题意：给定一棵n个点带点权的树，i号点的点定义f(i,j)为i到j路径上所有点的gcd，其中i是j的一个祖先，求所有f(i,j)之和mod1e9+7
2<=n<=1e5，0<=a[i]<=1e12
思路：从根往下直接暴力跑，每个点开个vector继承父节点的信息，取gcd之后再把值相等的合并
大胆猜想，不用求证
注意0还是要保留的
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned int uint; 5 typedef unsigned long long ull; 6 typedef pair<int,int> PII; 7 typedef pair<ll,ll> Pll; 8 typedef vector<int> VI; 9 typedef vector<PII> VII; 10 typedef pair<ll,ll>P; 11 #define N 200010 12 #define M 200010 13 #define fi first 14 #define se second 15 #define MP make_pair 16 #define pi acos(-1) 17 #define mem(a,b) memset(a,b,sizeof(a)) 18 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++) 19 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--) 20 #define lowbit(x) x&(-x) 21 #define Rand (rand()*(1<<16)+rand()) 22 #define id(x) ((x)<=B?(x):m-n/(x)+1) 23 #define ls p<<1 24 #define rs p<<1|1 25 26 const ll MOD=1e9+7,inv2=(MOD+1)/2; 27 double eps=1e-4; 28 int INF=1<<30; 29 ll inf=5e13; 30 int dx[4]={-1,1,0,0}; 31 int dy[4]={0,0,-1,1}; 32 33 int head[N],vet[N],nxt[N],fa[N],d[N],vis[N],tot; 34 ll a[N],ans; 35 P b[N]; 36 vector<P> c[N]; 37 38 int read() 39 { 40 int v=0,f=1; 41 char c=getchar(); 42 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 43 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 44 return v*f; 45 } 46 47 void add(int a,int b) 48 { 49 nxt[++tot]=head[a]; 50 vet[tot]=b; 51 head[a]=tot; 52 } 53 54 ll gcd(ll x,ll y) 55 { 56 if(!y) return x; 57 return gcd(y,x%y); 58 } 59 60 void dfs(int u,int fa) 61 { 62 //printf("u=%d fa=%d ",u,fa); 63 if(fa) 64 { 65 for(int i=0;i<c[fa].size();i++) c[u].push_back(c[fa][i]); 66 //c[u]=c[fa]; 67 } 68 69 for(int i=0;i<c[u].size();i++) c[u][i].fi=gcd(c[u][i].fi,a[u]); 70 int m=0; 71 for(int i=0;i<c[u].size();i++) b[++m]=c[u][i]; 72 sort(b+1,b+m+1); 73 c[u].clear(); 74 rep(i,1,m) 75 { 76 int now=(int)c[u].size()-1; 77 if(now==-1||b[i].fi!=c[u][now].fi) c[u].push_back(b[i]); 78 else c[u][now].se=(c[u][now].se+b[i].se)%MOD; 79 } 80 c[u].push_back(MP(a[u],1)); 81 for(int i=0;i<c[u].size();i++) ans=(ans+c[u][i].fi%MOD*c[u][i].se%MOD)%MOD; 82 int e=head[u]; 83 while(e) 84 { 85 int v=vet[e]; 86 if(v!=fa) dfs(v,u); 87 e=nxt[e]; 88 } 89 //printf("u=%d fa=%d ans=%I64d ",u,fa,ans); 90 } 91 92 93 94 int main() 95 { 96 //freopen("1.in","r",stdin); 97 int n=read(); 98 rep(i,1,n) scanf("%I64d",&a[i]); 99 tot=0; 100 rep(i,1,n) head[i]=0; 101 rep(i,1,n-1) 102 { 103 int x=read(),y=read(); 104 add(x,y); 105 add(y,x); 106 } 107 dfs(1,0); 108 printf("%I64d ",ans); 109 return 0; 110 }