HDU-2017 多校训练赛6-1007-GCDispower

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ACM模版

描述

题解

最讨厌莫比乌斯反演之类的问题了……不想说啥了，自己好笨的。

ZLH_HHHH 的博客，这里讲得还好了，可以看看。

莫比乌斯反演 + 树状数组 + 离线。

代码

#include <algorithm>#include <string.h>#include <stdio.h>#include <vector>using namespace std;typedef long long ll;const int MAXN = 110005;struct array        //  树状数组{    ll A[MAXN];    int n;    void clear()    {        fill(A, A + n + 5, 0);    }    array()    {        memset(A, 0, sizeof A);        n = 0;    }    int lowbit(int x)    {        return x & (-x);    }    void add(int x, ll key)    {        while (x <= n)        {            A[x] += key;            x += lowbit(x);        }    }    ll sum(int x)   //  查询前x个元素    {        ll tmp = 0;        while (x > 0)        {            tmp += A[x];            x -= lowbit(x);        }        return tmp;    }    ll get(int L, int R)    {        return sum(R) - sum(L - 1);    }} A;struct node{    int L, R, id;    node(int L, int R, int id) : L(L), R(R), id(id) {}    node()    {        *this = node(0, 0, 0);    }    bool operator < (const node &A) const    {        return R < A.R;    }} Q[MAXN];              //  离线问题int miu[MAXN];          //  莫比乌斯函数vector<int> V[MAXN];    //  用于存储因式分解。void get_miu(){    int n = MAXN;    for (int i = 1; i < n; i++)    {        int tar = (i == 1) ? 1 : 0;        int tmp = tar - miu[i];        miu[i] = tmp;        for (int j = i + i; j < n; j += i)        {            miu[j] += tmp;        }    }}void init(){    get_miu();    for (int i = 1; i < MAXN; i++)    {        for (int j = i; j < MAXN; j += i)        {            V[j].push_back(i);        }    }}int N, M, L, R;int F[MAXN];int X[MAXN];int P[MAXN];int D[MAXN];ll ans[MAXN];void solve(){    int cnt = 0;    for (int i = 1; i < N; i++)    {        int size = N / P[i] + 1;        fill(F, F + size + 1, 0);        int deep = 0;        for (int j = P[i]; j < N; j += P[i])        {            if (X[j] > i)            {                continue;            }            D[deep++] = X[j];        }        sort(D, D + deep);        deep--;        while ((--deep) > -1)        {            int x = D[deep];            int t = P[x] / P[i];            ll res = 0;            for (int d = 0; d < V[t].size(); d++)            {                res += (ll)miu[V[t][d]] * F[V[t][d]];            }            A.add(x, res * P[i]);            for (int d = 0; d < V[t].size(); d++)            {                F[V[t][d]]++;            }        }        while (Q[cnt].R == i && cnt < M)        {            ans[Q[cnt].id] = A.get(Q[cnt].L, Q[cnt].R);            cnt++;        }    }}int main (){    init();    int T;    scanf("%d", &T);    while (T--)    {        A.clear();        scanf("%d%d", &N, &M);        A.n = N;        N++;        for (int i = 1; i < N; i++)        {            scanf("%d", P + i);        }        for (int i = 1; i < N; i++)        {            X[P[i]] = i;        }        for (int i = 0; i < M; i++)        {            scanf("%d%d", &L, &R);            Q[i] = node(L, R, i);        }        sort(Q, Q + M);        solve();        for (int i = 0; i < M; i++)        {            printf("%lld\n", ans[i]);        }    }    return 0;}

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