Educational Codeforces Round 41

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 "
"
#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 "
"
#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 "
"
#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(sumcdotsum_{i=1}^{n-k+1}icdot binom{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)cdotoperatorname{stirling}(n-1,k))，这时候答案就是(sumcdot (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 "
"
#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;
}