A - 等火車

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0) #pragma GCC optimize(2) #include<set> #include<map> #include<cmath> #include<queue> #include<string> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<unordered_map> using namespace std; typedef long long ll; typedef pair<int,int> pii; const int mod=1e9+7; int main() {IO;int T=1;//cin>>T;while(T--){int s,q;cin>>s>>q;while(q--){int t;cin>>t;if(t%s) cout<<s-t%s<<'\n';else cout<<0<<'\n';}}return 0;}

B - 數(shù)字游戲

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0) #pragma GCC optimize(2) #include<set> #include<map> #include<cmath> #include<queue> #include<string> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<unordered_map> using namespace std; typedef long long ll; typedef pair<int,int> pii; const int mod=1e9+7; int main() {IO;int T=1;//cin>>T;while(T--){ll n;cin>>n;cout<<n-1<<'\n';}return 0;}

C - 最大生成樹

貪心直接求即可

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0) #pragma GCC optimize(2) #include<set> #include<map> #include<cmath> #include<queue> #include<string> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<unordered_map> using namespace std; typedef long long ll; typedef pair<int,int> pii; const ll mod=998244353; int main() {IO;int T=1;//cin>>T;while(T--){ll n;cin>>n;ll a=n/2;a%=mod;cout<<((3*a*a%mod-3*a%mod+1)%mod+mod)%mod;}return 0;}

E - 樹

真就化簡式子唄？？？
$$dist(x,y)=dep(x)+dep(y)?2dep(lca(x,y))$$
于是可知
$$\sum _{x=1}^n\sum _{y=1}^{n}dis{t}^2(x,y)=\sum _{x=1}^n\sum _{y=1}^{n}de{p}^2(x)+de{p}^2(y)+2dep(x)dep(y)+4de{p}^2(lca(x,y))?4dep(lca(x,y))\times (dep(x)+dep(y))$$
對于前三項
$$\sum _{x=1}^n\sum _{y=1}^{n}de{p}^2(x)+de{p}^2(y)=2\sum _{x=1}^{n}de{p}^2(x)$$
$$2\sum _{x=1}^n\sum _{y=1}^{n}dep(x)dep(y)=2\sum _{x=1}^{n}dep(x)\sum _{y=1}^{n}dep(y)=2(\sum _{x=1}^{n}dep(x){)}^2$$
對于后兩項可以枚舉公共祖先
$$\sum _{x=1}^n\sum _{y=1}^{n}de{p}^2(lca(x,y))=\sum _{u=lca(x,y)}de{p}^2(u)\times (s{z}^2(u)?\sum _{j\in son(u)}s{z}^2(j))$$
$$\sum _{x=1}^n\sum _{y=1}^{n}dep(lca(x,y))\times (dep(x)+dep(y))=\sum _{u=lca(x,y)}[2\sum _{j\in son(u)}sum(j)(sz(u)?sz(j))+2dep(u)sz(u)]$$

#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0) #pragma GCC optimize(2) #include<set> #include<map> #include<cmath> #include<queue> #include<string> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<unordered_map> using namespace std; typedef long long ll; typedef pair<int,int> pii; const int N=1000010; const ll mod=998244353; int h[N],e[2*N],ne[2*N],idx; ll dep[N],sum[N],sz[N]; ll res; int n; void add(int a,int b) {e[idx]=b;ne[idx]=h[a];h[a]=idx++; } void dfs1(int u,int fa) {dep[u]=dep[fa]+1;sz[u]=1;sum[u]=dep[u];for(int i=h[u];i!=-1;i=ne[i]){int j=e[i];if(j==fa) continue;dfs1(j,u);sz[u]+=sz[j];sum[u]=(sum[u]+sum[j])%mod;} } void dfs2(int u,int fa) {ll s1=sz[u]*sz[u]%mod,s2=2*dep[u]*sz[u]%mod;for(int i=h[u];i!=-1;i=ne[i]){int j=e[i];if(j==fa) continue;dfs2(j,u);s1=(s1-sz[j]*sz[j]%mod)%mod;s2=(s2+2*sum[j]*(sz[u]-sz[j])%mod)%mod;}res=(res+4*s1*dep[u]%mod*dep[u])%mod;res=(res-4*s2*dep[u]%mod)%mod; } int main() {IO;int T=1;//cin>>T;while(T--){memset(h,-1,sizeof h);cin>>n;for(int i=1;i<n;i++){int a,b;cin>>a>>b;add(a,b),add(b,a);}dfs1(1,0);ll s=0;for(int i=1;i<=n;i++) {s=(s+dep[i])%mod;res=(res+2ll*n*dep[i]%mod*dep[i])%mod;}res=(res+2*s*s%mod)%mod;dfs2(1,0);res=(res%mod+mod)%mod;cout<<res<<'\n';}return 0; }

這就是數(shù)學(xué)的魅力嗎？愛了愛了
要加油哦~

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