Educational Codeforces Round 163 (Rated for Div. 2)
Preface
这周很奇怪，连着计网、数据库、组合数学的课都莫名其妙不上了，突然变得很空闲了的说
正好有空赶紧补补题，不然接下来有很多造题/比赛的任务搁置着就忘记了
A. Special Characters
签到，\(n\)是偶数才有解，构造的话注意到一个AAB可以产生\(2\)的贡献，把\(\frac{n}{2}\)个AAB拼起来即可
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
int t,n;
int main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (scanf("%d",&t);t;--t)
{
if (scanf("%d",&n),n%2) { puts("NO"); continue; }
puts("YES"); for (RI i=1;i<=n/2;++i) printf("AAB"); putchar('\n');
}
return 0;
}
B. Array Fix
从前往后贪心，能分裂就尽量分裂
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=55;
int t,n,a[N];
int main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (scanf("%d",&t);t;--t)
{
RI i; for (scanf("%d",&n),i=1;i<=n;++i) scanf("%d",&a[i]);
bool flag=1; int pre=0; for (i=1;i<=n;++i)
{
if (a[i]<pre) { flag=0; break; }
if (a[i]>9&&a[i]/10<=a[i]%10&&pre<=a[i]/10) pre=a[i]%10; else pre=a[i];
}
puts(flag?"YES":"NO");
}
return 0;
}
C. Arrow Path
直接在上面跑个BFS看看哪些点能走到即可
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=200005,dx[4]={0,1,0,-1},dy[4]={1,0,-1,0};
int t,n,vis[2][N]; char s[2][N];
int main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (scanf("%d",&t);t;--t)
{
RI i; for (scanf("%d",&n),i=1;i<=n;++i) vis[0][i]=vis[1][i]=0;
scanf("%s%s",s[0]+1,s[1]+1); vis[0][1]=1;
queue <pi> q; q.push({0,1});
while (!q.empty())
{
auto [x,y]=q.front(); q.pop();
for (i=0;i<4;++i)
{
int nx=x+dx[i],ny=y+dy[i];
if (nx<0||nx>1||ny<1||ny>n) continue;
if (s[nx][ny]=='<') --ny; else ++ny;
if (!vis[nx][ny]) vis[nx][ny]=1,q.push({nx,ny});
}
}
puts(vis[1][n]?"YES":"NO");
}
return 0;
}
D. Tandem Repeats?
首先枚举答案的一半长度\(len\)，如果位置\(i\)上的字符可以与位置\(i+len\)上的字符匹配，则我们令位置\(i\)的权值为\(1\)；否则为\(0\)
不难发现此时要有长度为\(2\times len\)的答案串，当且仅当存在长度为\(len\)的连续一段权值为\(1\)的区间，然后就可以\(O(n^2)\)做完这道题了
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=5005;
int t,n; char s[N];
int main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (scanf("%d",&t);t;--t)
{
RI len,i; scanf("%s",s+1); n=strlen(s+1);
for (len=n/2;len>=0;--len)
{
int cur=0,mx=0;
auto cmp=[&](const char& a,const char& b)
{
return a==b||a=='?'||b=='?';
};
for (i=1;i+len<=n;++i)
if (cmp(s[i],s[i+len])) ++cur; else mx=max(mx,cur),cur=0;
mx=max(mx,cur);
if (mx>=len) { printf("%d\n",len*2); break; }
}
}
return 0;
}
E. Clique Partition
一眼构造题
稍微手玩一下会发现，对于固定的某个\(k\)，它最多只能bound住大小为\(k\)的Clique
而构造方法也很简单，假设当前存在编号为\(1\sim n\)的Clique需要被bound在一起
且我们存在以下构造方式：先设\(a_i=i\)，令\(m=\lfloor\frac{1+n}{2}\rfloor\)，然后翻转区间\([1,m],[m+1,n]\)即可
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=45;
int t,n,k,a[N];
int main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
for (scanf("%d",&t);t;--t)
{
RI i; for (scanf("%d%d",&n,&k),i=1;i<=n;++i) a[i]=i;
for (i=1;i<=n;i+=k)
{
int l=i,r=min(i+k-1,n),mid=l+r>>1;
reverse(a+i,a+mid+1); reverse(a+mid+1,a+r+1);
}
for (i=1;i<=n;++i) printf("%d%c",a[i]," \n"[i==n]);
for (printf("%d\n",(n+k-1)/k),i=1;i<=n;++i)
printf("%d%c",(i-1)/k+1," \n"[i==n]);
}
return 0;
}
F. Rare Coins
唉又是TM的组合数化简，感觉我的组合数学课再上一次课就能讲到这些组合恒等式了（然而这周课没了）
首先假设区间内有\(g_1\)枚金币，\(s_1\)枚银币；区间外有\(g_2\)枚金币，\(s_2\)枚银币
不妨假设区间内有\(a\)枚银币变成\(1\)，区间外有\(b\)枚银币变成\(0\)（这样定义的好处是防止下标出负数）
则需要满足\(g_1+a>g_2+(s_2-b)\)，即\(a+b\ge g_2+s_2-g_1+1\)，令\(M=g_2+s_2-g_1+1\)
因此答案的表达式就很简单了：\(\frac{1}{2^{s_1+s_2}}\times \sum_{a+b\ge M} C_{s_1}^a\times C_{s_2}^b\)，然后又遇到经典的组合数相乘了
根据范德蒙德卷积，上面的式子等价于\(\frac{1}{2^{s_1+s_2}}\times \sum_{x\ge M} C_{s_1+s_2}^x\)，因此给\(C_{s_1+s_2}^x\)预处理一个后缀和即可
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=1e6+5,mod=998244353;
int n,q,m,pfx[2][N],fact[N],ifac[N],g[N];
inline int quick_pow(int x,int p=mod-2,int mul=1)
{
for (;p;p>>=1,x=1LL*x*x%mod) if (p&1) mul=1LL*mul*x%mod; return mul;
}
inline void init(CI n)
{
RI i; for (fact[0]=i=1;i<=n;++i) fact[i]=1LL*fact[i-1]*i%mod;
for (ifac[n]=quick_pow(fact[n]),i=n-1;i>=0;--i) ifac[i]=1LL*ifac[i+1]*(i+1)%mod;
}
inline int C(CI n,CI m)
{
return 1LL*fact[n]*ifac[m]%mod*ifac[n-m]%mod;
}
int main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
RI i,j; for (scanf("%d%d",&n,&q),j=0;j<2;++j)
for (i=1;i<=n;++i) scanf("%d",&pfx[j][i]),pfx[j][i]+=pfx[j][i-1];
for (init(m=pfx[1][n]),i=0;i<=m;++i) g[i]=C(m,i);
for (i=m-1;i>=0;--i) (g[i]+=g[i+1])%=mod;
int prob=quick_pow(quick_pow(2,m));
for (i=1;i<=q;++i)
{
int l,r; scanf("%d%d",&l,&r);
int g1=pfx[0][r]-pfx[0][l-1],g2=pfx[0][n]-g1;
int s1=pfx[1][r]-pfx[1][l-1],s2=pfx[1][n]-s1;
printf("%d ",1LL*prob*g[min(m+1,max(0,g2+s2-g1+1))]%mod);
}
return 0;
}
G. MST with Matching
刚开始没看数据范围以为是个什么玄学奇妙题，后面发现妈的\(n\le 20\)
提到匹配会想到什么，对于二分图来说，匹配和点覆盖是等价的，而巧了树也是一种二分图（没有环当然没有奇环咯）
因此不妨直接\(2^n\)大力枚举生成树的一个点覆盖，根据规约方式我们知道，此时能用的边只有那些两端有至少一个在点覆盖解集里的
然后再对合法的边跑一个MST即可，注意用克鲁斯卡尔实现的话要注意常数，先在外面把边排序好不然可能会T
#include<cstdio>
#include<iostream>
#include<utility>
#include<vector>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<set>
#include<array>
#include<random>
#include<bitset>
#include<ctime>
#include<limits.h>
#include<assert.h>
#include<unordered_set>
#include<unordered_map>
#define RI register int
#define CI const int&
#define mp make_pair
#define fi first
#define se second
#define Tp template <typename T>
using namespace std;
typedef long long LL;
typedef long double LDB;
typedef unsigned long long u64;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=25;
int n,c,w[N][N],fa[N],ans=1e9;
inline int getfa(CI x)
{
return fa[x]!=x?fa[x]=getfa(fa[x]):x;
}
int main()
{
//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
vector <array<int,3>> edges;
RI i,j; for (scanf("%d%d",&n,&c),i=0;i<n;++i)
for (j=0;j<n;++j) scanf("%d",&w[i][j]);
for (i=0;i<n;++i) for (j=i+1;j<n;++j)
if (w[i][j]) edges.push_back({w[i][j],i,j});
sort(edges.begin(),edges.end());
for (RI mask=0;mask<(1<<n);++mask)
{
for (i=0;i<n;++i) fa[i]=i; int cs=0,val=0;
for (auto [w,x,y]:edges)
{
if (cs==n-1) break;
if (!(((mask>>x)&1)||((mask>>y)&1))) continue;
if (getfa(x)!=getfa(y)) fa[getfa(x)]=getfa(y),++cs,val+=w;
}
if (cs==n-1) ans=min(ans,val+c*__builtin_popcount(mask));
}
return printf("%d",ans),0;
}
Postscript
你说的对但是今天云顶S11更新了，牢闪不得不回归云顶一段时间力
辣鸡老年选手AFO在即
