hdu5468 Puzzled Elena

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。