bzoj4869

http://www.lydsy.com/JudgeOnline/problem.php?id=4869

终于A了。。。参考了下dalao的代码。。。

拓展欧几里得定理，改了几次就不变了，但是用的时候要在快速幂里判是不是要用。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = ;
int n, m, cnt;
ll p, c;
ll phi[N], table[N];
namespace seg // n^x = n^(x % phi[x] + phi[x])
{
    struct data {
        ll ans, mn;
    } tree[N << ];
    inline ll getphi(ll x)
    {
        ll ret = x, lim = x;
        for(ll i = ; i * i <= lim; ++i) if(x % i == )
        {
            ret = ret * (i - ) / i;
            while(x % i == ) x /= i;
        }
//      printf("ret=%d\n", ret);
        if(x > ) ret = ret * (x - ) / x;
        return ret;
    }
    inline ll power(ll x, ll t, ll p, bool &flag)
    {
        bool big = false;
        ll ret = ;
        for(; t; t >>= )
        {
            if(t & )
            {
                ret = ret * x ;
                flag |= big | (ret >= p);
                ret %= p;
            }
            x = x * x; if(x >= p) big = true, x %= p;
        }
        return ret;
    }
    ll calc(ll x, int t)
    {
        if(x >= phi[t]) x = x % phi[t] + phi[t];
        for(int i = t - ; i >= ; --i)
        {
            bool flag = false;
            x = power(c, x, phi[i], flag);
            if(flag) x += phi[i];
        }
        return x % phi[];
    }
    inline void build(int l, int r, int x)
    {
        if(l == r) {
            tree[x].ans = table[l];
            return;
        }
        int mid = (l + r) >> ;
        build(l, mid, x << );
        build(mid + , r, x <<  | );
        tree[x].ans = (tree[x << ].ans
                     + tree[x <<  | ].ans) % phi[];
    }
    inline void update(int l, int r, int x, int a, int b)
    { //如果这次的幂和上次一样就不变了
        if(tree[x].mn >= cnt) return;
        if(l > b || r < a) return;
        if(l == r)
        {
            ++tree[x].mn;
            tree[x].ans = calc(table[l], tree[x].mn);
            return;
        }
        int mid = (l + r) >> ;
        update(l, mid, x << , a, b);
        update(mid + , r, x <<  | , a, b);
        tree[x].mn = min(tree[x << ].mn,
                         tree[x <<  | ].mn);
        tree[x].ans = (tree[x << ].ans +
                       tree[x <<  | ].ans) % phi[];
    }
    inline ll query(int l, int r, int x, int a, int b)
    {
        if(l > b || r < a) return ;
        if(l >= a && r <= b) return tree[x].ans % phi[];
        int mid = (l + r) >> , ret = ;
        ret = (ret + query(l, mid, x << , a, b)) % phi[];
        ret = (ret + query(mid + , r, x <<  | , a, b)) % phi[];
        return ret;
    }
} using namespace seg;
int main()
{
    scanf("%d%d%lld%lld", &n, &m, &p, &c);
    phi[] = p;
    ll P = p;
    while(P != ) phi[++cnt] = P = getphi(P);
    phi[++cnt] = ;
    for(int i = ; i <= n; ++i) scanf("%lld", &table[i]);
    build(, n, );
    while(m--)
    {
        int opt, l, r; scanf("%d", &opt);
        if(opt == )
        {
            scanf("%d%d", &l, &r);
            update(, n, , l, r);
        }
        if(opt == )
        {
            scanf("%d%d", &l, &r);
            printf("%lld\n", query(, n, , l, r));
        }
    }
    return ;
}