XVI Open Cup named after E.V. Pankratiev. GP of Ekaterinburg.

贼惨 130/186

B Black Widow

简单题

#include <bits/stdc++.h>
const long long mod = 1e9+7;
const double ex = 1e-10;
#define inf 0x3f3f3f3f
using namespace std;
map <int,int> M;
int a[1010];
int b[100010];
int main()
{
    int N;
    scanf("%d",&N);
    int t = 0;
    for (int i = 1; i<=N;i++){
        scanf("%d",&a[i]);
        t = max(t,a[i]);
        for (int j = 1; j*j<=a[i];j++)
            if (a[i]%j==0) {
                    b[j] = 1;
                if (a[i]/j > 100000)M[a[i]/j]++;
                else b[a[i]/j] = 1;
            }
    }
    for (int i = 2 ;i<=1000000001 ;i++)
        if ((i<=100000&&b[i]==0)||(i>100000&&M[i]==0))
        {
            printf("%d
",i);
            return 0;
        }
    return 0;
}

C Chitauri

海盗分金，按照海盗分金的贪心策略递推的求。

#include <bits/stdc++.h>
using namespace std;
int n,k;
int vis[3005],money[3005],last;
int main()
{
    scanf("%d%d",&n,&k);
    money[1] = k;
    last = 1;
    for (int i=2;i<=n;++i) {
            int cnt = 0;
            int gg = 0;
            for (int j = 1;j<=i-1;j++){
                if (money[j] == 0)
                    cnt++;
                if (money[j] == -1)
                    gg++;
            }
            if (2*(min(cnt,k)+gg+1) < i ) {
                money[i] = -1;
                continue;
            }
            last = i;
            int tk=k;
            if (k-((i+1)/2 - gg -1) > 0) money[i] =k-((i+1)/2 - gg -1),tk-=money[i];
            for (int j=i-1;j>=1;--j) {
                if (money[j]==-1)
                    money[j]=0;
                else if (money[j]==0) {
                    if (tk>0) {
                        money[j]=1;
                        --tk;
                    }
                } else
                    money[j]=0;
            }

        }
            for (int z=n;z>=1;--z)
            printf("%d%c",money[z],z==1?'
':' ');
    return 0;
}

 D Dr. Banner

简单DP

#include <bits/stdc++.h>
using namespace std;
const long long mod=1e9+7;
int n;
long long dp[21][100005];
int main()
{
    scanf("%d",&n);
    dp[0][n]=1;
    for (int i=1;i<=20;++i) {
        for (int j=1;j<=n;++j)
            dp[i-1][j]=(dp[i-1][j]+dp[i-1][j-1])%mod;
        for (int j=1;j*2<=n;++j)
            dp[i][j]=(dp[i-1][n]-dp[i-1][2*j-1]+mod)%mod;
    }
    long long res=0;
    for (int i=0;i<=20;++i)
        res=(res+dp[i][n])%mod;
    cout<<res<<endl;
    return 0;
}

E Egocentric Loki (补)

题目很长，其实就是枚举每个三角形，判断是否有点处于这个三角形和它的外接圆之间。

精度的要求还是比较低的，就是输入数据范围有误，，心态崩了。

#include <bits/stdc++.h>
using namespace std;
const long double eps=1e-8;
int sgn ( long double x ) {
    return ( x > eps ) - ( x < -eps ) ;
}
struct Point{
    long double x,y;
    Point(){}
    Point(long double _x,long double _y){
        x=_x;y=_y;
    }
    Point operator +(const Point &b)const{
        return Point(x+b.x,y+b.y);
    }
    Point operator /(const long double &k)const{
        return Point(x/k,y/k);
    }
    Point operator -(const Point &b)const {
        return Point(x-b.x,y-b.y);
    }
    Point operator *(const long double &rhs)const{
        return Point(x*rhs,y*rhs);
    }
    long double operator ^(const Point &b) const {
        return x*b.y-y*b.x;
    }
    long double distance(Point p) {
        return (x-p.x)*(x-p.x)+(y-p.y)*(y-p.y);
    }
    Point rotleft(){
        return Point(-y,x);
    }

};
struct Line{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e){
        s=_s;
        e=_e;
    }
    /*
    Point line_intersection (Line v) {//( P a , P b , P p , P q  ) {

        double U = cross ( p - a , q - p ) ;
        double D = cross ( b - a , q - p ) ;
        return a + ( b - a ) * ( U / D ) ;
    }
    */
    Point crosspoint(Line v) {
        long double U=(v.s-s)^(v.e-v.s);
        long double D=(e-s)^(v.e-v.s);
        return s+(e-s)*(U/D);
    }
};
struct Circle{
    Point p;long double r;
    Circle(){}
    Circle(Point a,Point b,Point c){
        Line u=Line((a+b)/2.0,(a+b)/2.0+(a-b).rotleft());
        Line v=Line((b+c)/2.0,(b+c)/2.0+(c-b).rotleft());
        p=u.crosspoint(v);
        r=p.distance(a);
    }
    int relation(Point b){
        long double dst=b.distance(p);
        if(sgn(dst-r)<0) return 2;
        else if(sgn(dst-r)==0) return 1;
        return 0;
    }
};
inline double area(double x1,double y1,double x2,double y2,double x,double y) {
    double X1=x1-x,Y1=y1-y,X2=x2-x,Y2=y2-y;
    return fabs(X1*Y2-X2*Y1);
}
inline bool intriangle(double x1,double y1,double x2,double y2,double x3,double y3,double x,double y) {
    int flag=sgn(area(x1,y1,x2,y2,x3,y3)-area(x1,y1,x2,y2,x,y)-area(x1,y1,x3,y3,x,y)-area(x2,y2,x3,y3,x,y));
    return flag==0;
}
int n;
long double xx1[10010],yy1[10010],xx2[10010],yy2[10010],xx3[10010],yy3[10010];
int main()
{
    while (scanf("%d",&n)==1) {
        for(int i=1;i<=n;i++)
            cin>>xx1[i]>>yy1[i]>>xx2[i]>>yy2[i]>>xx3[i]>>yy3[i];
        int flag=0;
        for(int i=1;i<=n;i++) {
            Circle now=Circle(Point(xx1[i],yy1[i]),Point(xx2[i],yy2[i]),Point(xx3[i],yy3[i]));
            //cout<<now.r<<" "<<now.p.x<<" "<<now.p.y<<endl;

            for(int j=1;j<=n;j++) {
                if(j==i) continue;
              //  cout<<now.relation(Point(x1[j],y1[j]))<<endl;
                if(!intriangle(xx1[i],yy1[i],xx2[i],yy2[i],xx3[i],yy3[i],xx1[j],yy1[j])&&(now.relation(Point(xx1[j],yy1[j]))>=1))
                    flag=1;
                if(!intriangle(xx1[i],yy1[i],xx2[i],yy2[i],xx3[i],yy3[i],xx2[j],yy2[j])&&(now.relation(Point(xx2[j],yy2[j]))>=1))
                    flag=1;
                if(!intriangle(xx1[i],yy1[i],xx2[i],yy2[i],xx3[i],yy3[i],xx3[j],yy3[j])&&(now.relation(Point(xx3[j],yy3[j]))>=1))
                    flag=1;
                if(flag)
                    break;
            }
            if(flag)
                break;
        }
        if(flag) printf("NO
");
        else printf("YES
");
    }
    return 0;
}

F Fury

缩点 ，然后DAG上贪心

#include <bits/stdc++.h>
const long long mod = 1e9+7;
const double ex = 1e-10;
#define REP(i,x,y) for(int i=x;i<(y);i++)
#define RREP(i,x,y) for(int i=x;i>(y);i--)
#define inf 0x3f3f3f3f
#define maxn 310
using namespace std;
int n,m,pre[maxn],low[maxn],sccno[maxn],dfs_clock,scc_cnt;
vector<int>e[maxn];
stack<int>S;
void dfs(int st) {
    pre[st]=low[st]=(++dfs_clock);
    S.push(st);
    REP(i,0,e[st].size()) {
        int nxt=e[st][i];
        if(!pre[nxt]) {
            dfs(nxt);
            low[st]=min(low[st],low[nxt]);
        }
        else if(!sccno[nxt]) {
            low[st]=min(low[st],pre[nxt]);
        }
    }
    if(low[st]==pre[st]) {
        scc_cnt++;
        while(1) {
            int now=S.top();S.pop();
            sccno[now]=scc_cnt;
            if(now==st) break;

        }
    }
}
void Tarjan() {
    dfs_clock=scc_cnt=0;
    memset(sccno,0,sizeof(sccno));
    memset(pre,0,sizeof(pre));
    for(int i=1;i<=n;i++) if(!pre[i]) dfs(i);
}
struct edge{
    int v,f,u,vv;
    edge(){}
    edge(int _v,int _f,int u,int v):v(_v),f(_f),u(u),vv(v){}
};
vector<edge>DAG[maxn];
int col[maxn],DAG_cnt;
void gao() {
    REP(i,1,n+1) {
        int u=sccno[i];
        REP(j,0,e[i].size()) {
            int v=sccno[e[i][j]];
            if(v==u) continue;
            DAG_cnt++;
            DAG[u].push_back(edge(v,0,i,e[i][j]));
        }
    }
}
int num[maxn];
int dfs2(int rt) {
    num[rt]++;
    for(int i=0;i<DAG[rt].size();i++) {
        int nxt=DAG[rt][i].v;
        if(DAG[rt][i].f==1) continue;
        if(num[nxt]>=1) {num[nxt]++;continue;}
        dfs2(nxt);
    }
}
vector<int>SCC[maxn];
typedef pair<int,int> pii;
vector<pii>Ans;
int main()
{
    scanf("%d %d",&n,&m);
    REP(i,1,m+1) {
        int u,v;scanf("%d %d",&u,&v);
        e[u].push_back(v);
    }
    Tarjan();
    int ans=0;
    for(int i=1;i<=n;i++)
        SCC[sccno[i]].push_back(i);
    for(int i=1;i<=scc_cnt;i++){
        if(SCC[i].size()==1) continue;
        for(int j=0;j<SCC[i].size();j++){
            Ans.push_back(make_pair(SCC[i][j],SCC[i][(j+1)%SCC[i].size()]));
        }
    }
   // REP(i,0,Ans.size()) cout<<Ans[i].first<<" "<<Ans[i].second<<endl;
    for(int i=1;i<=n;i++) col[sccno[i]]++;
    for(int i=1;i<=scc_cnt;i++) if(col[i]>1) ans+=col[i];
    gao();
    for(int i=1;i<=scc_cnt;i++) {
        for(int i=1;i<=scc_cnt;i++) num[i]=0;
        dfs2(i);
        for(int j=0;j<DAG[i].size();j++) {
            if(num[DAG[i][j].v]<=1) continue;
            DAG_cnt--;
            num[DAG[i][j].v]--;
            DAG[i][j].f=1;
        }
        for(int j=0;j<DAG[i].size();j++) {
            if(DAG[i][j].f==1) continue;
            Ans.push_back(make_pair(DAG[i][j].u,DAG[i][j].vv));
        }
    }
   /* for(int i=1;i<=scc_cnt;i++) {
        printf("%d: %d
",i,DAG[i].size());
        for(int j=0;j<DAG[i].size();j++) {
            printf("%d ",DAG[i][j].v);
        }
        puts("");
    }*/

    printf("%d %d
",n,DAG_cnt+ans);
    REP(i,0,Ans.size()) {
        printf("%d %d
",Ans[i].first,Ans[i].second);
    }
    return 0;
}

G Groot

签到

#include <bits/stdc++.h>
const long long mod = 1e9+7;
const double ex = 1e-10;
#define inf 0x3f3f3f3f
using namespace std;
int main()
{
    string  s;
    cin >> s ;
    cin >> s ;
    cin >> s ;
    if (s.length() == 5) puts("Pfff");
        else
    {
        cout <<'W';
        for (int i = 1; i<=s.length()-5;i++) cout <<'o';
        puts("w");
    }
    return 0;
}

J 待补