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+, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}

ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',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[][][][], g[][][][];

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<||y<) return ;

    memset(f,,sizeof f);

    memset(g,,sizeof g);

    f[][][][] = ;

    PER(i,,)REP(a1,,)REP(a2,,)REP(a3,,) {

        int &r = f[i+][a1][a2][a3];

        if (!r) continue;

        int l1=a1&&!(x>>i&),l2=a2&&!(y>>i&),l3=a3&&!(k>>i&);

        REP(b1,,)REP(b2,,) {

            if (l1&&b1) continue;

            if (l2&&b2) continue;

            if (l3&&(b1^b2)) continue;

            int c1=a1&&b1>=(x>>i&),c2=a2&&b2>=(y>>i&),c3=a3&&(b1^b2)>=(k>>i&);

            add(f[i][c1][c2][c3],r);

            add(g[i][c1][c2][c3],g[i+][a1][a2][a3]*2ll+(b1^b2)*r);

        }

    }

    int ans = ;

    REP(a1,,)REP(a2,,)REP(a3,,) add(ans,g[][a1][a2][a3]+f[][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<) 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+, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}

ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}

//head

const int N = 1e6+;

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[] = h[] = ;

    for (int i=; i<N; ++i) {

        if (!vis[i]) p[++cnt]=i,phi[i]=i-,h[i]=inv(i-);

        for (int j=,t; j<=cnt&&i*p[j]<N; ++j) {

            vis[t=i*p[j]]=;

            if (i%p[j]==) {phi[t]=phi[i]*p[j],h[t]=;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]=,fa[x]=f;

    for (int y:g[x]) if (y!=f) {

        dfs(y,x,d+),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,,sz-) a.pb(lca(a[i],a[i-]));

    a.pb();

    sort(a.begin(),a.end(),cmp);

    a.erase(unique(a.begin(),a.end()),a.end());

    s[cnt=]=a[],sz=a.size();

    REP(i,,sz-) {

        while (cnt>=) {

            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();

    for (auto &x:a) gg[x].clear(),f[x]=sum[x]=;

    return t;

}

int main() {

    init();

    scanf("%d",&n);

    REP(i,,n) scanf("%d",a+i),b[a[i]]=i;

    REP(i,,n) {

        int u, v;

        scanf("%d%d",&u,&v);

        g[u].pb(v),g[v].pb(u);

    }

    dfs(,,),dfs(,);

    int ans = ;

    REP(T,,n) {

        int ret = , sum = ;

        REP(i,,n/T) sum = (sum+phi[i*T])%P;

        vector<int> v;

        REP(i,,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-)%P;

    if (ans<) ans += P;

    printf("%d\n",ans);

}

巴特西

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

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