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


    题目链接

    传送门

    题面

    思路

    (x=prodlimits_{i=l}^{r}a_i)=(prodlimits_{i=1}^{n}p_i^{c_i})
    由欧拉函数是积性函数得:

    [egin{aligned} phi(x)&=phi(prodlimits_{i=1}^{n}p_i^{c_i})&\ &=prodlimits_{i=1}^{n}phi(p_i^{c_i})&\ &=prodlimits_{i=1}^{n}p_i^{c_i} imes frac{p_i-1}{p_i}&\ &=x imes prodlimits_{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("*********
    ")
    #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
    ", 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("*********
    ")
    #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
    ", ans);
            }
        }
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/Dillonh/p/11177076.html
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