Educational Codeforces Round 6

620A - Professor GukiZ's Robot    20171122

(ans=max(left | x2-x1 ight |,left | y2-y1 ight |))

#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int X[2],Y[2],x,y;
int main()
{
    scanf("%d%d%d%d",&X[0],&Y[0],&X[1],&Y[1]);
    x=abs(X[1]-X[0]),y=abs(Y[1]-Y[0]);
    printf("%d
",max(x,y));
    return 0;
}

620B - Grandfather Dovlet’s calculator    20171122

预处理每个字符的花费，按照题意模拟即可

#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int ans,a,b,f[10];
int cal(int k)
{
    int _=0;
    while(k)
      _+=f[k%10],k/=10;
    return _;
}
int main()
{
    scanf("%d%d",&a,&b);
    f[0]=6,f[1]=2,f[2]=5,f[3]=5,f[4]=4;
    f[5]=5,f[6]=6,f[7]=3,f[8]=7,f[9]=6;
    for(int i=a;i<=b;i++)
      ans+=cal(i);
    printf("%d
",ans);
    return 0;
}

620C - Pearls in a Row    20171122

开个set记录当前已在区间类的珍珠种类，O(n)扫一遍即可

#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#include<set>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
vector<int>l,r;
int n,x,t=1;
set<int>s;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
      {
      if(t==i)s.clear();
      scanf("%d",&x);
      if(s.count(x))
        l.push_back(t),r.push_back(i),t=i+1;
      else s.insert(x);
      }
    if(!l.size())return printf("-1
"),0;
    r[r.size()-1]=n;
    printf("%d
",l.size());
    for(int i=0;i<l.size();i++)
      printf("%d %d
",l[i],r[i]);
    return 0;
}

620D - Professor GukiZ and Two Arrays    20171122

暴力分类讨论(k=0,1,2)的所有情况就好了

#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#define N 2001
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define LL long long
struct rua
{
    LL i,j,v;
}f[N*N];
LL n,m,d,sa,sb,a[N],b[N],ans[3],v[N*N],_,a1,b1,b2;
bool cmp(rua x,rua y){return x.v<y.v;}
void cdx(LL i,LL _)
{
    if(!_)return;
    if(f[_].i==f[i].i || f[_].j==f[i].j)return;
    if(ans[2]>abs(d+v[_]))
      ans[2]=abs(d+v[_]),b1=i,b2=_;
}
int main()
{
    ans[2]=(1ll<<62);
    scanf("%I64d",&n);
    for(LL i=1;i<=n;i++)
      scanf("%I64d",&a[i]),sa+=a[i];
    scanf("%I64d",&m);
    for(LL i=1;i<=m;i++)
      scanf("%I64d",&b[i]),sb+=b[i];
    d=sa-sb;
    for(LL i=1;i<=n;i++)
      for(LL j=1;j<=m;j++)
        {
        f[i*m-m+j].v=2*(b[j]-a[i]);
        f[i*m-m+j].i=i;
        f[i*m-m+j].j=j;
        }
    ans[0]=abs(d);
    sort(f+1,f+n*m+1,cmp);
    for(LL i=1;i<=n;i++)
      for(LL j=1;j<=m;j++)
        v[i*m-m+j]=f[i*m-m+j].v;
    _=lower_bound(v+1,v+n*m+1,-d)-v;
    if(_==n*m+1)ans[1]=abs(d+v[_-1]),a1=_-1;else
    if(_==1)ans[1]=abs(d+v[1]),a1=_;else
    if(abs(d+v[_])<abs(d+v[_-1]))ans[1]=abs(d+v[_]),a1=_;
    else ans[1]=abs(d+v[_-1]),a1=_-1;
    for(int i=1;i<n*m;i++)
      {
      d+=v[i];
      _=lower_bound(v+1,v+n*m+1,-d)-v;
      if(_==n*m+1)cdx(i,_-1);
      else if(_==1)cdx(i,1-(i==1));else
      cdx(i,_+(i==_)),cdx(i,_-1-(i==_-1));
      d-=v[i];
      }
    if(ans[0]<=ans[1] && ans[0]<=ans[2])
      return printf("%I64d
0
",ans[0]),0;
    if(ans[1]<=ans[0] && ans[1]<=ans[2])
      return printf("%I64d
1
%I64d %I64d
",ans[1],f[a1].i,f[a1].j),0;
    printf("%I64d
2
",ans[2]);
    printf("%I64d %I64d
",f[b1].i,f[b1].j);
    printf("%I64d %I64d
",f[b2].i,f[b2].j);
    return 0;
}

620E - New Year Tree    20180919

预处理每个点的DFS序，可以得出每棵子树对应的DFS序的范围。由于(c_i)不超过60，故可将修改操作转换为：把子树内所有点的价值改为(2^{c})，将询问操作转换为：询问子树内所有点价值进行或运算的结果在二进制中1的个数，用线段树做就好了

#include<bits/stdc++.h>
using namespace std;
#define N 400001
#define LL long long
vector<int>d[N];
int n,m,t,x,y,c[N],l[N],r[N];
struct rua{int l,r;LL f,w;}tr[N<<2];
LL a[N];
void update(int x)
{
    int lson=x*2,rson=x*2+1;
    tr[x].w=tr[lson].w|tr[rson].w;
}
void down(int x)
{
    LL c=tr[x].w;tr[x].f=0;
    int lson=x*2,rson=x*2+1;
    tr[lson].f=tr[lson].w=c;
    tr[rson].f=tr[rson].w=c;
}
void Build(int l,int r,int x)
{
    tr[x]={l,r,0,0};
    if(l==r){tr[x].w=a[l];return;}
    int mid=l+r>>1;
    Build(l,mid,x*2);
    Build(mid+1,r,x*2+1);
    update(x);
}
void Change(int L,int R,LL c,int x)
{
    int l=tr[x].l,r=tr[x].r;
    int mid=l+r>>1;
    if(L<=l && r<=R){tr[x].f=tr[x].w=c;return;}
    if(tr[x].f)down(x);
    if(L<=mid)Change(L,R,c,x*2);
    if(R>mid)Change(L,R,c,x*2+1);
    update(x);
}
LL Query(int L,int R,int x)
{
    int l=tr[x].l,r=tr[x].r;
    int mid=l+r>>1;
    LL res=0;
    if(L<=l && r<=R)return tr[x].w;
    if(tr[x].f)down(x);
    if(L<=mid)res|=Query(L,R,x*2);
    if(R>mid)res|=Query(L,R,x*2+1);
    update(x);
    return res;
}
void dfs(int cur,int pre)
{
    l[cur]=++x;
    a[x]=1ll<<c[cur];
    for(auto nxt:d[cur])if(nxt!=pre)dfs(nxt,cur);
    r[cur]=x;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
      scanf("%d",&c[i]);
    for(int i=2;i<=n;i++)
      scanf("%d%d",&x,&y),
      d[x].push_back(y),
      d[y].push_back(x);
    x=0,dfs(1,0);
    Build(1,n,1);
    for(int i=1;i<=m;i++)
      {
      scanf("%d",&t);
      if(t==1)
        scanf("%d%d",&x,&y),
        Change(l[x],r[x],1ll<<y,1);
      else
        scanf("%d",&x),
        printf("%d
",(int)__builtin_popcountll(Query(l[x],r[x],1))); 
      }
}

620F - Xors on Segments    20171122

这题...预处理从1异或到n的值（即前缀异或和），(O(n^{2}))莽一下就过了...CF的评测机是跑得真快

#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#define N 50001
#define M 1000001 
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,a[N],l[N],r[N],t[N],ans[N],f[M];
int main()
{
    for(int i=1;i<M;i++)
      f[i]=f[i-1]^i;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
      scanf("%d",&a[i]);
    for(int i=1;i<=m;i++)
      scanf("%d%d",&l[i],&r[i]);
    for(int i=1;i<=n;i++)
      {
      int _=0;t[i-1]=0;
      for(int j=i;j<=n;j++)
        _=max(_,f[a[i]]^f[a[j]]^min(a[i],a[j])),t[j]=_;
      for(int j=1;j<=m;j++)
        if(l[j]<=i && r[j]>=i)
          ans[j]=max(ans[j],t[r[j]]);
      }
    for(int i=1;i<=m;i++)printf("%d
",ans[i]);return 0;
}