Codeforces Round #485 (Div. 1) E. Prince's Problem

原题链接.

题目大意：

给出一棵树。

每个点有一个值a。

每次询问给出u,v,x,问$\prod_{p在u到v的路径上}~~~~~~gcd(a[p],x)$

题解：

显然每个质因子互不影响，所以分开考虑。

路径在lca处拆掉没毛病。

gcd就是指数取min。

发现不好考虑。

排序啊，一起做，用个树状数组搞搞dfs序就行了。

说得简单，打起来hhh…

Code:

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#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
#define low(a) ((a) & -(a))
#define fo(i, x, y) for(int i = x; i <= y; i ++)
#define fd(i, x, y) for(int i = x; i >= y; i --)
using namespace std;

const ll mo = 1e9 + 7;

const int N = 2e5 + 5;

int n, x, y;
struct node {
    int final[N], to[N], next[N], tot;
    void link(int x, int y) {
        next[++ tot] = final[x], to[tot] = y, final[x] = tot;
    }
} e;

int bz[N], f[18][N / 2], dep[N], p[N], q[N], tp;
void dg(int x) {
    bz[x] = 1;
    p[x] = ++ tp;
    for(int i = e.final[x]; i; i = e.next[i]) {
        int y = e.to[i]; if(bz[y]) continue;
        f[0][y] = x; dep[y] = dep[x] + 1;
        dg(y);
    }
    q[x] = tp;
    bz[x] = 0;
}

int lca(int x, int y) {
    if(dep[x] < dep[y]) swap(x, y);
    fd(i, 17, 0) if(dep[f[i][x]] >= dep[y]) x = f[i][x];
    if(x == y) return x;
    fd(i, 17, 0) if(f[i][x] != f[i][y]) x = f[i][x], y = f[i][y];
    return f[0][x];
}

const int M = 1e7;
int bx[M + 5], pe[M], t[M + 5];
void Shai() {
    fo(i, 2, M) {
        if(!bx[i]) pe[++ pe[0]] = i, t[i] = i;
        fo(j, 1, pe[0]) {
            int k = i * pe[j];
            if(k > M) break;
            bx[k] = 1; t[k] = pe[j];
            if(i % pe[j] == 0) break;
        }
    }
}

int u[N], v[N];
void fen(int x) {
    u[0] = 0;
    while(x > 1) {
        u[++ u[0]] = t[x]; v[u[0]] = 0;
        int y = t[x]; while(x % y == 0) v[u[0]] ++, x /= y;
    }
}

int a[N], Q;

struct Ask {
    int x, y, z;
} b[N];

const int C = 5000005;
struct zhi {
    int final[M], to[C], len[C], next[C], bl[C], le[C], tot;
    void link(int x, int y, int z, int u, int v) {
        if(y == 0) return;
        next[++ tot] = final[x], to[tot] = y, len[tot] = z, bl[tot] = u, le[tot] = v, final[x] = tot;
    }
} c;

struct node2 {
    int len, x, bl, le;
} d[M];

int cmp(node2 a, node2 b) {
    return a.len < b.len;
}

int g[N];

ll ksm(ll x, ll y) {
    ll s = 1;
    for(; y; y /= 2, x = x * x % mo)
        if(y & 1) s = s * x % mo;
    return s;
}

ll ans[N];

int main() {
//  freopen("e.in", "r", stdin);
    scanf("%d", &n);
    fo(i, 1, n - 1) {
        scanf("%d %d", &x, &y);
        e.link(x, y); e.link(y, x);
    }
    dep[1] = 1; dg(1);
    fo(i, 1, 17) fo(j, 1, n) f[i][j] = f[i - 1][f[i - 1][j]];

    Shai();
    fo(i, 1, n) {
        scanf("%d", &a[i]);
        fen(a[i]);
        fo(j, 1, u[0]) {
            c.link(u[j], i, v[j], 0, 0);
        }
    }
    scanf("%d", &Q);
    fo(i, 1, Q) {
        ans[i] = 1;
        scanf("%d %d %d", &b[i].x, &b[i].y, &b[i].z);
        fen(b[i].z);
        int z = lca(b[i].x, b[i].y);

        fo(j, 1, u[0]) {
            c.link(u[j], b[i].x, v[j], i, 1);
            c.link(u[j], b[i].y, v[j], i, 1);
            c.link(u[j], z, v[j], i, -1);
            c.link(u[j], f[0][z], v[j], i, -1);
        }
    }
    fo(ii, 2, M) if(t[ii] == ii) {
        int d0 = 0;
        for(int j = c.final[ii]; j; j = c.next[j]) {
            d[++ d0].len = c.len[j];
            d[d0].x = c.to[j];
            d[d0].bl = c.bl[j];
            d[d0].le = c.le[j];
        }
        sort(d + 1, d + d0 + 1, cmp);
        fo(i, 1, d0) {
            if(!d[i].bl) {
                x = p[d[i].x];
                while(x <= n) g[x] += d[i].len, x += low(x);
                x = q[d[i].x] + 1;
                while(x <= n) g[x] -= d[i].len, x += low(x);
            } else {
                x = p[d[i].x];
                int s = 0;
                while(x) s += g[x], x -= low(x);
                if(d[i].le == 1) ans[d[i].bl] = ans[d[i].bl] * ksm(ii, s) % mo; else ans[d[i].bl] = ans[d[i].bl] * ksm(ksm(ii, s), mo - 2) % mo;
            }
        }
        fo(i, 1, d0) if(!d[i].bl) {
            x = p[d[i].x];
            while(x <= n) g[x] -= d[i].len, x += low(x);
            x = q[d[i].x] + 1;
            while(x <= n) g[x] += d[i].len, x += low(x);
        }
        fd(i, d0, 1) {
            int s = 0;
            if(!d[i].bl) {
                x = p[d[i].x];
                while(x <= n) g[x] ++, x += low(x);
                x = q[d[i].x] + 1;
                while(x <= n) g[x] --, x += low(x);
            } else {
                x = p[d[i].x];
                int s = 0;
                while(x) s += g[x], x -= low(x);
                if(d[i].le == 1) ans[d[i].bl] = ans[d[i].bl] * ksm(ii, s * d[i].len) % mo; else ans[d[i].bl] = ans[d[i].bl] * ksm(ksm(ii, s * d[i].len), mo - 2) % mo;
            }
        }
        fd(i, d0, 1) if(!d[i].bl) {
            x = p[d[i].x];
            while(x <= n) g[x] --, x += low(x);
            x = q[d[i].x] + 1;
            while(x <= n) g[x] ++, x += low(x);
        }
    }
    fo(i, 1, Q) printf("%I64d\n", ans[i]);
}