Codeforces Round #441 (Div. 1, by Moscow Team Olympiad)

A Classroom Watch

显然sum of digit 不会超过9*9 暴力枚举

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} int main(){//  freopen("A.in","r",stdin);//  freopen(".out","w",stdout);    int n=read();    vi v;    Fork(i,max(n-1000,1),n) {        ll p=i,q=0;        while(p) q+=p%10,p/=10;        if (q+i==n) v.pb(i);    }    sort(ALL(v));    cout<<SI(v)<<endl;    Rep(i,SI(v)) cout<<v[i]<<endl;    return 0;}

B Sorting the Coins

冒泡排序每次都能冒泡一个数

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} ll a[312345];int main(){    freopen("B.in","r",stdin);//  freopen(".out","w",stdout);    int n=read();    MEM(a)    putchar('1');    int l=n+1,r=n;    For(i,n) {        int p=read();        gmin(l,p)        while(a[r]) --r;        cout<<' '<<r-l+1-i;    }    return 0;}

C National Property

2-sat 模型很明显了

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (200000+10)int cnt[MAXN]={},l[MAXN]={},r[MAXN]={};int a[MAXN]={};bool fl=1;int cap[MAXN]={};class SSC  {  public:      int n,b[MAXN],num[MAXN];      vector<int> G[MAXN],rG[MAXN]; //ͼ£¬·´ÏòºóµÄͼ       vector<int> vs; //ºóÐø±éÀú¶¥µãÁÐ±í       void mem(int _n)      {          n=_n; MEM(num)          For(i,n) G[i].clear(),rG[i].clear();          vs.clear();      }      void addedge(int u,int v)      {          G[u].push_back(v);          rG[v].push_back(u);      }      void dfs(int x)      {          b[x]=1;          Rep(i,G[x].size())          {              if (!b[G[x][i]]) dfs(G[x][i]);          }          vs.push_back(x);      }      void rdfs(int x,int k)      {          b[x]=1;num[x]=k;          Rep(i,rG[x].size())          {              if (!b[rG[x][i]]) rdfs(rG[x][i],k);          }      }      int ssc()      {          MEM(b)           For(i,n) if (!b[i]) dfs(i);          MEM(b) int k=0;          RepD(i,vs.size()-1) if (!b[vs[i]]) rdfs(vs[i],++k);          return k;      }  };  class Two_Sat //(a||!b)&&(c||b)&&...{public:    int n;    SSC S;  //1..n    void mem(int _n)    {        n=_n;        S.mem(2*n);    }    int no(int u){if (u>0) return u;else return n-u; }    void addedge(int u,int v) //((b¡Å ~c) : b -c    {        S.addedge(no(-u),no(v));        S.addedge(no(-v),no(u));    }    bool work(vi &v) {        S.ssc();        For(i,n) {            if (S.num[i]==S.num[i+n]) return 0;            else if (S.num[i]>S.num[i+n]) v.pb(i);        }        return 1;    }}S;void check(int l,int r){    while(a[l]&&a[r]) {        if (a[l]!=a[r]) {            if (a[l]<a[r]) S.addedge(-a[r],a[l]);            if (a[l]>a[r]) S.addedge(a[l],a[l]),S.addedge(-a[r],-a[r]);            return;        }        ++l,++r;    }    if (a[l]&&!a[r]) {        fl=0;return ;    }}int main(){//  freopen("C.in","r",stdin);//  freopen(".out","w",stdout);    int n=read(),m=read();    S.mem(m);    l[0]=r[0]=0;    For(i,n) {        cnt[i]=read();        l[i]=r[i-1]+1;        r[i]=l[i];        For(j,cnt[i]) {            a[r[i]++]=read();        }        a[r[i]]=0;    }    For(i,n-1) {        check(l[i],l[i+1]);    }    vi v;    if (!fl|| !S.work(v)) puts("No");    else {        puts("Yes");        int sz=SI(v);        printf("%d\n",sz);        if (sz) {            printf("%d\n",v[0]);            For(i,sz-1) printf(" %d",v[i]);        }    }    return 0;}

D High Cry

给一个数列，问有多少子串满足or和>max值

枚举左端点，or值最多有Log(Ai)个，可以暴力在每段2分。

E Delivery Club

F Royal Questions

10w点2分图带权匹配，注意某一边点度数均为2。

题目可以转化为，最大环加外向树生成树，类似MST。

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (200100)int fa[MAXN],s[MAXN];int getfa(int x) {return (fa[x]==x)?x:(fa[x]=getfa(fa[x]));}pair< int ,pi > p[MAXN];int main(){//  freopen("F.in","r",stdin);//  freopen(".out","w",stdout);    int n=read(),m=read();    For(i,n) fa[i]=i;    For(i,m) p[i].se.fi=read(),p[i].se.se=read(),p[i].fi=read();    sort(p+1,p+1+m);    int ans=0;    ForD(i,m) {        int x=p[i].se.fi,y=p[i].se.se,w=p[i].fi;        x=getfa(x),y=getfa(y);        if( (x^y) && (!s[y]||!s[x]) ) {            fa[x]=y;            s[y]+=s[x];            ans+=w;        } else if (!s[x]){            ans+=w;            s[x]=1;        }    }    cout<<ans<<endl;    return 0;}