UVA 11174 Stand in a Line——组合数+取模

要求（b / c） % mod,

令a = b / c;则a % mod = b * (c ^ (mod - 2)) % mod;

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>using namespace std;const long long mod = 1000000007;const int maxn = 400005;int T, n, m, tot, head[maxn], num[maxn], fa[maxn];long long f[maxn];struct Edge {    int to, next;}edge[maxn<<1];void init() {    tot = 0; memset(head, -1, sizeof(head));}void addedge(int u, int v) {    edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++;}void dfs(int u, int p) {    for (int i = head[u]; ~i; i = edge[i].next) {        int v = edge[i].to;        if (v == p) continue;        dfs(v, u);        num[u] += num[v];    }}long long mpow(long long x, long long y) {    long long ans = 1;    while (y) {        if (y & 1) ans = ans * x % mod;        x = x * x % mod;        y >>= 1;    }    return ans;}int main() {    f[0] = 1;    for (int i = 1; i < maxn; i++) {        f[i] = (f[i-1] * i) % mod;    }    scanf("%d", &T);    for (int kase = 1; kase <= T; kase++) {        scanf("%d %d", &n, &m);        init();        num[0] = 0;        for (int i = 1; i <= n; i++) num[i] = 1;        memset(fa, -1, sizeof(fa));        for (int i = 1; i <= m; i++) {            int u, v; scanf("%d %d", &u, &v);            fa[u] = v;            addedge(v, u);        }        for (int i = 1; i <= n; i++) {            if (fa[i] == -1) addedge(0, i);        }        dfs(0, -1);        long long ans = 1;        for (int i = 1; i <= n; i++) {            ans = (ans * num[i]) % mod;        }        ans = (f[n] * mpow(ans, mod-2)) % mod;        printf("%lld\n", ans);    }    return 0;}