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

Posted 2020-10-01

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\n",i);
            return 0;
        }
    return 0;
}

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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?‘\n‘:‘ ‘);
    return 0;
}

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 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;
}

View Code

 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\n",i,DAG[i].size());
        for(int j=0;j<DAG[i].size();j++) {
            printf("%d ",DAG[i][j].v);
        }
        puts("");
    }*/

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

View Code

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;
}

View Code

J 待补