传送门：https://www.nowcoder.com/acm/contest/203/B
思路及参考：https://blog.csdn.net/u013534123/article/details/82934820
这篇blog写得非常详细，但是我不会他说的立flag法，就学了其他同学的做法，如果不能做除法，就直接计数。我想了比较久明白的写在注释里啦。
//#pragma GCC optimize(3) //#pragma comment(linker, "/STACK:102400000,102400000") //c++ #include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <iomanip> #include <cstdlib> #include <cstdio> #include <string> #include <vector> #include <bitset> #include <cctype> #include <queue> #include <cmath> #include <list> #include <map> #include <set> using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int ,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' #define OKC ios::sync_with_stdio(false);cin.tie(0) #define FT(A,B,C) for(int A=B;A <= C;++A) //用来压行 #define REP(i , j , k) for(int i = j ; i < k ; ++i) //priority_queue<int ,vector<int>, greater<int> >que; const ll mos = 0x7FFFFFFFLL; //2147483647 const ll nmos = 0x80000000LL; //-2147483648 const int mod = 1e9+7; const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18 const double PI=acos(-1.0); template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } // #define _DEBUG; //*// #ifdef _DEBUG freopen("input", "r", stdin); // freopen("output.txt", "w", stdout); #endif /*-----------------------show time----------------------*/ const int maxn = 1e6+9; vector<int>mp[maxn]; ll dp[maxn],ans[maxn],pa[maxn],la[maxn]; ll ksm(ll a, ll b){ ll res = 1; while(b > 0){ if(b&1) res = res * a % mod; a = a * a % mod; b >>= 1; } return res; } void dfs1(int u,int fa){ dp[u] = 1; pa[u] = fa; for(int i=0; i<mp[u].size(); i++){ int v = mp[u][i]; if(v == fa)continue; dfs1(v,u); dp[u] = 1ll*dp[u] * (dp[v]+1) % mod; } } void dfs2(int u,int fa){ if(fa == -1)ans[u] = dp[u]; else { if((dp[u]+1)%mod == 0){ //这里是暴力计算的。 ans[u] = la[fa] + 1; //la数组记录，通过fa到u但不包括fa及子树的点的影响。 for(int i=0; i<mp[fa].size(); i++){ int v = mp[fa][i]; if(v == u || v == pa[fa])continue; ans[u] = 1ll*ans[u] * (dp[v] + 1) % mod; } } else ans[u] = 1ll*ans[fa] * ksm(dp[u]+1,mod-2) % mod; la[u] = ans[u]; ans[u] = 1ll*(ans[u]+1) * dp[u] % mod; } for(int i=0; i<mp[u].size(); i++){ int v = mp[u][i]; if(fa == v)continue; dfs2(v, u); } } int main(){ int n; scanf("%d", &n); for(int i=1; i<n; i++){ int u,v; scanf("%d%d", &u, &v); mp[u].pb(v); mp[v].pb(u); } dfs1(1,-1); dfs2(1,-1); for(int i=1; i<=n; i++){ printf("%lld\n", ans[i]); } return 0; }