hdu5468（容斥原理）-2015 ACM/ICPC Asia Regional Shanghai Online

题意：给一颗以1为根的树，以每一个节点为子树，计算该子树的所有节点value与该节点value互质的节点个数。

假设第i个节点value为vi：

1、由莫比乌斯函数容斥求互质个数，显然需要预处理vi的无平方素数的因子（miu[vi] !=0的因子）；

ans[i] = sigma( miu[d] * count[d] )，d为vi的无平方素数因子，count[d]为以第i个节点为根的子树中value为d的倍数的个数.

2、由上面的公式，容斥的时候需要知道count[d], 所以在遍历当前子树儿子节点之前统计所有的count1[d],在遍历完儿子节点之后统计所有的count2[d]，

显然，count2[d]-count1[d]即为上式中的count[d].

ps:子树中vi自身也算，当且仅当vi=1的时候成立.

参考代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<vector>#include<set>#include<map>#include<queue>#include<sstream>#include<string>#include<bitset>using namespace std;typedef long long LL;const LL LINF = (1LL <<63);const int INF = 1 << 31;const int NS = 100010;const int MS = 19;const int MOD = 1000000007;const int EDGE_MAX = NS;struct graphEdge{    int pst;    int next;};struct ForwardStart{    int top;    int head[EDGE_MAX];    graphEdge edge[EDGE_MAX << 1];    void init(int len)    {        top = 0;        memset(head, -1, sizeof(int) * len);    }    void addEdge(int u, int v)    {        edge[top].pst = v;        edge[top].next = head[u];        head[u] = top++;    }    void printAll()    {        printf("top = %d\n", top);        for(int i = 1; i < EDGE_MAX; i++)        {            if(-1 != head[i])            {                printf("head[%2d]'son:%2d", i, edge[head[i]].pst);                for(int j = edge[head[i]].next; j != -1; j = edge[j].next)                {                    printf(",%2d", edge[j].pst);                }                puts("");            }        }    }}cTree;bitset<NS> isPrime;vector<int> fac[NS];int miu[NS];void prepare(){    isPrime.set();    isPrime[1] = false;    miu[1] = 1;    for(int i = 2; i < NS; i++)    {        if(isPrime[i])        {            for(int j = i; j < NS; j+=i)            {                isPrime[j] = false;                int k = j / i;                if(k % i)                {                    miu[j] = -miu[k];                }                else                {                    miu[j] = 0;                }                fac[j].push_back(i);            }        }        else        {            if(miu[i] != 0)            {                for(int j = i; j < NS; j+=i)                {                    fac[j].push_back(i);                }            }        }    }}int n;int val[NS];int ans[NS];int sz[NS];int dp[NS];void dfs(int rt, int fa){    vector<int> temp;    sz[rt] = 1;    int value = val[rt];    int len = fac[value].size();    for(int i = 0; i < len; i++)    {        int d = fac[value][i];        int cnt = dp[d];        temp.push_back(cnt);        dp[d] += 1;    }    for(int i = cTree.head[rt]; i != -1; i = cTree.edge[i].next)    {        int cson = cTree.edge[i].pst;        if(cson == fa) continue;        dfs(cson, rt);        sz[rt] += sz[cson];    }    ans[rt] = sz[rt];    for(int i = 0; i < len; i++)    {        int d = fac[value][i];        int cnt = dp[d] - temp[i];        if(cnt > 0)        {            ans[rt] += miu[d] * cnt;        }    }    return ;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);#endif    prepare();    int nCase = 1;    while(~scanf("%d", &n))    {        cTree.init(n + n + 2);        int u, v;        for(int i = 1; i < n; i++)        {            scanf("%d %d", &u, &v);            cTree.addEdge(u, v);            cTree.addEdge(v, u);        }        for(int i = 1; i <= n; i++)        {            scanf("%d", &val[i]);            if(val[i] < 0)            {                val[i] = - val[i];            }        }        memset(dp, 0, sizeof(dp));        dfs(1, -1);        printf("Case #%d:", nCase++);        for(int i = 1; i <= n; i++)        {            printf(" %d", ans[i]);        }        printf("\n");    }    return 0;}