首先赞一下题目， 好题

题意：

Marjar University has decided to upgrade the infrastructure of school intranet by using fiber-optic technology. There are N buildings in the school. Each building will be installed with one router. These routers are connected by optical cables in such a way that there is exactly one path between any two routers.

Each router should be initialized with an operating frequency F_i before it starts to work. Due to the limitations of hardware and environment, the operating frequency should be an integer number within [L_i, R_i]. In order to reduce the signal noise, the operating frequency of any two adjacent routers should be co-prime.

Edward is the headmaster of Marjar University. He is very interested in the number of different ways to initialize the operating frequency. Please write a program to help him! To make the report simple and neat, you only need to calculate the sum of F_i (modulo 1000000007) in all solutions for each router.

英文自己看， 大概意思就是一棵树上每个点i可以给一个L[i] ~ R[i]的值， 相邻的两个点的值要互质， 问每个点的所有情况的值的和， mod 1000000007

大概思路就是枚举一个点， 然后枚举这个点上的值(i)， 然后求出这样的情况有多少种(dp[i])， 那么这个点上的答案就是 dp[1] * 1 + dp[2] * 2 + dp[3] * 3 + .........

然后就是树形dp了， 转移方程就是 dp[i][j] =π{(∑{dp[t][k] | gcd(j, k) == 1}) | i 和 t 相邻}

但是这样转移的复杂度是50 * 50000 * 50000， 就算15秒时限也会超时

所以我们可以考虑用不互质的来转移；

设s[i] = ∑dp[i][j]

那么转移方程就是   dp[i][j] =π{(s[i] - ∑{dp[t][k] | gcd(j, k) != 1}) | i 和 t 相邻}

对于dp[i][j]，我们可以把 j 质因数分解， 假设 j = p1^e1 * p2^e2 * p3^e3， 50000以内的数最多有6个不同的质因数；

然后我们记录一下 div[i][k] = ∑{ dp[i][j] | j 是 k 的倍数}， 这个可以nlogn的复杂度处理出来；

这样  ∑{dp[t][k] | gcd(j, k) != 1} = div[t][p1] + div[t][p2] + div[t][p3] - div[t][p1 * p2] - div[t][p1 * p3] - div[t][p2 * p3] + div[t][p1 * p2 * p3]， 这样可以用容斥原理算了， 复杂度最多为2 ^ 6；

这样dp一次复杂度大概就是 50 * 50000 * (log50000 + 2^6)；

要算50个点的话， 还是会超时；

但是这是一颗树， 对每个点都dp一次的话算了很多重复的东西， 所以我们不要每次都去全部dp一次， 例如算完i点的了， 要去算j点的， 假设i j相邻， 那么在dp数组中只有i和j的值有变化， 我们就只要再算这两个点的dp转移就够了。

更多细节请看代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <vector>

using namespace std;

typedef long long LL;
const int N = 50009;
const LL M = 1000000007;

inline void addIt(int &a, int b)
{
    a += b;
    if(a >= M) a -= M;
}

inline int sub(int a, int b)
{
    a -= b;
    if(a < 0) a += M;
    if(a >= M) a -= M;
    return a;
}

struct Num
{
    int p[11];
    int allp;
}num[N];

struct Data
{
    int dp[N], div[N], all;
}data[55], fb[55];

int L[55], R[55];
int ans[55];
int n;
vector<int> e[55];
bool vis[55];
int rcn, rcv;


void print()
{
    for(int i = 0; i < n; i++)
    {
        printf("i = %d\n", i);
        for(int j = 0; j < 6; j++) printf("dp[%d] = %d\n", j, data[i].dp[j]);
    }
}

int rc(int i, int xs)
{
    int t, r = 0;
    for(; i < num[rcn].allp; i++)
    {
        if(xs * num[rcn].p[i] <= R[rcv]) t = data[rcv].div[xs * num[rcn].p[i]];
        else t = 0;
        //printf("***%d, t = %d\n", xs * num[rcn].p[i], t);
        addIt(r, sub(t, rc(i + 1, xs * num[rcn].p[i])));
    }
    //printf("r = %d\n", r);
    return r;
}

void dfsTree(int pre, int now)
{
    if(vis[now]) return;
    vis[now] = true;
    int i, siz = e[now].size();
    for(i = 0; i < siz; i++) dfsTree(now, e[now][i]);
    if(pre >= 0 && siz <= 1) for(i = L[now]; i <= R[now]; i++) data[now].dp[i] = 1;
    else for(i = L[now]; i <= R[now]; i++)
    {
        data[now].dp[i] = 1;
        for(int j = 0; j < siz; j++) if(e[now][j] != pre)
        {
            rcn = i;
            rcv = e[now][j];
            data[now].dp[i] = (LL)(data[now].dp[i]) * sub(data[e[now][j]].all, rc(0, 1)) % M;
        }
    }
    data[now].all = 0;
    for(i = 1; i <= R[now]; i++)
    {
        data[now].div[i] = 0;
        for(int j = i; j <= R[now]; j += i) if(j >= L[now]) addIt(data[now].div[i], data[now].dp[j]);
        if(i >= L[now]) addIt(data[now].all, data[now].dp[i]);
    }
}

void dfs(int pre, int now, int deep)
{
    dfsTree(-1, now);
    //if(now == 1) print();
    int i, siz = e[now].size();
    ans[now] = 0;
    for(i = L[now]; i <= R[now]; i++) addIt(ans[now], (LL)data[now].dp[i] * i % M);
    //fb[deep] = data[now];
    //vis[now] = false;
    for(i = 0; i < siz; i++) if(e[now][i] != pre)
    {
        vis[e[now][i]] = false;
        fb[deep] = data[e[now][i]];
        vis[now] = false;
        dfs(now, e[now][i], deep + 1);
        data[e[now][i]] = fb[deep];
        vis[e[now][i]] = true;
    }
}

void init()
{
    int i, j, k;
    for(i = 0; i < N; i++) num[i].allp = 0;
    for(i = 2; i < N; i++) if(num[i].allp == 0) for(j = i; j < N; j += i) num[j].p[num[j].allp++] = i;
}

int main()
{
    //freopen("13F.in", "r", stdin);
    init();
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        int i, j, k;
        for(i = 0; i < n; i++)
        {
            scanf("%d", &L[i]);
        }
        for(i = 0; i < n; i++)
        {
            scanf("%d", &R[i]);
            e[i].clear();
        }
        for(i = 0; i < n - 1; i++)
        {
            scanf("%d %d", &j, &k);
            j--;
            k--;
            e[j].push_back(k);
            e[k].push_back(j);
        }
        memset(vis, false, sizeof(vis));
        memset(data, 0, sizeof(data));
        dfsTree(-1, 0);
        //print();
        dfs(-1, 0, 0);
        for(i = 0; i < n - 1; i++) printf("%d ", ans[i]); printf("%d\n", ans[i]);
    }
    return 0;
}

2014牡丹江 现场赛 F zoj 3824 Fiber-optic Network