2017-05-25T20:24:07+08:00 Typecho /feed/atom/category/cf/ /archives/236/ 2017-05-25T20:24:07+08:00 2017-05-25T20:24:07+08:00 admin

###Description 给一棵树，每个点点权 $$a_i$$，保证 $$a_i$$各不相同，现在随机选两个点 $$u,v$$，求 $$f(u,v)=\phi(a_u\cdot a_v)\cdot dis(u,v)$$ 的期望，$$mod\ 10^9+7$$。 链接：[http://codeforces.com/contest/809/problem/E](http://codeforces.com/contest/809/problem/E "http://codeforces.com/contest/809/problem/E") ###Solution 首先，有一个结论：$$\phi(a\cdot b)=\frac{\phi(a)\cdot\phi(b)\cdot g}{\phi(g)},g=gcd(a,b)$$ 证明：考虑找一个 $$x$$，使得 $$x|b,gcd(\frac{b}{x},a)=1,gcd(x,\frac{b}{x})=1$$ 且 $$a$$ 包含 $$x$$ 中所有质因子。 那么 $$\phi(a\cdot b)=\phi(a)\cdot x\cdot\phi(\frac{b}{x})=\frac{\phi(a)\cdot\phi(b)\cdot x}{\phi(x)}$$ 显然，$$x$$ 是 $$g$$ 的倍数，所以 $$\frac{x}{g}=\frac{\phi(x)}{\phi(g)}$$，所以$$\phi(a\cdot b)=\frac{\phi(a)\cdot\phi(b)\cdot g}{\phi(g)}$$ 有了这个结论之后可以考虑点分，那么设 $$dis(x)$$ 表示 $$x$$ 到当前根的距离。 设 $$h(x)=\frac{x}{\phi(x)}$$，那么$$f(u,v)=\phi(a_u)\cdot\phi(a_v)\cdot h(g)\cdot(dis(u)+dis(v)),g=gcd(a_u,a_v)$$ 考虑当 $$u$$ 确定时，需要知道 $$\sum\phi(a_v)\cdot h(gcd(a_u,a_v))$$ 和 $$\sum\phi(a_v)\cdot h(gcd(a_u,a_v))\cdot dis(v)$$。 设 $$g(x)=h(x)-\sum[y|x,y\neq x]g(y)$$，那么 $$h(gcd(a_u,a_v))=\sum[x|a_u,x|a_v]g(x)$$ 这样就可以直接枚举约数算贡献了。 ###Code ```c++ #include<bits/stdc++.h> typedef long long ll; inline void repr(int&a,int b){if(a<b)a=b;} #define mset(a,b) memset(a,b,sizeof(a)) #define fo0(i,n) for(int i=0,i##end=n;i<i##end;i++) #define fo1(i,n) for(int i=1,i##end=n;i<=i##end;i++) #define fo(i,a,b) for(int i=a,i##end=b;i<=i##end;i++) #define foe(i,x)for(__typeof(x.end())i=x.begin();i!=x.end();++i) const int N=200007,P=1000000007; struct edge { int to;edge*ne; }_e[N*2],*e=_e,*p[N]; inline void add(int a,int b) { *e=(edge){b,p[a]};p[a]=e++; } std::vector<int>v[N]; int n,pr[N],pm,phi[N],inv[N],f[N],w[N],sz[N],rsz,nro,nsz,g[N],h[N],q[N],qe,ans; bool vis[N]; inline void dfs1(int x,int fa) { sz[x]=1; for(edge*i=p[x];i;i=i->ne) if(i->to!=fa&&!vis[i->to]) { dfs1(i->to,x); sz[x]+=sz[i->to]; } } inline void dfs2(int x,int fa) { int t=rsz-sz[x]; for(edge*i=p[x];i;i=i->ne) if(i->to!=fa&&!vis[i->to]) { repr(t,sz[i->to]); dfs2(i->to,x); } if(t<nsz)nsz=t,nro=x; } inline void set(int x,int y,int z) { if(!~g[x])q[qe++]=x,g[x]=h[x]=0; g[x]=(g[x]+(ll)y*f[x])%P; h[x]=(h[x]+(ll)z*f[x])%P; } inline void get(int x,int&y,int&z) { if(~g[x])y=g[x],z=h[x]; else y=z=0; } inline void dfs3(int x,int fa,int dis) { int a=phi[w[x]],b=(ll)a*dis%P; foe(i,v[w[x]])set(*i,a,b); for(edge*i=p[x];i;i=i->ne) if(i->to!=fa&&!vis[i->to]) dfs3(i->to,x,dis+1); } inline void dfs4(int x,int fa,int dis) { int a=phi[w[x]],b=(ll)a*dis%P,A,B; foe(i,v[w[x]]) { get(*i,A,B); ans=(ans+(ll)a*B+(ll)b*A)%P; } for(edge*i=p[x];i;i=i->ne) if(i->to!=fa&&!vis[i->to]) dfs4(i->to,x,dis+1); } inline void work(int x) { dfs1(x,0); rsz=nsz=sz[x],nro=x; dfs2(x,0); vis[x=nro]=1; foe(i,v[w[x]])set(*i,phi[w[x]],0); for(edge*i=p[x];i;i=i->ne) if(!vis[i->to]) { dfs4(i->to,x,1); dfs3(i->to,x,1); } fo0(i,qe)g[q[i]]=h[q[i]]=-1; qe=0; for(edge*i=p[x];i;i=i->ne) if(!vis[i->to])work(i->to); } int main() { fo1(i,N-1)for(int j=i;j<N;j+=i)v[j].push_back(i); fo(i,2,N-1) { if(!vis[i])pr[pm++]=i,phi[i]=i-1; for(int j=0;i*pr[j]<N;j++) { vis[i*pr[j]]=1; if(i%pr[j]==0) { phi[i*pr[j]]=phi[i]*pr[j]; break; } phi[i*pr[j]]=phi[i]*(pr[j]-1); } } phi[1]=1; inv[1]=1; fo(i,2,N-1)inv[i]=(ll)(P-P/i)*inv[P%i]%P; fo1(i,N-1) { f[i]=(ll)i*inv[phi[i]]%P; foe(j,v[i])if(i^*j)(f[i]+=P-f[*j])%=P; } scanf("%d",&n); fo1(i,n)scanf("%d",w+i); fo0(i,n-1) { int x,y; scanf("%d%d",&x,&y); add(x,y); add(y,x); } mset(vis,0); mset(g,0xff); mset(h,0xff); work(1); ans=(ll)ans*inv[n]%P*inv[n-1]%P*2%P; printf("%d",ans); } ``` ]]> /archives/47/ 2016-10-03T22:55:00+08:00 2016-10-03T22:55:00+08:00 admin

###Description There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads. The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it. 有一个无向图，n个点，m条边，无自环，无重边，现在要给每个边定方向，使得入度等于出度的点最多 ###Input The first line contains a positive integer t (1 ≤ t ≤ 200) — the number of testsets in the input. Each of the testsets is given in the following way. The first line contains two integers n and m (1 ≤ n ≤ 200, 0 ≤ m ≤ n·(n - 1) / 2) — the number of cities and the number of roads in Berland. The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 ≤ u, v ≤ n) — the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities. It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200. Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one. ###Output For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it. In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once. ###Example input 2 5 5 2 1 4 5 2 3 1 3 3 5 7 2 3 7 4 2 output 3 1 3 3 5 5 4 3 2 2 1 3 2 4 3 7 ###Solution 将度数为奇的边间连上边，然后每次从任一点出发，走过一条边就定为走的方向，直到回到出发点。当所有边被定向后停止操作 那么对于度数为偶数的边，其入度一定等于出度，输出时去掉新加的边 ###Code ```c++ #include<bits/stdc++.h> struct edge { int to,ne; bool v; }e[50000]; int p[201],em=2,deg[201],n,m; bool ok[50000]; inline void add(int a,int b,bool v) { e[em].to=b,e[em].v=v,e[em].ne=p[a],p[a]=em++; } inline void solve() { scanf("%d%d",&n,&m); em=2; memset(p,0,n*4+4); memset(deg,0,n*4+4); while(m--) { int a,b; scanf("%d%d",&a,&b); add(a,b,1); add(b,a,1); deg[a]++; deg[b]++; } int lst=0,ans=n; for(int i=1;i<=n;i++) if(deg[i]&1) { if(lst) add(i,lst,0),add(lst,i,0),deg[i]++,deg[lst]++; else lst=i; ans--; } memset(ok,0,em); lst=1; printf("%d\n",ans); while(1) { while(lst<=n&&!deg[lst])lst++; if(lst>n)break; int k=lst; while(1) { bool f=1; for(int i=p[k];i;i=e[i].ne) if(!ok[i]) { f=0,ok[i]=1,ok[i^1]=1; if(e[i].v)printf("%d %d\n",k,e[i].to); deg[k]--; deg[e[i].to]--; k=e[i].to; break; } if(f)break; } } } int main() { int T; scanf("%d",&T); while(T--)solve(); } ``` ]]>