Codechef August Challenge 2019 Division 2
Preface
老年菜鸡终于开始打CC了，由于他太弱了所以只能打Div2
因为台风的原因challenge并没有写，所以水了个Rank7
A Football
SB模拟题不解释
#include<cstdio>
#include<iostream>
#define RI register int
using namespace std;
const int N=155;
int t,n,a[N],b[N],ans;
int main()
{
for (scanf("%d",&t);t;--t)
{
RI i; for (ans=0,scanf("%d",&n),i=1;i<=n;++i) scanf("%d",&a[i]);
for (i=1;i<=n;++i) scanf("%d",&b[i]),ans=max(ans,20*a[i]-10*b[i]);
printf("%d\n",ans);
}
return 0;
}
B Distribute Apples
题意杀，简单分析之后发现如果\(k^2|n\)就是NO，否则就是YES
#include<cstdio>
#include<iostream>
#define RI register int
using namespace std;
int t; long long n,k;
int main()
{
for (scanf("%d",&t);t;--t)
{
scanf("%lld%lld",&n,&k);
if (n/k<k) { puts("YES"); continue; }
puts(n%(k*k)?"YES":"NO");
}
return 0;
}
C Dilemma
首先不考虑有\(0\)的情况，我们发现长度为奇数的\(1\)就是合法的
考虑在两段\(1\)中间插入一段\(0\)，手玩一下发现加入多少都是没有影响的
因此我们按照连续的\(1\)的奇偶性个数讨论，如果奇数的连续的\(1\)有奇数段就是合法的，否则不合法
#include<cstdio>
#include<cstring>
#define RI register int
using namespace std;
const int N=100005;
int t,n; char s[N];
int main()
{
for (scanf("%d",&t);t;--t)
{
scanf("%s",s+1); n=strlen(s+1); int num=0,len=0;
for (RI i=1;i<=n;++i) if (s[i]=='1')
{
if (s[i-1]=='1') ++len; else num+=len&1,len=1;
}
num+=len&1; puts(num&1?"WIN":"LOSE");
}
return 0;
}
D Zombie and the Caves
刚开始看错题了，以为要重排这个序列来构造方案
没想到是直接模拟判断是否可行即可，区间加单点求值差分一下即可
#include<cstdio>
#include<cctype>
#include<iostream>
#include<algorithm>
#define RI register int
#define CI const int&
#define Tp template <typename T>
using namespace std;
const int N=100005;
int t,n,x,h[N],dlt[N];
class FileInputOutput
{
private:
static const int S=1<<21;
#define tc() (A==B&&(B=(A=Fin)+fread(Fin,1,S,stdin),A==B)?EOF:*A++)
char Fin[S],*A,*B;
public:
Tp inline void read(T& x)
{
x=0; char ch; while (!isdigit(ch=tc()));
while (x=(x<<3)+(x<<1)+(ch&15),isdigit(ch=tc()));
}
#undef tc
}F;
int main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (F.read(t);t;--t)
{
RI i; for (F.read(n),i=1;i<=n;++i)
F.read(x),++dlt[max(i-x,1)],--dlt[min(i+x,n)+1];
for (i=1;i<=n;++i) dlt[i]+=dlt[i-1],F.read(h[i]);
sort(dlt+1,dlt+n+1); sort(h+1,h+n+1);
bool flag=1; for (i=1;i<=n;++i)
if (dlt[i]!=h[i]) { flag=0; break; }
for (i=1;i<=n+1;++i) dlt[i]=0;
puts(flag?"YES":"NO");
}
return 0;
}
E Guddu and his Mother
我们发现如果一个区间\([l,r]\)内所有数异或起来是\(0\)那么中间的点可以取区间长度-1的位置
因此我们统计一下异或和为\(0\)的区间长度和即可，用树状数组统计一下即可
#include<cstdio>
#include<cctype>
#include<vector>
#include<algorithm>
#define int long long
#define RI register int
#define CI const int&
#define Tp template <typename T>
using namespace std;
typedef vector <int>::iterator VI;
const int N=100005;
int t,n,x,ret,stk[N],top; bool vis[N*10];
vector <int> L[N*10],R[N*10]; long long ans;
class FileInputOutput
{
private:
static const int S=1<<21;
#define tc() (A==B&&(B=(A=Fin)+fread(Fin,1,S,stdin),A==B)?EOF:*A++)
char Fin[S],*A,*B;
public:
Tp inline void read(T& x)
{
x=0; char ch; while (!isdigit(ch=tc()));
while (x=(x<<3)+(x<<1)+(ch&15),isdigit(ch=tc()));
}
#undef tc
}F;
class Tree_Array
{
private:
int bit[N];
public:
#define lowbit(x) x&-x
inline void add(RI x,CI y)
{
for (;x<=n;x+=lowbit(x)) bit[x]+=y;
}
inline int get(RI x,int ret=0)
{
for (;x;x-=lowbit(x)) ret+=bit[x]; return ret;
}
#undef lowbit
}NUM,SUM;
signed main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (F.read(t);t;--t)
{
RI i,j; for (F.read(n),vis[stk[top=1]=ret=0]=i=1;i<=n;++i)
{
L[ret].push_back(i); F.read(x);
ret^=x; R[ret].push_back(i);
if (!vis[ret]) stk[++top]=ret,vis[ret]=0;
}
for (ans=0,i=1;i<=top;++i)
{
for (VI it=L[stk[i]].begin();it!=L[stk[i]].end();++it)
NUM.add(*it,1),SUM.add(*it,*it);
for (VI it=R[stk[i]].begin();it!=R[stk[i]].end();++it)
ans+=(*it)*NUM.get(*it)-SUM.get(*it);
for (VI it=L[stk[i]].begin();it!=L[stk[i]].end();++it)
NUM.add(*it,-1),SUM.add(*it,-(*it));
L[stk[i]].clear(); R[stk[i]].clear(); vis[stk[i]]=0;
}
printf("%lld\n",ans);
}
return 0;
}
F Encoding
看到数据范围直接数位DP一下就好了
定义\(f_{i,j}\)表示当前处理完了前\(i\)位，上一位是\(j\)时的状态
状态我们要记录两个，一个是合法的状态数，用来累计这一位的贡献，另一个就是总贡献数
#include<cstdio>
#include<cctype>
#define RI register int
#define CI const int&
#define Tp template <typename T>
using namespace std;
const int N=100005,mod=1e9+7;
int t,n,a[N],pw[N]; char ch;
struct data
{
int num,sum;
inline data(CI Num=-1,CI Sum=-1)
{
num=Num; sum=Sum;
}
}f[N][10];
inline bool operator ~ (const data& x)
{
return ~x.num&&~x.sum;
}
class FileInputOutput
{
private:
static const int S=1<<21;
#define tc() (A==B&&(B=(A=Fin)+fread(Fin,1,S,stdin),A==B)?EOF:*A++)
char Fin[S],*A,*B;
public:
Tp inline void read(T& x)
{
x=0; char ch; while (!isdigit(ch=tc()));
while (x=(x<<3)+(x<<1)+(ch&15),isdigit(ch=tc()));
}
inline void get_digit(char& ch)
{
while (!isdigit(ch=tc()));
}
#undef tc
}F;
inline void inc(int& x,CI y)
{
if ((x+=y)>=mod) x-=mod;
}
inline data MDFS(CI nw,CI lst,CI cl)
{
if (!~nw) return data(1,0); if (!cl&&~f[nw][lst]) return f[nw][lst];
data ret(0,0); int lim=cl?a[nw]:9; for (RI i=0;i<=lim;++i)
{
data nxt=MDFS(nw-1,i,cl&&(i==a[nw]));
inc(ret.num,nxt.num); inc(ret.sum,nxt.sum);
if (i!=lst) inc(ret.sum,1LL*i*pw[nw]%mod*nxt.num%mod);
}
if (cl) return ret; return f[nw][lst]=ret;
}
signed main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
RI i,j; for (pw[0]=i=1;i<N;++i) pw[i]=10LL*pw[i-1]%mod;
for (F.read(t);t;--t)
{
for (F.read(n),i=n-1;~i;--i)
F.get_digit(ch),a[i]=ch&15;
for (--a[i=0];a[i]<0;++i) a[i]+=10,--a[i+1];
if (!a[n-1]) --n; for (i=0;i<=n;++i) for (j=0;j<10;++j)
f[i][j]=data(); int dlt=MDFS(n-1,0,1).sum;
for (F.read(n),i=n-1;~i;--i)
F.get_digit(ch),a[i]=ch&15;
for (i=0;i<=n;++i) for (j=0;j<10;++j)
f[i][j]=data(); int ret=MDFS(n-1,0,1).sum;
printf("%d\n",(ret-dlt+mod)%mod);
}
return 0;
}
G Chef and Gordon Ramsay
首先有一个很套路的统计方法，我们考虑将每个点作为中间点时计算答案
那么我们分类讨论一下三种情况，发现其实只需要统计出子树内小于（大于）它的数的个数和子树外小于（大于）它的数的个数
子树内我们套路的线段树合并一下就能求出来了，然后减一减就得到了外面的，将方案乘起来就好了
注意中间点为\(1\)或\(3\)时算点对要除以\(2\)
#include<cstdio>
#include<cctype>
#define RI register int
#define CI const int&
#define Tp template <typename T>
using namespace std;
const int N=100005;
struct edge
{
int to,nxt;
}e[N<<1]; int t,head[N],n,p1,p2,p3,cnt,x,y,rt[N]; long long ans;
class FileInputOutput
{
private:
static const int S=1<<21;
#define tc() (A==B&&(B=(A=Fin)+fread(Fin,1,S,stdin),A==B)?EOF:*A++)
char Fin[S],*A,*B;
public:
Tp inline void read(T& x)
{
x=0; char ch; while (!isdigit(ch=tc()));
while (x=(x<<3)+(x<<1)+(ch&15),isdigit(ch=tc()));
}
#undef tc
}F;
inline void addedge(CI x,CI y)
{
e[++cnt]=(edge){y,head[x]}; head[x]=cnt;
e[++cnt]=(edge){x,head[y]}; head[y]=cnt;
}
class Segment_Tree
{
private:
static const int P=20;
struct segment
{
int ch[2],sum;
}node[N*P<<1]; int tot;
public:
#define TN CI l=1,CI r=n
#define LS l,mid
#define RS mid+1,r
#define lc(x) node[x].ch[0]
#define rc(x) node[x].ch[1]
#define S(x) node[x].sum
inline void insert(int& now,CI pos,TN)
{
if (!now) now=++tot; ++S(now); if (l==r) return; int mid=l+r>>1;
if (pos<=mid) insert(lc(now),pos,LS); else insert(rc(now),pos,RS);
}
inline int merge(CI x,CI y,TN)
{
if (!x||!y) return x|y; int now=++tot; S(now)=S(x)+S(y);
if (l==r) return now; int mid=l+r>>1;
lc(now)=merge(lc(x),lc(y),LS); rc(now)=merge(rc(x),rc(y),RS); return now;
}
inline int query(CI now,CI beg,CI end,TN)
{
if (!now) return 0; if (beg<=l&&r<=end) return S(now);
int mid=l+r>>1,ret=0; if (beg<=mid) ret+=query(lc(now),beg,end,LS);
if (end>mid) ret+=query(rc(now),beg,end,RS); return ret;
}
inline void clear(void)
{
for (RI i=1;i<=tot;++i) lc(i)=rc(i)=S(i)=0; tot=0;
}
#undef TN
#undef LS
#undef RS
#undef lc
#undef rc
}SEG;
#define to e[i].to
inline void DFS(CI now=1,CI fa=0)
{
RI i; SEG.insert(rt[now],now); for (i=head[now];i;i=e[i].nxt)
if (to!=fa) DFS(to,now),rt[now]=SEG.merge(rt[now],rt[to]);
if (p2==1)
{
int mxnum=SEG.query(rt[now],now+1,n); long long ret=0;
for (i=head[now];i;i=e[i].nxt) if (to!=fa)
{
int cur=SEG.query(rt[to],now+1,n);
ret+=1LL*cur*(mxnum-cur);
}
ans+=(ret>>1LL)+1LL*mxnum*(n-now-mxnum);
} else if (p2==2)
{
int mxnum=SEG.query(rt[now],now+1,n),minum=SEG.query(rt[now],1,now-1);
for (i=head[now];i;i=e[i].nxt) if (to!=fa)
ans+=1LL*SEG.query(rt[to],now+1,n)*(minum-SEG.query(rt[to],1,now-1));
ans+=1LL*mxnum*(now-1-minum); ans+=1LL*minum*(n-now-mxnum);
} else
{
int minum=SEG.query(rt[now],1,now-1); long long ret=0;
for (i=head[now];i;i=e[i].nxt) if (to!=fa)
{
int cur=SEG.query(rt[to],1,now-1);
ret+=1LL*cur*(minum-cur);
}
ans+=(ret>>1LL)+1LL*minum*(now-1-minum);
}
}
#undef to
int main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (F.read(t);t;--t)
{
RI i; F.read(n); F.read(p1); F.read(p2); F.read(p3);
for (ans=0,i=1;i<n;++i) F.read(x),F.read(y),addedge(x,y);
for (DFS(),printf("%lld\n",ans),cnt=0,i=1;i<=n;++i) head[i]=0;
for (SEG.clear(),i=1;i<=n;++i) rt[i]=0;
}
return 0;
}
Postscript
讲道理Div2的题目还是偏简单的，全部做下来两三个小时就好了
希望到时候打Div1别被完虐吧QWQ
辣鸡老年选手AFO在即
