ASC1 做题记录

A.

考虑相邻两个质数的差不会太大，于是我们从n/2开始暴力枚举即可，需要大数运算

def gcd(a,b):
    if b==0:
        return a
    else:
        return gcd(b,a%b)
 
f=open('china.in','r')
n=int(f.read())
f.close()
for i in range(0,n//2):
    if gcd(n//2-i,n)==1:
        f=open('china.out','w')
        f.write(str(n//2-i))
        f.close()
        break

B.

上下界可行流裸题，一条边拆三条建图跑一下

#include<bits/stdc++.h>
#define maxn 205
using namespace std;
typedef long long ll;
const int inf = 2e9+7;
int head[maxn],p,s,t;
struct edge
{
    int to,next,id,f;
}e[maxn*maxn*4];
void addedge(int u,int v,int f,int id)
{
    e[p].to=v;e[p].id=0;e[p].f=f;e[p].next=head[u];head[u]=p++;
    e[p].to=u;e[p].id=id;e[p].f=0;e[p].next=head[v];head[v]=p++;
}
int dis[maxn];
bool bfs(int s,int t)
{
    memset(dis,0,sizeof(dis));
    queue<int> q;
    q.push(s);
    dis[s]=1;
    while(!q.empty())
    {
        int u=q.front();q.pop();
        for(int i=head[u];i!=-1;i=e[i].next)
        {
            int v=e[i].to;
            int f=e[i].f;
            if(f&&!dis[v])dis[v]=dis[u]+1,q.push(v);
        }
    }
    return dis[t]!=0;
}
int dfs(int u,int maxf,int t)
{
    if(u==t)return maxf;
    int tmp=0;
    for(int i=head[u];i!=-1&&tmp<maxf;i=e[i].next)
    {
        int v=e[i].to;
        int f=e[i].f;
        if(f&&dis[v]==dis[u]+1)
        {
            int minn=min(maxf-tmp,f);
            f=dfs(v,minn,t);
            tmp+=f;
            e[i].f-=f;e[i^1].f+=f;
        }
    }
    if(!tmp)dis[u]=inf;
    return tmp;
}
int dinic(int s,int t)
{
    int ans=0;
    while(bfs(s,t))ans+=dfs(s,inf,t);
    return ans;
}
int n,m;
int L[maxn*maxn];
int main()
{
    freopen("cooling.in","r",stdin);
    freopen("cooling.out","w",stdout);
    scanf("%d%d",&n,&m);
    s=201;t=202;p=0;
    memset(head,-1,sizeof(head));
    for(int i=1;i<=m;++i)
    {
        int u,v,l,r;
        scanf("%d%d%d%d",&u,&v,&l,&r);
        L[i]=l;
        addedge(s,v,l,0);
        addedge(u,t,l,0);
        addedge(u,v,r-l,i);
    }
    dinic(s,t);
    bool yes=1;
    for(int i=head[s];~i;i=e[i].next)if(e[i].f)yes=0;
    for(int i=head[t];~i;i=e[i].next)if(e[i^1].f)yes=0;
    if(yes)
    {
        puts("YES");
        for(int i=0;i<p;++i)L[e[i].id]+=e[i].f;
        for(int i=1;i<=m;++i)printf("%d\n",L[i]);
    }
    else
    {
        puts("NO");
    }
}

C.

类似树上最大独立集的DP，记一下方案，注意卡空间

#include<bits/stdc++.h>
#define maxn 500005
using namespace std;
int n;
int head[maxn],c;
int to[maxn],nxt[maxn];
void add(int u,int v)
{
    to[++c]=v;nxt[c]=head[u];head[u]=c;
}
int pre[maxn];
int dp[maxn][2];
int Ans[maxn],cnt;
pair<int,int> tmp[maxn];
int p;
void dfs(int u)
{
    dp[u][0]=0;dp[u][1]=1;
    for(int i=head[u];i;i=nxt[i])dfs(to[i]);
    p=0;
    for(int i=head[u];i;i=nxt[i])
    {
        int v=to[i];
        dp[u][1]+=dp[v][0];
        dp[u][0]+=dp[v][0];
        tmp[++p]=make_pair(dp[v][1]-dp[v][0],v);
    }
    sort(tmp+1,tmp+p+1);
    if(p>0)
    {
        int k=tmp[p].second;
        if(dp[k][1]-dp[k][0]>0)
        {
            dp[u][0]+=dp[k][1]-dp[k][0];
            pre[u]=k;
        }
    }
}
void print(int u,int k)
{
    if(k==1)
    {
        Ans[++cnt]=u;
        for(int i=head[u];i;i=nxt[i])print(to[i],0);
    }
    else
    {
        if(pre[u])print(pre[u],1);
        for(int i=head[u];i;i=nxt[i])if(to[i]!=pre[u])print(to[i],0);
    }
}
int main()
{
    freopen("grant.in","r",stdin);
    freopen("grant.out","w",stdout);
    scanf("%d",&n);
    for(int i=2;i<=n;++i)
    {
        int x;
        scanf("%d",&x);
        add(x,i);
    }
    dfs(1);
    printf("%d\n",dp[1][0]*1000);
    print(1,0);
    sort(Ans+1,Ans+cnt+1);
    for(int i=1;i<=cnt;++i)printf("%d%c",Ans[i],(i==cnt)?'\n':' ');
}

D.

发现就是找朝外走两步的方案数，朝外走一步就是度数，朝外走两步就是相邻点度数求和

#include<bits/stdc++.h>
#define maxn 200005
using namespace std;
typedef long long ll;
int n,m;
vector<int> g[maxn];
int d[maxn];
int main()
{
    freopen("matrix.in","r",stdin);
    freopen("matrix.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;++i)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        g[u].push_back(v);
        g[v].push_back(u);
        d[u]++;d[v]++;
    }
    ll ans=0;
    for(int i=1;i<=n;++i)
        for(int v:g[i])ans+=d[v];
    cout<<ans<<endl;
}

E.

状压矩阵快速幂，这里可以直接十进制快速幂

#include<bits/stdc++.h>
using namespace std;
char N[105];
int m,mod;
struct Matrix
{
    int a[35][35];
    void clear()
    {
        memset(a,0,sizeof(a));
    }
    void init()
    {
        clear();
        for(int i=0;i<(1<<m);++i)a[i][i]=1;
    }
};
Matrix operator * (Matrix &A,Matrix &B)
{
    Matrix C;C.clear();
    for(int j=0;j<(1<<m);++j)
        for(int k=0;k<(1<<m);++k)if(B.a[k][j])
            for(int i=0;i<(1<<m);++i)
                C.a[i][j]=(C.a[i][j]+A.a[i][k]*B.a[k][j])%mod;
    return C;
}
Matrix fastpow(Matrix a,char *N)
{
    int len=strlen(N);
    Matrix ans;ans.init();
    for(int i=len-1;i>=0;--i)
    {
        int d=N[i]-'0';
        for(int j=1;j<=d;++j)ans=ans*a;
        Matrix tmp=a;
        for(int j=1;j<=9;++j)tmp=tmp*a;
        a=tmp;
    }
    return ans;
}
int main()
{
    freopen("nice.in","r",stdin);
    freopen("nice.out","w",stdout);
    scanf("%s",N);
    scanf("%d%d",&m,&mod);
    Matrix A;A.clear();
    for(int i=0;i<(1<<m);++i)
        for(int j=0;j<(1<<m);++j)
        {
            int yes=1;
            int si=(1<<m)-1-i,sj=(1<<m)-1-j;
            for(int k=1;k<m;++k)if((i&(1<<k))&&(j&(1<<k))&&(i&(1<<(k-1)))&&(j&(1<<(k-1))))yes=0;
            for(int k=1;k<m;++k)if((si&(1<<k))&&(sj&(1<<k))&&(si&(1<<(k-1)))&&(sj&(1<<(k-1))))yes=0;
            A.a[i][j]=yes;
        }
    int Len=strlen(N);
    N[Len-1]--;
    for(int i=Len-1;i>=0;--i)if(N[i]<'0')N[i]+=10,N[i-1]--;
    A=fastpow(A,N);
    int ans=0;
    for(int i=0;i<(1<<m);++i)
        for(int j=0;j<(1<<m);++j)ans=(ans+A.a[i][j])%mod;
    printf("%d\n",ans);
}

F.

把船的圆的半径加到岛上，那么就转化成一个点是否被一堆圆围住

有一个很容易观察到的转化是在有公共部分的两圆圆心之间连线，转化成点是否在一个多边形内

这里多边形不止一个，因此可以通过角度和判断，把边权赋为有向角度，那么角度和为负等价于有负环，Floyd跑一下就行了

#include<bits/stdc++.h> 
using namespace std;
//LOOP
#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
//OTHER
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define all(x) (x).begin(),(x).end()
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RS(s) scanf("%s", s)
//OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(n) printf("%s\n", n)
//debug
//#define online_judge
#ifndef online_judge
#define debugt(a) cout << (#a) << "=" << a << " ";
#define debugI(a) debugt(a) cout << endl
#define debugII(a, b) debugt(a) debugt(b) cout << endl
#define debugIII(a, b, c) debugt(a) debugt(b) debugt(c) cout << endl
#define debugIV(a, b, c, d) debugt(a) debugt(b) debugt(c) debugt(d) cout << endl
#else
#define debugI(v)
#define debugII(a, b)
#define debugIII(a, b, c)
#define debugIV(a, b, c, d)
#endif
typedef long long LL;
typedef unsigned long long ULL;
typedef vector <int> VI;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const int MOD = 100000007;
const int MAXN = 1000010;
const double PI = acos(-1.0);


double torad(double deg)
{
    return deg / 180 * PI;
}
inline int dcmp(double x)
{
    if(fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}
struct Point
{
    double x, y;
    Point(double x=0, double y=0):x(x),y(y) { }
    inline void read()
    {
        scanf("%lf%lf", &x, &y);
    }
};
typedef vector<Point> Polygon;
typedef Point Vector;

inline Vector operator + (Vector A, Vector B)
{
    return Vector(A.x+B.x, A.y+B.y);
}
inline Vector operator - (Point A, Point B)
{
    return Vector(A.x-B.x, A.y-B.y);
}
inline Vector operator * (Vector A, double p)
{
    return Vector(A.x*p, A.y*p);
}
inline Vector operator / (Vector A, double p)
{
    return Vector(A.x/p, A.y/p);
}
inline bool operator < (Point a, Point b)
{
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}


inline bool operator == (Point a, Point b)
{
    return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}


inline double Dot(Vector A, Vector B)
{
    return A.x*B.x + A.y*B.y;
}
inline double Length(Vector A)
{
    return sqrt(Dot(A, A));
}
inline double Angle(Vector A, Vector B)
{
    return acos(Dot(A, B) / Length(A) / Length(B));
}
inline double angle(Vector v)
{
    return atan2(v.y, v.x);
}
inline double Cross(Vector A, Vector B)
{
    return A.x*B.y - A.y*B.x;
}
inline Vector Unit(Vector x)
{
    return x / Length(x);
}
inline Vector Normal(Vector x)
{
    return Point(-x.y, x.x) / Length(x);
}
inline Vector Rotate(Vector A, double rad)
{
    return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));
}
inline double Area2(Point A, Point B, Point C)
{
    return Cross(B-A, C-A);
}
template <class T> T sqr(T x)
{
    return x * x ;
}

struct Circle
{
    Point c;
    double r;
    Circle() {}
    Circle(Point c, double r):c(c), r(r) {}
    inline Point point(double a)
    {
        return Point(c.x+cos(a)*r, c.y+sin(a)*r);
    }
    inline void read()
    {
        scanf("%lf%lf%lf", &c.x, &c.y, &r);
    }
};

bool havecommonpoint(Circle A,Circle B)
{
    return dcmp(sqr(A.c.x-B.c.x)+sqr(A.c.y-B.c.y)-sqr(A.r+B.r))<0;
}
#define maxn 305
int n;
Circle a[maxn];
Point O;
double R;
double dis[maxn][maxn];
int main()
{
    freopen("out.in","r",stdin);
    freopen("out.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;++i)a[i].read();
    O.read();
    scanf("%lf",&R);
    for(int i=1;i<=n;++i)a[i].r+=R;
    for(int i=1;i<=n;++i)
        for(int j=1;j<=n;++j)dis[i][j]=1e18;
    for(int i=1;i<=n;++i)
        for(int j=1;j<=n;++j)if(i!=j&&havecommonpoint(a[i],a[j]))
        {
            double t=Angle(a[i].c-O,a[j].c-O);
            if(dcmp(Cross(a[i].c-O,a[j].c-O))<0)t=-t;
            dis[i][j]=min(dis[i][j],t);
        }
    for(int k=1;k<=n;++k)
        for(int i=1;i<=n;++i)
            for(int j=1;j<=n;++j)dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]+eps);
    bool yes=1;
    for(int i=1;i<=n;++i)
    {
        if(dcmp(dis[i][i])<=0)yes=0;
    }
    if(yes)puts("YES");
    else puts("NO");
}

G.

类似LIS，线段树优化记录方案，注意细节

#include<bits/stdc++.h>
#define maxn 100005
using namespace std;
int n;
struct Point
{
    int x,y,id;
}a[maxn];
int dy[maxn],d;
bool operator < (Point A,Point B)
{
    return A.x<B.x||(A.x==B.x&&A.y>B.y);
}
bool operator == (Point A,Point B)
{
    return A.x==B.x&&A.y==B.y;
}
int maxv[maxn<<2],maxid[maxn<<2];
void pushup(int rt)
{
    if(maxv[rt<<1]>maxv[rt<<1|1])maxv[rt]=maxv[rt<<1],maxid[rt]=maxid[rt<<1];
    else maxv[rt]=maxv[rt<<1|1],maxid[rt]=maxid[rt<<1|1];
}
void update(int rt,int l,int r,int pos,int v,int id)
{
    if(l==r)
    {
        if(v>maxv[rt])
        {
            maxv[rt]=v;
            maxid[rt]=id;
        }
        return;
    }
    int mid=(l+r)>>1;
    if(pos<=mid)update(rt<<1,l,mid,pos,v,id);
    else update(rt<<1|1,mid+1,r,pos,v,id);
    pushup(rt);
}
pair<int,int> query(int rt,int l,int r,int ql,int qr)
{
    if(ql>qr)return make_pair(0,0);
    if(ql<=l&&r<=qr)return make_pair(maxv[rt],maxid[rt]);
    int mid=(l+r)>>1;
    if(qr<=mid)return query(rt<<1,l,mid,ql,qr);
    else if(ql>mid)return query(rt<<1|1,mid+1,r,ql,qr);
    else
    {
        pair<int,int> ans1=query(rt<<1,l,mid,ql,qr),ans2=query(rt<<1|1,mid+1,r,ql,qr);
        if(ans1.first>=ans2.first)return ans1;
        else return ans2;
    }
}
int dp[maxn],pre[maxn];
void print(int u)
{
    if(!u)return;
    printf("%d ",a[u].id);
    print(pre[u]);
}
int main()
{
    freopen("people.in","r",stdin);
    freopen("people.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;++i)scanf("%d%d",&a[i].x,&a[i].y),a[i].id=i,dy[++d]=a[i].y;
    sort(a+1,a+n+1);
    sort(dy+1,dy+d+1);
    d=unique(dy+1,dy+d+1)-dy-1;
    for(int i=1;i<=n;++i)a[i].y=lower_bound(dy+1,dy+d+1,a[i].y)-dy;
    n=unique(a+1,a+n+1)-a-1;
    for(int i=1;i<=n;++i)
    {
        pair<int,int> t=query(1,1,n,1,a[i].y-1);
        dp[i]=t.first+1;pre[i]=t.second;
        update(1,1,n,a[i].y,dp[i],i);
    }
    int ans=0;
    for(int i=1;i<=n;++i)ans=max(ans,dp[i]);
    for(int i=1;i<=n;++i)if(dp[i]==ans)
    {
        printf("%d\n",ans);
        print(i);
        break;
    }
}

H.

可以转化为计数一些01串xor为0的方案数

这个东西就是选一个线性基，假设大小为r，那么剩下n-r个随便选，在线性基里总对应唯一方案，答案就是\(2^{n-r}\)

#include<bits/stdc++.h>
using namespace std;
int m,n;
int prime[5005],cnt;
bool isprime(int x)
{
    for(int i=2;i*i<=x;++i)if(x%i==0)return 0;
    return 1;
}
void get(int N)
{
    for(int i=2;i<=N;++i)if(isprime(i))prime[cnt++]=i;
}
bitset<100> a[105];
void D(int id,int x)
{
    for(int i=0;i<m;++i)if(x%prime[i]==0)
    {
        while(x%prime[i]==0)
        {
            x/=prime[i];
            a[id][i]=a[id][i]^1;
        }
    }
}
struct LinearBasis
{
    bitset<100> basis[105];
    int r;
    void clear()
    {
        r=0;
        for(int i=0;i<105;++i)basis[i].reset();
    }
    inline void insert(bitset<100> val)
    {
        for(int i=99;i>=0;i--)if(val[i])
        {
            if(basis[i].count()==0)
            {
                basis[i]=val;
                r++;
                break;
            }
            val^=basis[i];
        }
    } 
}A;
int out[105];
void print(__int128 x)
{
    if(!x)
    {
        puts("0");
        return;
    }
    int p=0;
    while(x)
    {
        out[++p]=x%10;
        x/=10;
    }
    for(int i=p;i>=1;--i)printf("%d",out[i]);
}
int main()
{
    freopen("rsa.in","r",stdin);
    freopen("rsa.out","w",stdout);
    scanf("%d%d",&m,&n);
    get(5000);
    A.clear();
    for(int i=1;i<=n;++i)
    {
        int x;
        scanf("%d",&x);
        D(i,x);
        A.insert(a[i]);
    }
    int r=A.r;
    __int128 ans=1;
    for(int i=1;i<=n-r;++i)ans=ans*2;
    ans-=1;
    print(ans);
}