hdu5468 Puzzled Elena(容斥 莫比乌斯反演)

hdu5468 Puzzled Elena

题意

求一棵子树内与它互质的点个数

解法

容斥

我们先求出与它不互质的数的个数，再用总数减去就好。

#include <cstdio>#include <cstring>#include <iostream>#include <vector>using namespace std;namespace Input {    int a; char c; bool sign;    inline int geti() {        sign = false;        while ((c = getchar()) < '0' || c > '9') sign |= c == '-';        a = c - '0';        while ((c = getchar()) >= '0' && c <= '9') a = (a << 3) + (a << 1) + c - '0';        return sign ? -a : a;    }}const int N = 1e5 + 5;vector<int> edge[N], Num[N], ty[N];int Cnt[N], Val[N], ans[N], ch[N][70];void init() {    memset(Cnt, 1, sizeof Cnt);    int i, j, cnt, len, t, k; cnt = 0;    for (i = 0; i < N; ++i) Num[i].clear(), ty[i].clear();    for (i = 2; i < N; ++i) {        if (Cnt[i]) {            for (j = i; j < N; j += i)              Cnt[j] = 0, Num[j].push_back(i), cnt += j == 4;        }    }    vector<int>tmp;    for (i = 2; i < N; ++i) {        tmp.clear();        for (j = 0; j < Num[i].size(); ++j)          tmp.push_back(Num[i][j]);        len = tmp.size(); Num[i].clear();        for (j = 1; j < (1 << len); ++j) {            cnt = 0, t = 1;            for (k = 0; k < len; ++k)              if (j & (1 << k)) {                  ++cnt; t *= tmp[k];              }            if (cnt & 1) ty[i].push_back(-1);            else ty[i].push_back(1);            Num[i].push_back(t);        }    }}int dfs(int u, int fa) {    int si = 0, va = Val[u], i, v;    for (i = 0; i < Num[va].size(); ++i)      ch[u][i] = Cnt[Num[va][i]];    for (i = 0; i < edge[u].size(); ++i) {        v = edge[u][i];        if (v == fa) continue;        si += dfs(v, u);    }    ans[u] = si;    for (i = 0; i < Num[va].size(); ++i)      ans[u] += Cnt[Num[va][i]] - ch[u][i];    for (i = 0; i < Num[va].size(); ++i)      Cnt[Num[va][i]] += ty[va][i];    if (va == 1) ++ans[u];    return si + 1;}int main() {    init();    int Case = 0, n, u, v, i;    while (scanf("%d", &n) ^ EOF) {        for (i = 1; i <= n; ++i) edge[i].clear();        for (i = 1; i <  n; ++i) {            u = Input::geti(), v = Input::geti();            edge[u].push_back(v), edge[v].push_back(u);        }        memset(Cnt, 0, sizeof Cnt);        for (i = 1; i <= n; ++i) Val[i] = Input::geti();        dfs(1, 0);        printf("Case #%d:", ++Case);        for (i = 1; i <= n; ++i)          printf(" %d", ans[i]);        puts("");    }    return 0;}

莫比乌斯反演

此题其实也可以用莫比乌斯反演做，不过其实与容斥差不多，因为mu[i]其实与ty[i]是一样的。
代码就不贴了，其实比较像。