, int y) {
	if(x <= l && r <= y) return sum[i];
	int mid = (l + r) >> 1, re = 0; pd(i);
	if(x <= mid) re = (re + query(i<<1, l, mid, x, y)) % mod;
	if(mid < y) re = (re + query(i<<1|1, mid+1, r, x, y)) % mod;
	return re;
}

int main() {
	scanf("%d%d", &n, &mod); pre(mod);
	build(1, 1, n);
	scanf("%d", &q);
	int op, l, r, x;
	int num; node tmp;
	while(q--) {
		scanf("%d", &op);
		if(op == 1) {
			scanf("%d%d%d", &l, &r, &x);
			num = tmp.init(x);
			multi(1, 1, n, l, r, num, tmp, x);
		}
		else if(op == 2) {
			scanf("%d%d", &l, &x);
			num = tmp.init(x);
			divide(1, 1, n, l, num, tmp);
		}
		else {
			scanf("%d%d", &l, &r);
			printf("%d
", query(1, 1, n, l, r));
		}
	}
}