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Please, another Queries on Array?(Codeforces Round #538 (Div. 2)F+线段树+欧拉函数+bitset）

2019-07-12 17:03:04 作者：互联网

题目链接

题面

思路

设\(x=\prod\limits_{i=l}^{r}a_i\)=\(\prod\limits_{i=1}^{n}p_i^{c_i}\)
由欧拉函数是积性函数得：
\[ \begin{aligned} \phi(x)&=\phi(\prod\limits_{i=1}^{n}p_i^{c_i})&\\ &=\prod\limits_{i=1}^{n}\phi(p_i^{c_i})&\\ &=\prod\limits_{i=1}^{n}p_i^{c_i}\times \frac{p_i-1}{p_i}&\\ &=x\times \prod\limits_{i=1}^{n}\frac{p_i-1}{p_i}& \end{aligned} \]
因此对于此题我们用线段树来维护区间乘积\(x\)和每个素数的是否存在。
由于小于等于\(300\)的素数只有\(62\)个，因此我们可以用一个\(long\) \(long\)变量来存，也可以用\(bitset\)写，第一次写这题的时候用的\(long\) \(long\)变量，这次暑训专题里面又遇到这个题目就用\(bitset\)写了一下当作是学习\(bitset\)的用法，发现\(bitset\)是真的省空间昂。

代码实现如下

\(long\) \(long\)变量写法

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
 
typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
 
#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
 
const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 4e5 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
 
char op[15];
LL inv[305];
LL ans1, ans2;
int p[305], isp[63];
int n, q, m, l, r, x;
 
LL qpow(LL x, int n) {
    LL res = 1;
    while(n) {
        if(n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}
 
void init() {
    for(int i = 2; i <= 300; i++) p[i] = 1;
    for(int i = 2; i * i <= 300; i++) {
        if(p[i]) {
            for(int j = i * i; j <= 300; j += i) {
                p[j] = 0;
            }
        }
    }
    for(int i = 2; i <= 300; i++) if(p[i]) isp[m++] = i;
    for(int i = 0; i < 62; i++) inv[i] = qpow(isp[i], mod - 2);
}
 
struct node {
    int l, r;
    LL lazy1, lazy2, sum, pp;
}segtree[maxn<<2];
 
void push_up(int rt) {
    segtree[rt].sum = segtree[lson].sum * segtree[rson].sum % mod;
    segtree[rt].pp = segtree[lson].pp | segtree[rson].pp;
}
 
void push_down(int rt) {
    LL x = segtree[rt].lazy1;
    (segtree[lson].lazy1 *= x) %= mod;
    (segtree[rson].lazy1 *= x) %= mod;
    (segtree[lson].sum *= qpow(x, segtree[lson].r - segtree[lson].l + 1)) %= mod;
    (segtree[rson].sum *= qpow(x, segtree[rson].r - segtree[rson].l + 1)) %= mod;
    segtree[rt].lazy1 = 1;
    x = segtree[rt].lazy2;
    segtree[lson].lazy2 |= x;
    segtree[rson].lazy2 |= x;
    segtree[lson].pp |= x;
    segtree[rson].pp |= x;
    segtree[rt].lazy2 = 0;
}
 
void build(int rt, int l, int r) {
    segtree[rt].l = l, segtree[rt].r = r;
    segtree[rt].sum = segtree[rt].pp = 0;
    segtree[rt].lazy1 = 1, segtree[rt].lazy2 = 0;
    if(l == r) {
        scanf("%lld", &segtree[rt].sum);
        for(int i = 0; i < 62; i++) {
            if(segtree[rt].sum % isp[i] == 0) segtree[rt].pp |= (1LL<<i);
        }
        return;
    }
    int mid = (l + r) >> 1;
    build(lson, l, mid);
    build(rson, mid + 1, r);
    push_up(rt);
}
 
void update(int rt, int l, int r, int x) {
    if(segtree[rt].l == l && segtree[rt].r == r) {
        (segtree[rt].sum *= qpow(x, segtree[rt].r - segtree[rt].l + 1)) %= mod;
        (segtree[rt].lazy1 *= x) %= mod;
        for(int i = 0; i < 62; i++) {
            if(x % isp[i] == 0) segtree[rt].pp |= (1LL<<i), segtree[rt].lazy2 |= (1LL<<i);
        }
        return;
    }
    push_down(rt);
    int mid = (segtree[rt].l + segtree[rt].r) >> 1;
    if(r <= mid) update(lson, l, r, x);
    else if(l > mid) update(rson, l, r, x);
    else {
        update(lson, l, mid, x);
        update(rson, mid + 1, r, x);
    }
    push_up(rt);
}
 
void query(int rt, int l, int r) {
    if(segtree[rt].l == l && segtree[rt].r == r) {
        (ans1 *= segtree[rt].sum) %= mod;
        ans2 |= segtree[rt].pp;
        return;
    }
    push_down(rt);
    int mid = (segtree[rt].l + segtree[rt].r) >> 1;
    if(r <= mid) query(lson, l, r);
    else if(l > mid) query(rson, l, r);
    else {
        query(lson, l, mid);
        query(rson, mid + 1, r);
    }
}
 
int main(){
    init();
    scanf("%d%d", &n, &q);
    build(1, 1, n);
    while(q--) {
        scanf("%s%d%d", op, &l, &r);
        if(op[0] == 'T') {
            ans1 = 1, ans2 = 0;
            query(1, l, r);
            for(int i = 0; i < 62; i++) {
                if(ans2 & (1LL<<i)) {
                    ans1 = ans1 * (isp[i] - 1) % mod * inv[i] % mod;
                }
            }
            printf("%lld\n", ans1);
        } else {
            scanf("%d", &x);
            update(1, l, r, x);
        }
    }
    return 0;
}

\(bitset\)写法

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 1000000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

char op[20];
bool v[305];
bitset<64> pp;
int cnt, n, q, l, r, x;
int p[65], inv[65];

int qpow(int x, int n) {
    int res = 1;
    while(n) {
        if(n & 1) res = 1LL * res * x % mod;
        x = 1LL * x * x % mod;
        n >>= 1;
    }
    return res;
}

void init() {
    for(int i = 2; i <= 300; ++i) {
        if(!v[i]) {
            p[cnt++] = i;
        }
        for(int j = 0; j < cnt && i * p[j] <= 300; ++j) {
            v[i*p[j]] = 1;
            if(i % p[j] == 0) break;
        }
    }
    for(int i = 0; i < cnt; ++i) inv[i] = qpow(p[i], mod - 2);
}

struct node {
    int l, r, mul, lazy1;
    bitset<64> b, lazy2;
}segtree[maxn<<2];

void push_up(int rt) {
    segtree[rt].mul = 1LL * segtree[lson].mul * segtree[rson].mul % mod;
    segtree[rt].b = segtree[lson].b | segtree[rson].b;
}

void push_down(int rt) {
    segtree[lson].lazy1 = 1LL * segtree[lson].lazy1 * segtree[rt].lazy1 % mod;
    segtree[rson].lazy1 = 1LL * segtree[rson].lazy1 * segtree[rt].lazy1 % mod;
    segtree[lson].mul = 1LL * segtree[lson].mul * qpow(segtree[rt].lazy1, segtree[lson].r - segtree[lson].l + 1) % mod;
    segtree[rson].mul = 1LL * segtree[rson].mul * qpow(segtree[rt].lazy1, segtree[rson].r - segtree[rson].l + 1) % mod;
    segtree[rt].lazy1 = 1;
    segtree[lson].b |= segtree[rt].lazy2;
    segtree[rson].b |= segtree[rt].lazy2;
    segtree[lson].lazy2 |= segtree[rt].lazy2;
    segtree[rson].lazy2 |= segtree[rt].lazy2;
    segtree[rt].lazy2.reset();
}

void build(int rt, int l, int r) {
    segtree[rt].l = l, segtree[rt].r = r;
    segtree[rt].lazy1 = 1, segtree[rt].lazy2.reset();
    segtree[rt].b.reset();
    if(l == r) {
        scanf("%d", &segtree[rt].mul);
        int x = segtree[rt].mul;
        for(int i = 0; i < cnt; ++i) {
            if(x % p[i] == 0) segtree[rt].b.set(i);
        }
        return;
    }
    int mid = (l + r) >> 1;
    build(lson, l, mid);
    build(rson, mid + 1, r);
    push_up(rt);
}

void update(int rt, int l, int r, int x) {
    if(segtree[rt].l == l && segtree[rt].r == r) {
        segtree[rt].mul = 1LL * segtree[rt].mul * qpow(x, segtree[rt].r - segtree[rt].l + 1) % mod;
        segtree[rt].lazy1 = 1LL * segtree[rt].lazy1 * x % mod;
        for(int i = 0; i < cnt; ++i) {
            if(x % p[i] == 0) segtree[rt].lazy2.set(i), segtree[rt].b.set(i);
        }
        return;
    }
    push_down(rt);
    int mid = (segtree[rt].l + segtree[rt].r) >> 1;
    if(r <= mid) update(lson, l, r, x);
    else if(l > mid) update(rson, l, r, x);
    else {
        update(lson, l, mid, x);
        update(rson, mid + 1, r, x);
    }
    push_up(rt);
}

int query(int rt, int l, int r) {
    if(segtree[rt].l == l && segtree[rt].r == r) {
        pp |= segtree[rt].b;
        return segtree[rt].mul;
    }
    push_down(rt);
    int mid = (segtree[rt].l + segtree[rt].r) >> 1;
    if(r <= mid) return query(lson, l, r);
    else if(l > mid) return query(rson, l, r);
    else return 1LL * query(lson, l, mid) * query(rson, mid + 1, r) % mod;
}

int main() {
#ifndef ONLINE_JUDGE
    FIN;
#endif // ONLINE_JUDGE
    init();
    scanf("%d%d", &n, &q);
    build(1, 1, n);
    while(q--) {
        scanf("%s%d%d", op, &l, &r);
        if(op[0] == 'M') {
            scanf("%d", &x);
            update(1, l, r, x);
        } else {
            pp.reset();
            int ans = query(1, l, r);
            for(int i = 0; i < cnt; ++i) {
                if(pp[i]) ans = 1LL * ans * (p[i] - 1) % mod * inv[i] % mod;
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}

标签：rt,int,Please,segtree,Codeforces,long,mid,Div,include
来源： https://www.cnblogs.com/Dillonh/p/11177076.html