大意: 給定n元素序列, q個操作: (1)區間乘 (2)單點除(保證整除) (3)區間求和對m取模

要求回答所有操作(3)的結果

主要是除法難辦, 假設單點除$x$, $x$中與$m$互素的素因子可以直接費馬小定理求逆, 其余因子維護一個向量即可.

這種沙茶題結果各種細節出錯改了一個多小時......太菜了

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl ‘\n‘
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head

const int N = 1e5+10;

int n, m, q, sz, Phi;
vector<int> A;

int phi(int x) {
	int s = x, mx = sqrt(x+0.5);
	REP(i,2,mx) if (x%i==0) {
		s = s/i*(i-1);
		A.pb(i);
		while (x%i==0) x/=i;
	}
	if (x>1) A.pb(x),s=s/x*(x-1);
	sz = A.size();
	return s;
}

ll qpow(ll a,ll n) {ll r=1%m;for (a%=m;n;a=a*a%m,n>>=1)if(n&1)r=r*a%m;return r;}

int sum[N<<2], rtag[N<<2], tag[N<<2], res[N<<2];
int c[N<<2][10], tagc[N<<2][10];
int ans, ql, qr, qv[10];

void pd(int o) {
	if (tag[o]!=1) {
		tag[lc] = (ll)tag[lc]*tag[o]%m;
		tag[rc] = (ll)tag[rc]*tag[o]%m;
		sum[lc] = (ll)sum[lc]*tag[o]%m;
		sum[rc] = (ll)sum[rc]*tag[o]%m;
		tag[o] = 1;
	}
	if (rtag[o]!=1) {
		res[lc] = (ll)res[lc]*rtag[o]%m;
		res[rc] = (ll)res[rc]*rtag[o]%m;
		rtag[lc] = (ll)rtag[lc]*rtag[o]%m;
		rtag[rc] = (ll)rtag[rc]*rtag[o]%m;
		rtag[o] = 1;
	}
	REP(i,0,sz-1) { 
		c[lc][i]+=tagc[o][i],c[rc][i]+=tagc[o][i];
		tagc[lc][i]+=tagc[o][i],tagc[rc][i]+=tagc[o][i];
		tagc[o][i]=0;
	}
}

void pu(int o) {
	sum[o]=(sum[rc]+sum[lc])%m;
}

void build(int o, int l, int r) {
	sum[o]=res[o]=tag[o]=rtag[o]=1;
	if (l==r) {
		int t;
		scanf("%d", &t);
		sum[o] = t%m;
		REP(i,0,sz-1) {
			while (t%A[i]==0) t/=A[i],++c[o][i];
		}
		res[o] = t%m;
		return;
	}
	build(ls),build(rs);
	pu(o);
}

void mul(int o, int l, int r, int x, int R) {
	if (ql<=l&&r<=qr) {
		sum[o] = (ll)sum[o]*x%m;
		tag[o] = (ll)tag[o]*x%m;
		res[o] = (ll)res[o]*R%m;
		rtag[o] = (ll)rtag[o]*R%m;
		REP(i,0,sz-1) c[o][i]+=qv[i],tagc[o][i]+=qv[i];
		return;
	}
	pd(o);
	if (mid>=ql) mul(ls,x,R);
	if (mid<qr) mul(rs,x,R);
	pu(o);
}
void div(int o, int l, int r, int x, int v) {
	if (l==r) {
		REP(i,0,sz-1) {
			while (v%A[i]==0) v/=A[i],--c[o][i];
		}
		res[o] = (ll)res[o]*qpow(v,Phi-1)%m;
		sum[o] = res[o];
		REP(i,0,sz-1) sum[o]=(ll)sum[o]*qpow(A[i],c[o][i])%m;
		return;
	}
	pd(o);
	if (mid>=x) div(ls,x,v);
	else div(rs,x,v);
	pu(o);
}
void query(int o, int l, int r, int ql, int qr) {
	if (ql<=l&&r<=qr) return (ans+=sum[o])%=m,void();
	pd(o);
	if (mid>=ql) query(ls,ql,qr);
	if (mid<qr) query(rs,ql,qr);
}

int main() {
	scanf("%d%d", &n, &m);
	Phi = phi(m);
	build(1,1,n);
	scanf("%d", &q);
	while (q--) {
		int op, x, p;
		scanf("%d", &op);
		if (op==1) {
			scanf("%d%d%d",&ql,&qr,&x);
			int t = x;
			REP(i,0,sz-1) { 
				qv[i] = 0;
				while (t%A[i]==0) t/=A[i],++qv[i];
			}
			mul(1,1,n,x,t);
		} else if (op==2) {
			scanf("%d%d",&p,&x);
			div(1,1,n,p,x);
		} else {
			scanf("%d%d",&ql,&qr);
			ans = 0, query(1,1,n,ql,qr);
			printf("%d\n", ans);
		}
	}
}

Sasha and a Very Easy Test CodeForces - 1109E (數學,線段樹)