[HG]AK 题解

前言

什么鬼畜玩意，扶我起来，我要用__int128，这辈子都不珂能用龟速乘的...
真香。

题解

我们知道这个模数是个神奇的东西
(2305843008676823040 = 2^{29} imes 3 imes 5 imes 17 imes 65537)
所以(varphi(2305843008676823040) = 2^{28} imes 2 imes 2^2 imes 2^4 imes 2^{16} = 2^{51})
(a^{varphi(n)}=1 (mod n))，所以在(60)次以内，该数(mod 2305843008676823040)一定会变为(1)。
用线段树维护序列，再加一个龟速乘，解决问题

#include <cstdio>
#define ll long long

const ll MOD = 2305843008676823040ll;
ll s[1000001];
bool flg[1000001];

long long read(){
    long long x = 0; int zf = 1; char ch = ' ';
    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if (ch == '-') zf = -1, ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;
}

inline ll mul(ll x, ll y){ 
    ll res = 0;
    for ( ; y; y >>= 1ll, (x <<= 1ll) %= MOD)
        if(y & 1)
            (res += x) %= MOD;
    return res;
}

void build(int pos, int l, int r){
    if (l == r){
        s[pos] = read();
        flg[pos] = ((s[pos] == 1 || !s[pos]) ? 1 : 0);
        return ;
    }
    int mid = (l + r) >> 1;
    build(pos << 1, l, mid);
    build(pos << 1 | 1, mid + 1, r);
    flg[pos] = flg[pos << 1] & flg[pos << 1 | 1];
    s[pos] = (s[pos << 1] + s[pos << 1 | 1]) % MOD;
}

ll query(int pos, int l, int r, int x, int y){
    if (x <= l && r <= y && flg[pos])
        return s[pos];
    if (l == r){
        ll res = s[pos];
        ll tmp = mul(s[pos], s[pos]);
        if (tmp == s[pos])
            flg[pos] = 1;
        s[pos] = tmp;
        return res;
    }
    int mid = (l + r) >> 1;
    ll res;
    if (y <= mid)
        res = query(pos << 1, l, mid, x, y);
    else if (x > mid)
        res = query(pos << 1 | 1, mid + 1, r, x, y);
    else{
        res = query(pos << 1, l, mid, x, y);
        (res += query(pos << 1 | 1, mid + 1, r, x, y)) %= MOD;
    }
    s[pos] = (s[pos << 1] + s[pos << 1 | 1]) % MOD;
    flg[pos] = flg[pos << 1] & flg[pos << 1 | 1];
    return res
}

int main(){
    int n = read(), m = read();
    for (register int i = 1; i <= 1000000; ++i) flg[i] = 1;
    build(1, 1, n);
    while (m--){
        int l = read(), r = read();
        ll res = query(1, 1, n, l, r) % MOD;
        printf("%lld
", res);
    }
    return 0;
}