Codeforces Round #415 (Div. 1) (CDE)

1. CF 809C Find a car

大意: 给定一个$1e9\times 1e9$的矩阵$a$, $a_{i,j}$为它正上方和正左方未出现过的最小数, 每个询问求一个矩形内的和. 

可以发现$a_{i,j}=(i-1)\oplus (j-1)+1$, 暴力数位$dp$即可

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



//求 0<=i<=x, 0<=j<=y, 0<=i^j<=k的所有i^j的和
int f[33][2][2][2], g[33][2][2][2];
void add(int &a, ll b) {a=(a+b)%P;}
//f为个数, g为和
int calc(int x, int y, int k) {
    if (x<0||y<0) return 0;
    memset(f,0,sizeof f);
    memset(g,0,sizeof g);
    f[32][1][1][1] = 1;
    PER(i,0,31)REP(a1,0,1)REP(a2,0,1)REP(a3,0,1) { 
        int &r = f[i+1][a1][a2][a3];
        if (!r) continue;
        int l1=a1&&!(x>>i&1),l2=a2&&!(y>>i&1),l3=a3&&!(k>>i&1);
        REP(b1,0,1)REP(b2,0,1) {
            if (l1&&b1) continue;
            if (l2&&b2) continue;
            if (l3&&(b1^b2)) continue;
            int c1=a1&&b1>=(x>>i&1),c2=a2&&b2>=(y>>i&1),c3=a3&&(b1^b2)>=(k>>i&1);
            add(f[i][c1][c2][c3],r);
            add(g[i][c1][c2][c3],g[i+1][a1][a2][a3]*2ll+(b1^b2)*r);
        }
    }
    int ans = 0;
    REP(a1,0,1)REP(a2,0,1)REP(a3,0,1) add(ans,g[0][a1][a2][a3]+f[0][a1][a2][a3]);
    return ans;
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        int x1,y1,x2,y2,k;
        scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&k);
        --x1,--y1,--x2,--y2,--x1,--y1,--k;
        int ans = (calc(x2,y2,k)-calc(x1,y2,k)-calc(x2,y1,k)+calc(x1,y1,k))%P;
        if (ans<0) ans += P;
        printf("%d\n", ans);
    }
}

 

 

2. CF 809D

 

3. CF 809E Surprise me!

大意: 给定树, 点$i$权值$a_i$, $a$为$n$排列, $dis(x,y)$为$x,y$路径上的边数, 求$\frac{1}{n(n-1)}\sum\limits_{x\not = y} \varphi(a_xa_y)dis(x,y)$

设$b=a^{-1}$, 转化为求$\sum\limits_{i=1}^n\sum\limits_{j=1}^n \varphi(ij)dis(b_i,b_j)$

根据$\varphi(ij)=\frac{\varphi(i)\varphi(j)gcd(i,j)}{\varphi(gcd(i,j))}$

反演一下可以化为$\sum\limits_{T=1}^n\Bigg(\sum\limits_{d\mid T}\frac{d}{\varphi(d)}\mu(\frac{T}{d})\Bigg)\sum\limits_{i=1}^{\lfloor\frac{n}{T}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{n}{T}\rfloor}\varphi(iT)\varphi(jT)dis(b_{iT},b_{jT})$

考虑枚举$T$, 左边的$\sum\limits_{d\mid T}\frac{d}{\varphi(d)}\mu(\frac{T}{d})$可以线性筛预处理出来.

右边的就是$\sum\limits_{i=1}^{\lfloor\frac{n}{T}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{n}{T}\rfloor}\varphi(iT)\varphi(jT)({dep}_{b_{iT}}+{dep}_{b_{jT}}-2{dep}_{lca(b_{iT},b_{jT})}) $

预处理一下$\sum\limits_{i=1}^{\lfloor\frac{n}{T}\rfloor}\varphi(iT)$, 那么$\sum\limits_{i=1}^{\lfloor\frac{n}{T}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{n}{T}\rfloor}\varphi(iT)\varphi(jT)({dep}_{b_{iT}}+{dep}_{b_{jT}})$就很容易求出.

现在只需要考虑求$\sum\limits_{i=1}^{\lfloor\frac{n}{T}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{n}{T}\rfloor}\varphi(iT)\varphi(jT){dep}_{lca(b_{iT},b_{jT})}$

有效的点数是$O(\frac{n}{T})$的, 提出来建一棵虚树然后$DP$即可

那么总的复杂度就为$O(nlogn)$

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e6+50;
int n,dep[N],fa[N],son[N],top[N];
int phi[N],a[N],b[N],sz[N];
int p[N], cnt, vis[N];
int h[N],L[N],R[N],f[N],s[N],sum[N];
vector<int> g[N],gg[N];
void init() {
    phi[1] = h[1] = 1;
    for (int i=2; i<N; ++i) {
        if (!vis[i]) p[++cnt]=i,phi[i]=i-1,h[i]=inv(i-1);
        for (int j=1,t; j<=cnt&&i*p[j]<N; ++j) {
            vis[t=i*p[j]]=1;
            if (i%p[j]==0) {phi[t]=phi[i]*p[j],h[t]=0;break;}
            phi[t]=phi[i]*phi[p[j]];
            h[t]=(ll)h[i]*h[p[j]]%P;
        }
    }
}
void dfs(int x, int f, int d) {
    L[x]=++*L,dep[x]=d,sz[x]=1,fa[x]=f;
    for (int y:g[x]) if (y!=f) {
        dfs(y,x,d+1),sz[x]+=sz[y];
        if (sz[y]>sz[son[x]]) son[x]=y;
    }
    R[x]=*L;
}
void dfs(int x, int tf) {
    top[x]=tf;
    if (son[x]) dfs(son[x],tf);
    for (int y:g[x]) if (!top[y]) dfs(y,y);
}
int lca(int x, int y) {
    while (top[x]!=top[y]) {
        if (dep[top[x]]<dep[top[y]]) swap(x,y);
        x = fa[top[x]];
    }
    return dep[x]<dep[y]?x:y;
}
bool cmp(int x, int y) {
    return L[x]<L[y];
}

int DP(int x) {
    int ans = (ll)f[x]*f[x]%P*dep[x]%P;
    sum[x] = f[x];
    for (int y:gg[x]) {
        ans = (ans+DP(y))%P;
        ans = (ans+2ll*sum[y]*sum[x]%P*dep[x])%P;
        sum[x] = (sum[x]+sum[y])%P;
    }
    return ans;
}

int solve(vector<int> &a) {
    sort(a.begin(),a.end(),cmp);
    int sz = a.size();
    REP(i,1,sz-1) a.pb(lca(a[i],a[i-1]));
    a.pb(1);
    sort(a.begin(),a.end(),cmp);
    a.erase(unique(a.begin(),a.end()),a.end());
    s[cnt=1]=a[0],sz=a.size();
    REP(i,1,sz-1) {
        while (cnt>=1) {
            if (L[s[cnt]]<=L[a[i]]&L[a[i]]<=R[s[cnt]]) {
                gg[s[cnt]].pb(a[i]);
                break;
            }
            --cnt;
        }
        s[++cnt]=a[i];
    }
    int t = DP(1);
    for (auto &x:a) gg[x].clear(),f[x]=sum[x]=0;
    return t;
}

int main() {
    init();    
    scanf("%d",&n);
    REP(i,1,n) scanf("%d",a+i),b[a[i]]=i;
    REP(i,2,n) {
        int u, v;
        scanf("%d%d",&u,&v);
        g[u].pb(v),g[v].pb(u);
    }
    dfs(1,0,0),dfs(1,1);
    int ans = 0;
    REP(T,1,n) {
        int ret = 0, sum = 0;
        REP(i,1,n/T) sum = (sum+phi[i*T])%P;
        vector<int> v;
        REP(i,1,n/T) { 
            ret=(ret+2ll*sum*phi[i*T]%P*dep[b[i*T]])%P;
            v.pb(i*T);
        }
        for (auto &t:v) f[b[t]] = phi[t], t = b[t];
        ret = (ret-2ll*solve(v))%P;
        ans = (ans+(ll)h[T]*ret)%P;
    }
    ans = (ll)ans*inv(n)%P*inv(n-1)%P;
    if (ans<0) ans += P;
    printf("%d\n",ans);
}