Codeforces 1114F Please, another Queries on Array? 线段树

Please, another Queries on Array?

利用欧拉函数的计算方法， 用线段树搞一搞就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
 
using namespace std;
 
const int N = 4e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
const double PI = acos(-);
 
int n, q, tot, b[N], prime[], mul[];
LL S[];
char op[];
 
bool ok(int x) {
    for(int i = ; i * i <= x; i++)
        if(x % i == ) return false;
    return true;
}
 
int Power(int a, int b) {
    int ans = ;
    while(b) {
        if(b & ) ans = 1ll * ans * a % mod;
        a = 1ll * a * a % mod; b >>= ;
    }
    return ans;
}
 
int a[N << ], lazy[N << ];
LL mask[N << ], mlazy[N << ];
 
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
 
inline void pull(int rt) {
    a[rt] = 1ll * a[rt << ] * a[rt <<  | ] % mod;
    mask[rt] = mask[rt << ] | mask[rt <<  | ];
}
 
void push(int rt, int l, int mid, int r) {
    if(lazy[rt] != ) {
        a[rt << ] = 1ll * a[rt << ] * Power(lazy[rt], mid - l + ) % mod;
        a[rt <<  | ] = 1ll * a[rt <<  | ] * Power(lazy[rt], r - mid) % mod;
        lazy[rt << ] = 1ll * lazy[rt << ] * lazy[rt] % mod;
        lazy[rt <<  | ] = 1ll * lazy[rt <<  | ] * lazy[rt] % mod;
        lazy[rt] = ;
    }
    if(mlazy[rt] != ) {
        mask[rt << ] |= mlazy[rt]; mask[rt <<  | ] |= mlazy[rt];
        mlazy[rt << ] |= mlazy[rt]; mlazy[rt <<  | ] |= mlazy[rt];
        mlazy[rt] = ;
    }
}
 
void build(int l, int r, int rt) {
    lazy[rt] = ;
    mlazy[rt] = ;
    if(l == r) {
        a[rt] = b[l];
        mask[rt] = S[b[l]];
        return;
    }
    int mid = l + r >> ;
    build(lson); build(rson);
    pull(rt);
}
 
void update(int L, int R, int mul, int l, int r, int rt) {
    if(l >= L && r <= R) {
        a[rt] = 1ll * a[rt] * Power(mul, r - l + ) % mod;
        lazy[rt] = 1ll * lazy[rt] * mul % mod;
        mask[rt] |= S[mul];
        mlazy[rt] |= S[mul];
        return;
    }
    int mid = l + r >> ;
    push(rt, l, mid, r);
    if(L <= mid) update(L, R, mul, lson);
    if(R > mid) update(L, R, mul, rson);
    pull(rt);
}
 
PLI query(int L, int R, int l, int r, int rt) {
    if(l >= L && r <= R) return mk(mask[rt], a[rt]);
    int mid = l + r >> ;
    push(rt, l, mid, r);
    PLI ans = mk(, );
    if(L <= mid) {
        PLI tmp = query(L, R, lson);
        ans.fi |= tmp.fi;
        ans.se = 1ll * ans.se * tmp.se % mod;
    }
    if(R > mid) {
        PLI tmp = query(L, R, rson);
        ans.fi |= tmp.fi;
        ans.se = 1ll * ans.se * tmp.se % mod;
    }
    return ans;
}
 
int main() {
    for(int i = ; i <= ; i++)
        if(ok(i)) prime[tot++] = i;
    for(int i = ; i < tot; i++)
        mul[i] = ( - Power(prime[i], mod - ) + mod) % mod;
    for(int i = ; i <= ; i++)
        for(int j = ; j < tot && prime[j] <= i; j++)
            if(i % prime[j] == ) S[i] |= 1ll << j;
    scanf("%d%d", &n, &q);
    for(int i = ; i <= n; i++) scanf("%d", &b[i]);
    build(, n, );
    while(q--) {
        scanf("%s", op);
        if(op[] == 'T') {
            int L, R; scanf("%d%d", &L, &R);
            PLI ret = query(L, R, , n, );
            int ans = ret.se; LL mask = ret.fi;
            for(int i = ; i < tot; i++)
                if(mask >> i & ) ans = 1ll * ans * mul[i] % mod;
            printf("%d\n", ans);
        } else {
            int L, R, x;
            scanf("%d%d%d", &L, &R, &x);
            update(L, R, x, , n, );
        }
    }
    return ;
}
 
/*
*/