Educational Codeforces Round 41 

D. Pair Of Lines

考虑先把凸包找出来，如果凸包上的点数大于\(4\)显然不存在解，小于等于\(2\)必然存在解

否则枚举凸包上两个点连线，判断剩余点能否被一条线覆盖即可

view code 
#pragma GCC optimize("O3")

#pragma GCC optimize("Ofast,no-stack-protector")

#include<bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define endl "\n"

#define LL long long int

#define vi vector<int>

#define vl vector<LL>

#define all(V) V.begin(),V.end()

#define sci(x) scanf("%d",&x)

#define scl(x) scanf("%lld",&x)

#define scs(s) scanf("%s",s)

#define pii pair<int,int>

#define pll pair<LL,LL>

#ifndef ONLINE_JUDGE

#define cout cerr

#endif

#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))

#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))

#define debug(x)  cerr << #x << " = " << x << endl

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }

template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }

const int MAXN = 2e5+7;

pii V(const pii &A, const pii &B){ return make_pair(B.first-A.first,B.second-A.second); }

LL cross(const pii &A, const pii &B){ return 1ll * A.first * B.second - 1ll * A.second * B.first; }

int sgn(LL x){ return x > 0 ? 1 : x < 0 ? -1 : 0; }

void solve(){

    int n; sci(n);

    vector<pii> point(n);

    for(auto &p : point) sci(p.first), sci(p.second);

    sort(all(point),[&](pii &A, pii &B){ return A.first==B.first?A.second<B.second:A.first<B.first; });

    sort(point.begin()+1,point.end(),[&](pii &A, pii &B){

        return atan2(A.second-point[0].second,A.first-point[0].first) < atan2(B.second-point[0].second,B.first-point[0].first);

    });

    static pii stk[MAXN];

    static int top;

    top = 0;

    stk[++top] = point[0];

    for(int i = 1; i < n; i++){

        while(top>1 and sgn(cross(V(stk[top-1],stk[top]),V(stk[top-1],point[i])))<=0) top--;

        stk[++top] = point[i];

    }

    if(top>=3 and sgn(cross(V(stk[1],stk[top]),V(stk[1],stk[top-1])))==0) top--;

    if(top>4){

        cout << "NO" << endl;

        return;

    }

    if(top<=2){

        cout << "YES" << endl;

        return;

    }

    auto sameline = [&](const pii &A, const pii &B, const pii &C){

        return sgn(cross(V(A,B),V(A,C))) == 0;

    };

    auto check = [&](const pii &A, const pii &B){

        vector<bool> vis(n,false);

        for(int i = 0; i < n; i++){

            if(point[i]==A or point[i]==B) vis[i] = true;

            if(sameline(A,B,point[i])) vis[i] = true;

        }

        vector<pii> vec;

        for(int i = 0; i < n; i++) if(!vis[i]) vec << point[i];

        if(vec.size()<=2) return true;

        for(int i = 2; i < (int)vec.size(); i++) if(!sameline(vec[0],vec[1],vec[i])) return false;

        return true;

    };

    for(int i = 1; i < top; i++) for(int j = i + 1; j <= top; j++) if(check(stk[i],stk[j])){

        cout << "YES" << endl;

        return;

    }

    cout << "NO" << endl;

}

int main(){

    #ifndef ONLINE_JUDGE

    freopen("Local.in","r",stdin);

    freopen("ans.out","w",stdout);

    #endif

    solve();

    return 0;

}

E.Tufurama

问题等价于：对于每个\(i\)，我们需要知道\([i+1,a_i]\)之间有多少数大于等于\(i\)即可

由于\(a_i\)超过\(n\)没什么用，所以可以先把\(a_i\)和\(n\)取\(\min\)，然后用主席树搞一下就好了

view code 
#pragma GCC optimize("O3")

#pragma GCC optimize("Ofast,no-stack-protector")

#include<bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define endl "\n"

#define LL long long int

#define vi vector<int>

#define vl vector<LL>

#define all(V) V.begin(),V.end()

#define sci(x) scanf("%d",&x)

#define scl(x) scanf("%lld",&x)

#define scs(s) scanf("%s",s)

#define pii pair<int,int>

#define pll pair<LL,LL>

#ifndef ONLINE_JUDGE

#define cout cerr

#endif

#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))

#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))

#define debug(x)  cerr << #x << " = " << x << endl

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }

template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }

const int MAXN = 2e5+7;

struct SegmentTree{

    int root[MAXN], sum[MAXN<<5], ls[MAXN<<5], rs[MAXN<<5], tot;

    void modify(int &rt, int pre, int pos, int x, int l, int r){

        rt = ++tot;

        sum[rt] = sum[pre] + x; ls[rt] = ls[pre]; rs[rt] = rs[pre];

        if(l+1==r) return;

        int mid = (l + r) >> 1;

        if(pos<mid) modify(ls[rt],ls[pre],pos,x,l,mid);

        else modify(rs[rt],rs[pre],pos,x,mid,r);

    }

    int qsum(int Lrt, int Rrt, int L, int R, int l, int r){

        if(L>=r or l>=R) return 0;

        if(L<=l and r<=R) return sum[Rrt] - sum[Lrt];

        int mid = (l + r) >> 1;

        return qsum(ls[Lrt],ls[Rrt],L,R,l,mid) + qsum(rs[Lrt],rs[Rrt],L,R,mid,r);

    }

}ST;

void solve(){

    int n; sci(n);

    vi A(n);

    for(int &x : A) sci(x), cmin(x,n);

    LL ret = 0;

    for(int i = n; i; i--){

        int x = A[i-1];

        if(x>i) ret += ST.qsum(ST.root[x+1],ST.root[i+1],i,n+1,1,n+1);

        ST.modify(ST.root[i],ST.root[i+1],x,1,1,n+1);

    }

    cout << ret << endl;

}

int main(){

    #ifndef ONLINE_JUDGE

    freopen("Local.in","r",stdin);

    freopen("ans.out","w",stdout);

    #endif

    solve();

    return 0;

}

F. k-substrings

直接二分是不行的，没有单调性

考虑枚举前后缀的中点\(i\)，然后找以\(i\)和\(n-i+1\)为中点的最长相同串，可以发现这个长度是可以二分的，找到最长的相同串，这个用哈希就好了

view code 
#pragma GCC optimize("O3")

#pragma GCC optimize("Ofast,no-stack-protector")

#include<bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define endl "\n"

#define LL long long int

#define vi vector<int>

#define vl vector<LL>

#define all(V) V.begin(),V.end()

#define sci(x) scanf("%d",&x)

#define scl(x) scanf("%lld",&x)

#define scs(s) scanf("%s",s)

#define pii pair<int,int>

#define pll pair<LL,LL>

#ifndef ONLINE_JUDGE

#define cout cerr

#endif

#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))

#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))

#define debug(x)  cerr << #x << " = " << x << endl

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }

template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }

const int MAXN = 1e6+7;

const int MOD1 = 998244353;

const int MOD2 = 1004525809;

const int BASE1 = 233;

const int BASE2 = 1033;

int n, pw1[MAXN], pw2[MAXN], ret[MAXN];

pii hax[MAXN];

char s[MAXN];

pii calc(int l, int r){

    return make_pair((hax[r].first - 1ll * hax[l-1].first * pw1[r-l+1] % MOD1 + MOD1) % MOD1,

                     (hax[r].second - 1ll * hax[l-1].second * pw2[r-l+1] % MOD2 + MOD2) % MOD2);

}

void solve(){

    sci(n); scs(s+1);

    pw1[0] = pw2[0] = 1;

    for(int i = 1; i < MAXN; i++) pw1[i] = 1ll * pw1[i-1] * BASE1 % MOD1, pw2[i] = 1ll * pw2[i-1] * BASE2 % MOD2;

    for(int i = 1; i <= n; i++){

        hax[i].first = (1ll * hax[i-1].first * BASE1 + s[i]) % MOD1;

        hax[i].second = (1ll * hax[i-1].second * BASE2 + s[i]) % MOD2;

    }

    memset(ret,255,sizeof(ret));

    for(int i = 1; i <= (n >> 1); i++){

        int l = 1, r = i;

        while(l<=r){

            int mid = (l + r) >> 1;

            if(calc(i-mid+1,i+mid-1)==calc(n-i+1-mid+1,n-i+1+mid-1)) l = mid + 1;

            else r = mid - 1;

        }

        ret[i-r+1] = r * 2 - 1;

    }

    for(int i = 1; i <= (n + 1) / 2; i++) cout << (ret[i] = max(ret[i],ret[i-1] - 2)) << ' '; cout << endl;

}

int main(){

    #ifndef ONLINE_JUDGE

    freopen("Local.in","r",stdin);

    freopen("ans.out","w",stdout);

    #endif

    solve();

    return 0;

}

G. Partitions

每个值对答案的贡献次数是相同的，那么答案就是所有数的和\(sum\)乘上每个数的贡献数

第一种方法：

通过枚举指定数所在块的大小\(i\)，选出\(i-1\)个数和指定数在同一块中，然后剩下的\(n-i\)个数分成\(k-1\)个集合，答案应该是\(sum\cdot\sum_{i=1}^{n-k+1}i\cdot \tbinom{n-1}{i-1}\cdot \operatorname{stirling}(n-i,k-1)\)

这样需要用任意模数\(NTT\)来算第二类斯特林数的列，比较麻烦

第二种方法：

对于指定的数，自身对自身的贡献为\(\operatorname{stirling}(n,k)\)而其他每个数只要和指定数在一个块内，就能有一次的贡献，考虑枚举把指定数和其他数捆绑在一起的方案数是\((n-1)\cdot\operatorname{stirling}(n-1,k)\)，这时候答案就是\(sum\cdot (\operatorname{stirling}(n,k)+(n-1)\cdot \operatorname{stirling}(n-1,k))\)

view code 
#pragma GCC optimize("O3")

#pragma GCC optimize("Ofast,no-stack-protector")

#include<bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define endl "\n"

#define LL long long int

#define vi vector<int>

#define vl vector<LL>

#define all(V) V.begin(),V.end()

#define sci(x) scanf("%d",&x)

#define scl(x) scanf("%lld",&x)

#define scs(s) scanf("%s",s)

#define pii pair<int,int>

#define pll pair<LL,LL>

#ifndef ONLINE_JUDGE

#define cout cerr

#endif

#define cmax(a,b) ((a) = (a) > (b) ? (a) : (b))

#define cmin(a,b) ((a) = (a) < (b) ? (a) : (b))

#define debug(x)  cerr << #x << " = " << x << endl

function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};

template <typename T> vector<T>& operator << (vector<T> &__container, T x){ __container.push_back(x); return __container; }

template <typename T> ostream& operator << (ostream &out, vector<T> &__container){ for(T _ : __container) out << _ << ' '; return out; }

const int MAXN = 2e5+7;

const int MOD = 1e9+7;

int fac[MAXN], inv[MAXN], rfac[MAXN];

int ksm(int a, int b){

    int ret = 1;

    while(b){

        if(b&1) ret = 1ll * ret * a % MOD;

        b >>= 1;

        a = 1ll * a * a % MOD;

    }

    return ret;

}

int C(int n, int m){ return n < m ? 0 : 1ll * fac[n] * rfac[m] % MOD * rfac[n-m] % MOD; }

int stirling(int n, int m){

    int ret = 0;

    for(int i = 0; i < m; i++) ret = (ret + (i&1?-1ll:1ll) * C(m,i) % MOD * ksm(m-i,n) % MOD) % MOD;

    return (1ll * ret * rfac[m] % MOD + MOD) % MOD;

}

void solve(){

    int n, k;

    sci(n); sci(k);

    vi A(n); for(int &x : A) sci(x);

    int tot = accumulate(all(A),0ll) % MOD;

    if(n==k) cout << tot << endl;

    else cout << 1ll * tot * (stirling(n,k) + 1ll * (n-1) * stirling(n-1,k) % MOD) % MOD << endl;

}

int main(){

    #ifndef ONLINE_JUDGE

    freopen("Local.in","r",stdin);

    freopen("ans.out","w",stdout);

    #endif

    fac[0] = rfac[0] = inv[1] = 1;

    for(int i = 1; i < MAXN; i++) fac[i] = 1ll * fac[i-1] * i % MOD;

    for(int i = 2; i < MAXN; i++) inv[i] = 1ll * (MOD - MOD / i) * inv[MOD%i] % MOD;

    for(int i = 1; i < MAXN; i++) rfac[i] = 1ll * rfac[i-1] * inv[i] % MOD;

    solve();

    return 0;

}

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