2021江苏省赛A. Array-线段树，维护值域,欧拉降幂

题目链接:

https://codeforces.com/gym/102875/problem/A

题目大意:

区间加，区间乘，区间每个数取幂,查询区间幂次和，查询区间累乘。
模数$\le 30$

题目思路:

因为模数小，那么每个数也很小。所以直接维护区间某个值$x$的出现次数.就可以方便的回答查询。

问题在于维护修改:
由于我们转成值域问题了，所以修改一个区间等于轮换这个区间每个值。标记下放就类似于滚动数组。之后我们令$lz[i]$代表$i$要变成的数字.考虑标记如何叠加:
原本$i$需要变到$lz[i]$,现在来了最新的修改方案:$tr[i]$
那么$i$就变成了$tr[lz[i]]$了.

2.一个节点不用pushdown 当且仅当三十个$lz[i]=i$

然后这题线段树里快速幂会T。加欧拉降幂(注意严格遵循其公式)。然后再记忆化一下快速幂。可以过.

心得:
1.欧拉降幂需要严格按照其式子，不能擅自修改
2.打标记的时候一定要思考标记如何叠加.

AC代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define tl (t<<1)
#define tr (t<<1|1)
#define mid ((r + l) >> 1)
#define pii pair<int,int>
#define pb push_back
#define vi vector<int>
#define vll vector<ll>
#define fi first
#define se second
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
int sum[maxn << 2][32] , lz[maxn<<2][32] , a[maxn];
int p , n;
int phi[55] , g[55][55];
void pushup(int t)
{
    
    for (int i = 0 ; i < p ; i++)
        sum[t][i] = sum[tl][i] + sum[tr][i];
}
int mp[101][101][101];
int ksm (int a , int b , int mod){
    
    if (a == 0) return b == 0;
    if (a < 0 || a > 100 || b < 0 || b > 100 || mod < 0 || mod > 100){
    
        cout << "gg" << endl;
        exit(0);
    }
    if (~mp[a][b][mod]) return mp[a][b][mod];
    int ans = 1 , base = a;
    while (b){
    
        if (b & 1) ans = ans * base % mod;
        b >>= 1;
        base = base * base % mod;
    }
    return ans;
}
int jm (int a , int b , int p){
    
    if (a == 0) return b;
    if (g[a][p] == 1) return b % phi[p];
    if (b < phi[p]) return b;
    return b % phi[p] + phi[p];
}
int tmp[50];
void D (int t , int to[32])
{
    
    memset(tmp , 0 , sizeof tmp);
    for (int i = 0 ; i < p ; i++) tmp[to[i]] += sum[t][i];
    for (int i = 0 ; i < p ; i++) sum[t][i] = tmp[i];
    for (int i = 0 ; i < p ; i++) lz[t][i] = to[lz[t][i]];
}
void pushdown(int t)
{
    
    bool ok = false;
    for (int i = 0 ; i < p ; i++)
        if (lz[t][i] != i) ok = true;
    if (!ok) return ;
    D(tl,lz[t]);
    D(tr,lz[t]);
    for (int i = 0 ; i < p ; i++)
        lz[t][i] = i;
}
void build(int l,int r,int t)
{
    
    for (int i = 0 ; i < p ; i++) lz[t][i] = i;
    if(l == r)
    {
    
        sum[t][a[l]] = 1;
        return ;
    }
    build(l , mid , tl);
    build(mid + 1 , r , tr);
    pushup(t);
}
void modify(int L,int R,int l,int r, int t , int type , int c)
{
    
    if(L <= l && r <= R)
    {
    
        int to[32]; memset(to , 0 , sizeof to);
        if (type == 1)
            for (int i = 0 ; i < p ; i++) {
    
               // lz[t][i] = (lz[t][i] + c % p) % p;
                to[i] = (i + c % p) % p;
            }
        else if (type == 2)
            for (int i = 0 ; i < p ; i++) {
    
              // lz[t][i] = (lz[t][i] * (c % p)) % p;
                to[i] = (i * (c % p)) % p;
            }
        else
            for (int i = 0 ; i < p ; i++) {
    
              // lz[t][i] = ksm(lz[t][i] , jm(lz[t][i] , c , p) , p);
                to[i] = ksm(i , jm(i , c , p) , p);
            }
        D(t , to);
        return ;
    }
    pushdown(t);
    if(L <= mid)
        modify(L,R,l,mid,tl,type,c);
    if(R > mid)
        modify(L,R,mid+1,r,tr,type,c);
    pushup(t);
}
int ask_power (int L,int R,int l,int r,int t , int k)
{
    
    if(L <= l && r <= R)
    {
    
        int ans = 0;
        for (int i = 0 ; i < p ; i++)
            ans = (ans + sum[t][i] % p * ksm(i , jm(i , k , p) , p) % p) % p;
        return ans;
    }
    int ans = 0;
    pushdown(t);
    if(L <= mid)
        ans = (ans + ask_power(L,R,l,mid,tl,k)) % p;
    if(R > mid)
        ans  = (ans + ask_power(L,R,mid+1,r,tr,k)) % p;
    return ans;
}
int ask_mul (int L,int R,int l,int r,int t)
{
    
    if(L <= l && r <= R)
    {
    
        int ans = 1;
        for (int i = 0 ; i < p ; i++)
            ans = (ans * ksm(i , jm(i , sum[t][i] , p) , p)) % p;
        return ans;
    }
    int ans = 1;
    pushdown(t);
    if(L <= mid)
        ans = (ans * ask_mul(L,R,l,mid,tl)) % p;
    if(R > mid)
        ans  = (ans * ask_mul(L,R,mid+1,r,tr)) % p;
    return ans;
}
int b[maxn];
void bf_modify (int l , int r , int t , int k){
    
    if (t == 1){
    
        for (int i = l ; i <= r ; i++) b[i] = (b[i] + k) % p;
    }else if (t == 2){
    
        for (int i = l ; i <= r ; i++) b[i] = (b[i] * k) % p;
    }else {
    
        for (int i = l ; i <= r ; i++) b[i] = ksm(b[i] , k , p);
    }
}
int bf_ask (int l , int r , int t , int k){
    
    int ans;
    if (t == 4){
    
        ans = 0;
        for (int i = l ; i <= r ; i++) ans = (ans + ksm(b[i] , k , p)) % p;
    }else if (t == 5){
    
        ans = 1;
        for (int i = l ; i <= r ; i++) ans = (ans * b[i]) % p;
    }
    return ans;
}
mt19937 rnd(time(0));
void getrnd (int &t , int &l , int &r , int &k){
    
    t = rnd() % 5 + 1;
    if (t == 5) k = 0;
    else k = rnd() % 5000000;
    l = rnd() % n + 1;
    r = rnd() % n + 1;
    if (l > r) swap(l , r);
}
template <typename T>
void read(T & x){
     x = 0;T f = 1;char ch = getchar();while(!isdigit(ch)){
    if(ch == '-') f = -1;
ch = getchar();}while (isdigit(ch)){
    x = x * 10 + (ch ^ 48);ch = getchar();}x *= f;}
template <typename T>
void write(T x){
    if(x < 0){
    putchar('-');x = -x;}if (x > 9)write(x / 10);putchar(x % 10 + '0');}
int main()
{
    
    memset(mp , -1 , sizeof mp);
    for (int i = 2 ; i <= 30 ; i++)
        for (int j = 1 ; j <= i ; j++)
            if (__gcd(i , j) == 1) phi[i]++ , g[i][j] = g[j][i] = 1;
    read(n);read(p);
    for (int i = 1 ; i <= n ; i++){
    
        read(a[i]);
        a[i] %= p;
    }
    build(1 , n , 1);
    int m; read(m);
    int t , l , r , k;
    for (int i = 1 ; i <= m ; i++){
    
        read(t);read(l);read(r);read(k);
        if (t <= 3) {
    
            if (t <= 2) k %= p;
            modify(l , r , 1 , n , 1 , t , k);
            bf_modify(l , r , t , k);
        }
        else if (t == 4) {
    
            int res1 = ask_power (l , r , 1 , n , 1 , k);
            printf("%d\n" , res1);
        }
        else {
    
            int res1 = ask_mul(l , r , 1 , n , 1);
            printf("%d\n" , res1);
        }
    }
    return 0;
}