HDU Multi-University Training Contest 1

Sloved	A	B	C	D	E	F	G	H	I	J	K	L	M
6/13	·	Ø	·	Ø	O	!	.	Ø	!	·	Ø	!	Ø

• O for passing during the contest
• Ø for passing after the contest
• ! for attempted but failed
• · for having not attempted yet

upsolve:I
G题放弃

仅作记录

A.Blank

B.Operation

线性基

#include 
using namespace std;
#define rep(i,a,n) for (int i=a;i
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
const int maxn = 1e6+100;

struct LB
{
    ll p[33];
    int g[33];
    void ins(ll x,int pos)
    {
         per(i,0,30)
         {
             if((x>>i) & 1)
             {
                 if(p[i])
                 {
                     if(g[i] <= pos)
                     {
                         x ^= p[i];
                         p[i] ^= x;
                         swap(g[i],pos);
                     }
                     else
                         x ^= p[i];
                 }
                 else
                 {
                     p[i] = x;
                     g[i] = pos;
                     break;
                 }
             }
         }
    }    
    ll query(int l)
    {
        ll res = 0;
        per(i,0,30)
        {
            if(g[i] >= l)
            {
                res = max(res,res^p[i]);
            }
        }
        return res;
    }
} base[maxn];

int n,m;

int gao(int x,int lastans)
{
    return (x^lastans)%n+1;
}

int T;
int x;

int main(int argc, char const *argv[])
{
    // ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        rep(i,1,n+1)
        {
            scanf("%d",&x);
            base[i] = base[i-1];
            base[i].ins(x,i);
        }
        ll ans = 0;
        int l,r;
        while(m--)
        {
            int op;
            scanf("%d",&op);
            if(!op)
            {
                scanf("%d%d",&l,&r);
                l = gao(l,ans);r=gao(r,ans);
                if(l>r) swap(l,r);
                ans = base[r].query(l);
                printf("%lld\n",ans);
            }
            else
            {
                n++;
                scanf("%d",&l);
                base[n] = base[n-1];
                base[n].ins(l^ans,n);
            }
        }
    }
    return 0;
}

C.Milk

D. Vacation

贪心

#include 
using namespace std;
#define rep(i,a,n) for (int i=a;i
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
const int maxn = 1e5 + 100;

int n;
ll sum;
int l[maxn],s[maxn],v[maxn];
double ans;
int main(int argc, char const *argv[])
{
    while(scanf("%d",&n)!=EOF)
    {
        sum = 0;
        ans = 0;
        rep(i,0,n+1) scanf("%d",&l[i]),sum+=l[i];
        rep(i,0,n+1) scanf("%d",&s[i]);
        rep(i,0,n+1) scanf("%d",&v[i]);
        sum -= l[0];
        per(i,0,n+1)
        {
            ans = max(ans,(double)(sum+s[i])/v[i]);
            sum -= l[i];
        }
        printf("%.10f\n",ans);
    }
    return 0;
}

E. Path

最小割

F. Typewriter

K. Function

$$\sum _{i=1}^{n}gcd(\lfloor \sqrt[3]{i}\rfloor ,i)\%998244353$$

$$由x=\lfloor \sqrt[3]{i}\rfloor 显然分块\Rightarrow \sum _{i=1}^{\lfloor \sqrt[3]{n}\rfloor }\sum _{j=i^3}^{min(n,(i+1{)}^3+1)}gcd(i,j)$$

$$设f(i,j)=\sum _{x=1}^{m}gcd(i,x)\Rightarrow f(i,j)=\lfloor \frac{j}{i}\rfloor f(i,i)+f(i,j\%i)$$

$$易得出f(i,i)=\sum _{d|i}d\varphi (\frac{i}{d}),f(i,j)=\sum _{d|i}\frac{j}{d}\varphi (d)$$

$$对于\sum _{j=i^3}^{(i+1{)}^3+1}gcd(i,j),通过差分有f(i,(i+1{)}^3+1)-f(i,i^3)=3\ast (i+1)f(i,i)+gcd(i,i^3)=3\ast (i+1)f(i,i)+i,显然f(i,i)通过线性欧拉筛可以O(n+\sqrt{n}\log n)预处理.$$

$$所以显然只需要考虑上界为n时的情况,O(\sqrt{n})的跑一边f(i,n)即可$$

$$这题卡常很严，所以离线查询(\_\_int128长见识了)$$

#include 
using namespace std;
typedef long long ll;
const int N = 1e7 + 7;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
mt19937 mrand(random_device{}());
int rnd(int x) { return mrand() % x;}
ll gcd(ll a,ll b){return b ? gcd(b,a % b) : a;}
int lcm(int a,int b){return a / gcd(a,b) * b;}
ll fpow(ll a,ll b)
{
    ll res = 1;
    while(b){
        if(b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}
void read(__int128 &x) {
    static char ch;static bool neg;
    for(ch=neg=0;ch<'0' || '9'<ch;neg|=ch=='-',ch=getchar());
    for(x=0;'0'<=ch && ch<='9';(x*=10)+=ch-'0',ch=getchar());
    x=neg?-x:x;
}
int phi[N],prime[N],cnt;
bool isprime[N];
ll f[N];
void get_phi()
{
   phi[1] = 1;
   for(int i = 2;i < N;++i){
       if(!isprime[i]){
             prime[++cnt] = i;
             phi[i] = i - 1;
        }
       for(int j = 1;j <= cnt && i * prime[j] < N;++j){
          isprime[i * prime[j]] = 1;
          if(i % prime[j] == 0){
             phi[i * prime[j]] = phi[i] * prime[j];
             break;
          }
          else phi[i * prime[j]] = phi[i] * (prime[j] - 1);
       }
   }
   for(int i = 1;i * i < N;++i){
       int p = i * i;
       for(int j = i * i;j < N;j += i){
           f[j] = (f[j] + i * phi[j / i]) % mod;
           if(p < j) f[j] = (f[j] + (j / i) * phi[i]) % mod;
       }
   }
}
struct node{
    __int128 n;
    int id;
}a[20];
bool cmp(node a,node b)
{
    return a.n < b.n;
}
//ll n;
ll ans[20],pos[20];
int main()
{
    //freopen("in.txt","r",stdin);
    get_phi();
    int t;
    scanf("%d",&t);
    for(int i = 0;i < t;++i){
        read(a[i].n);
        a[i].id = i;
        //scanf("%lld",&n);
    }
    sort(a,a + t,cmp);
    for(__int128 i = 1,cnt = 0;i * i * i < a[t - 1].n;++i){
        if((i + 1) * (i + 1) * (i + 1) > a[cnt].n){
            if(cnt == t - 1){
                pos[cnt] = i;
                break;
            }
            while((i + 1) * (i + 1) * (i + 1) > a[cnt].n){
                pos[cnt] = i;
                cnt++;
                ans[cnt] = ans[cnt - 1];
            }

        }
        ans[cnt] = (ans[cnt] + 3LL * (i + 1) * f[i] + i) % mod;
        if((i + 1) * (i + 1) * (i + 1) == a[cnt].n) ans[cnt] = (ans[cnt] + i + 1) % mod,pos[cnt] = 0;
    }
    for(int jq = 0;jq < t;++jq){
        if(pos[jq]){
            ll rnt = a[jq].n % pos[jq],ret = a[jq].n / pos[jq] % mod;
            for(ll i = 1;i * i <= pos[jq];++i){
                ll l = pos[jq] / i;
                if(pos[jq] % i == 0){
                    ans[jq] = (ans[jq] + (rnt / i) * phi[i]) % mod;
                    if(i * i < pos[jq]) ans[jq] = (ans[jq] + (rnt / l) * phi[l]) % mod;
                }
                i = pos[jq] / l;
            }
            ans[jq] = (ans[jq] + (ret - (pos[jq] * pos[jq])) % mod * f[pos[jq]] % mod + pos[jq] + mod) % mod;
        }
    }
    for(int i = 0;i < t;++i){
        for(int j = 0;j < t;++j)
            if(a[j].id == i)printf("%lld\n",ans[j]);
    }
    return 0;
}

L. Sequence

M. Code

//二维凸包求交集，d<=2必然成立（之前没注意被坑了好久）
#include 
using namespace std;
typedef long long ll;
const int N = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-10;
mt19937 mrand(random_device{}());
int rnd(int x) { return mrand() % x;}
ll gcd(ll a,ll b){return b ? gcd(b,a % b) : a;}
int lcm(int a,int b){return a / gcd(a,b) * b;}
ll fpow(ll a,ll b)
{
    ll res = 1;
    while(b){
        if(b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}
int dcmp(double x){
    if(x > eps) return 1;
    if(x < -eps) return -1;
    return 0;
}
struct point{
    double x, y;
    point(double _x = 0, double _y = 0){
        x = _x; y = _y;
    }
    point operator + (const point &q){
        return point(x + q.x, y + q.y);
    }
    point operator - (const point &q){
        return point(x-q.x, y-q.y);
    }
    bool operator < (const point &q){
        if(x == q.x) return y < q.y;
        return x < q.x;
    }
    point operator *(const int rhq){
        return point(x*rhq,y*rhq);
    }
    bool operator == (const point &q){
        return dcmp(x-q.x) == 0 && dcmp(y -q.y) == 0;
    }
    point rotate(point p, double ang){
        point v = (*this) - p;
        double c = cos(ang), s = sin(ang);
        return point(p.x + v.x * c- v.y * s,p.y + v.x * s + v.y * c);
    }
    void read(){
        scanf("%lf%lf",&x,&y);
    }
};
double sqr(double a){
    return a * a;
}
double dot(point p, point q){
    return p.x * q.x + p.y * q.y;
}
double det(point p, point q){
    return p.x*q.y - p.y*q.x;
}
double cross(point p, point q){
    return det(p, q);
}
double cross(point p1, point p2, point q1, point q2){
    return det(p1 - p2,q1 - q2);
}
double dis(point p, point q){
    return sqrt(sqr(p.x - q.x) + sqr(p.y - q.y));
}
double dis2(point p, point q){
    return sqr(p.x-q.x) + sqr(p.y-q.y);
}
double ang(point p, point q){
    point t;
    return acos(dot(p,q)/dis(p,t)/dis(q,t));
}
int ConvexHull(point* P, int cnt, point* res) { //求凸包
    sort(P, P + cnt);//x1 < x1 || (x1 == x2 && y1 < y2)
    cnt = unique(P, P + cnt) - P;
    int m = 0;
    for(int i = 0; i < cnt; i++) {
        while (m > 1 && cross(res[m - 1] - res[m - 2], P[i] - res[m - 2]) <= 0) m--;
        res[m++] = P[i];
    }
    int k = m;
    for (int i = cnt - 2; i >= 0; i--) {
        while (m > k && cross(res[m - 1] - res[m - 2], P[i] - res[m - 2]) <= 0) m--;
        res[m++] = P[i];
    }
    if(cnt > 1) m--;
    return m;
}

bool OnSegment(point p, point &a1, point a2) {
    return dcmp(cross(a1 - p, a2 - p)) == 0 && dcmp(dot(a1 - p, a2 - p)) < 0;
}

bool Intersection(point a1, point a2, point b1, point b2) {
    double c1 = cross(a2 - a1, b1 - a1), c2 = cross(a2 - a1, b2 - a1);
    double c3 = cross(b2 - b1, a1 - b1), c4 = cross(b2 - b1, a2 - b1);
    return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}
int ispoint(point p, point* pp,int cnt)
{
    int wn = 0;
    for (int i = 0; i < cnt; i++) {
        point p1 = pp[i];
        point p2 = pp[(i + 1) % cnt];
        if (p1 == p || p2 == p || OnSegment(p, p1, p2)) return -1;
        int k = dcmp(cross(p2 - p1, p - p1));
        int d1 = dcmp(p1.y - p.y);
        int d2 = dcmp(p2.y - p.y);
        if (k > 0 && d1 <= 0 && d2 > 0) wn++;
        if (k < 0 && d2 <= 0 && d1 > 0) wn--;
    }
    if (wn != 0) return 1;
    return 0;
}

bool judge(point *p1,int c1,point *p2,int c2) {
    for (int i = 0; i < c1; i++) {
        if (ispoint(p1[i], p2,c2) != 0) return false;
    }
    for (int i = 0; i < c2; i++) {
        if (ispoint(p2[i], p1,c1) != 0) return false;
    }
    for (int i = 0; i < c1; i++) {
        for (int j = 0; j < c2; j++) {
            if (Intersection(p1[i], p1[(i + 1) % c1], p2[j], p2[(j + 1) % c2]))
                return false;
        }
    }
    return true;
}
point za[100],fa[100],tu1[100],tu2[100];
int a[N];
int main()
{
    //freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t--){
        int n,x1,y1,f,cnt1 = 0,cnt2 = 0,c1,c2;
        scanf("%d",&n);
        for(int i = 0;i < n;++i){
            scanf("%d%d%d",&x1,&y1,&f);
            if(f == 1)za[cnt1].x = x1,za[cnt1++].y = y1;
            else fa[cnt2].x = x1,fa[cnt2++].y = y1;
        }
        c1 = cnt1,c2 = cnt2;
        if(cnt1 > 1){
            c1 = ConvexHull(za,cnt1,tu1);
        }
        if(cnt2 > 1){
            c2 = ConvexHull(fa,cnt2,tu2);
        }
        if(judge(tu1,c1,tu2,c2) || c1 < 2 || c2 < 2) printf("Successful!\n");
        else printf("Infinite loop!\n");
    }
    return 0;
}