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 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int n, q, tot, b[N], prime[301], mul[301];
LL S[301];
char op[20];

bool ok(int x) {
    for(int i = 2; i * i <= x; i++)
        if(x % i == 0) return false;
    return true;
}

int Power(int a, int b) {
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1ll * ans * a % mod;
        a = 1ll * a * a % mod; b >>= 1;
    }
    return ans;
}

int a[N << 2], lazy[N << 2];
LL mask[N << 2], mlazy[N << 2];


#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1

inline void pull(int rt) {
    a[rt] = 1ll * a[rt << 1] * a[rt << 1 | 1] % mod;
    mask[rt] = mask[rt << 1] | mask[rt << 1 | 1];
}

void push(int rt, int l, int mid, int r) {
    if(lazy[rt] != 1) {
        a[rt << 1] = 1ll * a[rt << 1] * Power(lazy[rt], mid - l + 1) % mod;
        a[rt << 1 | 1] = 1ll * a[rt << 1 | 1] * Power(lazy[rt], r - mid) % mod;
        lazy[rt << 1] = 1ll * lazy[rt << 1] * lazy[rt] % mod;
        lazy[rt << 1 | 1] = 1ll * lazy[rt << 1 | 1] * lazy[rt] % mod;
        lazy[rt] = 1;
    }
    if(mlazy[rt] != 0) {
        mask[rt << 1] |= mlazy[rt]; mask[rt << 1 | 1] |= mlazy[rt];
        mlazy[rt << 1] |= mlazy[rt]; mlazy[rt << 1 | 1] |= mlazy[rt];
        mlazy[rt] = 0;
    }
}

void build(int l, int r, int rt) {
    lazy[rt] = 1;
    mlazy[rt] = 0;
    if(l == r) {
        a[rt] = b[l];
        mask[rt] = S[b[l]];
        return;
    }
    int mid = l + r >> 1;
    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 + 1) % mod;
        lazy[rt] = 1ll * lazy[rt] * mul % mod;
        mask[rt] |= S[mul];
        mlazy[rt] |= S[mul];
        return;
    }
    int mid = l + r >> 1;
    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 >> 1;
    push(rt, l, mid, r);
    PLI ans = mk(0, 1);
    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 = 2; i <= 300; i++)
        if(ok(i)) prime[tot++] = i;
    for(int i = 0; i < tot; i++)
        mul[i] = (1 - Power(prime[i], mod - 2) + mod) % mod;
    for(int i = 2; i <= 300; i++)
        for(int j = 0; j < tot && prime[j] <= i; j++)
            if(i % prime[j] == 0) S[i] |= 1ll << j;
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; i++) scanf("%d", &b[i]);
    build(1, n, 1);
    while(q--) {
        scanf("%s", op);
        if(op[0] == 'T') {
            int L, R; scanf("%d%d", &L, &R);
            PLI ret = query(L, R, 1, n, 1);
            int ans = ret.se; LL mask = ret.fi;
            for(int i = 0; i < tot; i++)
                if(mask >> i & 1) ans = 1ll * ans * mul[i] % mod;
            printf("%d
", ans);
        } else {
            int L, R, x;
            scanf("%d%d%d", &L, &R, &x);
            update(L, R, x, 1, n, 1);
        }
    }
    return 0;
}

/*
*/