题目链接

BZOJ4869

题解

这题调得我怀疑人生,,结果就是因为某些地方\(sb\)地忘了取模

前置题目:BZOJ3884
扩展欧拉定理:

\[c^a \equiv c^{a \mod \varphi(p) + [a \ge p]p} \pmod p \]

我们发现当我们进行\(0\)操作,就相当于在\(a\)底部添加一层\(c\)
当我们进行得足够多的时候,\(\varphi(p)\)就会取到\(1\),从而不再变化
所以每个位置操作次数其实是有限的,为\(O(logp)\)
为何是\(O(logp)\)次呢?
考虑欧拉函数:

\[\varphi(n) = n \prod\limits_{i = 1}^{k} \frac{p_i - 1}{p_i} \]

由于质数一定是奇数,所以\(p_i - 1\)一定为偶数,所以操作一次后的\(p\)一定为偶数
偶数有\(2\)这个质因子,式子中就会存在\(\frac{1}{2}\)这一项,所以至少减少\(\frac{1}{2}\)
所以到达\(1\)\(O(logp)\)

用线段树进行维护
每次修改重新计算\(O(log^2p)\),总复杂度\(O(nlognlog^3p)\),凭信仰可过

由于每次快速幂底都是\(c\),我们可以预处理\(c^x\),从而做到\(O(nlognlog^2p)\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define ls (u << 1)
#define rs (u << 1 | 1)
using namespace std;
const int maxn = 100005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int p[maxn],pi,isn[maxn];
void init(){
    for (int i = 2; i <= 10000; i++){
        if (!isn[i]) p[++pi] = i;
        for (int j = 1; j <= pi && i * p[j] <= 10000; j++){
            isn[i * p[j]] = true;
            if (i % p[j] == 0) break;
        }
    }
}
int n,m,P[30],Pi,c,a[maxn],cnt[maxn],low[30];
LL C[30][maxn],CC[30][maxn];
int phi(int x){
    int tmp = x,ans = x;
    for (int i = 1; i <= pi && p[i] <= tmp; i++){
        int v = p[i];
        if (tmp % v == 0){
            ans = ans / v * (v - 1);
            while (tmp % v == 0) tmp /= v;
        }
    }
    if (tmp - 1) ans = ans / tmp * (tmp - 1);
    return ans;
}
LL qpow(int b,int md){
    return CC[md][b / 10000] * C[md][b % 10000] % P[md];
}
int cal(int x,int t){
	LL last,tmp = x;
    if (tmp > P[t]) tmp = tmp % P[t] + P[t];
    for(int i = t; i; i--)
    {
        last = tmp;
        tmp = qpow(tmp,i - 1);
        if (last >= low[i - 1]){
			tmp += P[i - 1];
		}
    }
    return tmp;
}
int sum[maxn << 2],val[maxn << 2];
void upd(int u){
    sum[u] = (sum[ls] + sum[rs]) % P[0];
    val[u] = val[ls] | val[rs];
}
void build(int u,int l,int r){
    val[u] = 1;
    if (l == r){sum[u] = a[l] % P[0]; return;}
    int mid = l + r >> 1;
    build(ls,l,mid);
    build(rs,mid + 1,r);
    upd(u);
}
void change(int u,int l,int r){
    if (!val[u]) return;
    if (l == r){
        cnt[l]++;
        if (cnt[l] >= Pi) val[u] = 0;
        sum[u] = cal(a[l],cnt[l]) % P[0];
        return;
    }
    int mid = l + r >> 1;
    change(ls,l,mid);
    change(rs,mid + 1,r);
    upd(u);
}
void modify(int u,int l,int r,int L,int R){
    if (!val[u]) return;
    if (l >= L && r <= R){
        change(u,l,r);
        return;
    }
    int mid = l + r >> 1;
    if (mid >= L) modify(ls,l,mid,L,R);
    if (mid < R) modify(rs,mid + 1,r,L,R);
    upd(u);
}
int query(int u,int l,int r,int L,int R){
    if (l >= L && r <= R) return sum[u] % P[0];
    int mid = l + r >> 1;
    if (mid >= R) return query(ls,l,mid,L,R);
    if (mid < L) return query(rs,mid + 1,r,L,R);
    return (query(ls,l,mid,L,R) + query(rs,mid + 1,r,L,R)) % P[0];
}
int main(){
    init();
    n = read(); m = read(); P[0] = read(); c = read();
    REP(i,n) a[i] = read();
    while (P[Pi] != 1){
        ++Pi;
        P[Pi] = phi(P[Pi - 1]);
    }
	P[++Pi] = 1;
    for (register int t = 0; t <= Pi; t++){
        C[t][0] = CC[t][0] = 1 % P[t];
		low[t] = 0;
        for (register int i = 1; i <= 10000; i++){
			C[t][i] = C[t][i - 1] * c;
			if (C[t][i] >= P[t] && !low[t]) low[t] = i;
			C[t][i] %= P[t];
		}
		CC[t][1] = C[t][10000];
		for (register int i = 2; i <= 10000; i++)
			CC[t][i] = CC[t][i - 1] * C[t][10000] % P[t];
    }
    build(1,1,n);
    int opt,l,r;
    while (m--){
        opt = read(); l = read(); r = read();
        if (!opt) modify(1,1,n,l,r);
        else printf("%d\n",query(1,1,n,l,r));
    }
    return 0;
}