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Educational Codeforces Round 154 (Rated for Div. 2)

1|0Preface


太FW了现在，纯纯给队伍拖后腿，马上要成为我们队CF Rating最低的了

但换句话说徐神和祁神都这么猛，我直接躺着被嘎嘎带飞好像也很爽啊

不管怎么样还是要多练，不过接下来可能要按专题重点突破了，明天队里开个会确定下大家的主攻方向再说

2|0A. Prime Deletion


因为13和31都是质数，因此判断这两个数的位置关系即可

#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_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 __int128 i128;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=15;
int t,pos[N]; char s[N];
int main()
{
	//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
	for (scanf("%d",&t);t;--t)
	{
		RI i; for (scanf("%s",s+1),i=1;i<=9;++i) pos[s[i]-'0']=i;
		if (pos[1]<pos[3]) puts("13"); else puts("31");
	}
	return 0;
}

3|0B. Two Binary Strings


由于开头和结尾确定，因此我们可以枚举两个位置i,j，当ai=bi=0∧aj=bj=1∧a[i+1,j−1]=b[i+1,j−1]时一定有解

当然其实可以直接枚举i,i+1，比赛时候铸币了看数据范围就滚去写O(n2)的了，没写O(n)

#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_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 __int128 i128;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=5005;
int t,n,f[N][N]; char a[N],b[N];
int main()
{
	//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
	for (scanf("%d",&t);t;--t)
	{
		RI i,j; scanf("%s%s",a+1,b+1); n=strlen(a+1);
		for (i=1;i<=n;++i) for (j=1;j<=n;++j) f[i][j]=0;
		for (i=1;i<=n;++i)
		{
			if (!(f[i][i]=a[i]==b[i])) continue;
			for (j=i+1;j<=n;++j)
			if (!(f[i][j]=a[j]==b[j])) break;
		}
		bool flag=0; for (i=1;i<=n;++i) for (j=i+1;j<=n;++j)
		if (a[i]=='0'&&b[i]=='0'&&a[j]=='1'&&b[j]=='1'&&(i+1<=j-1?f[i+1][j-1]:1)) flag=1;
		puts(flag?"YES":"NO");
	}
	return 0;
}

4|0C. Queries for the Array


被C题狠狠地腐乳了，还是靠ztc教我才会的，被狠狠地克制了

考虑序列的形态形如这样，先是若干个已知的位置，然后是若干个未知的位置

当遇到0时就把未知的位置的最后一个改成不合法的标记，然后如果有添加操作就在后面再加上若干个未知的位置

然后就是分情况讨论清楚即可，应该有更简单的做法的说

#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_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 __int128 i128;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=200005;
int t,n,cnt; char s[N]; set <int> mx[2];
int main()
{
	//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
	for (scanf("%d",&t);t;--t)
	{
		RI i,j; scanf("%s",s+1); n=strlen(s+1);
		bool flag=1,sign=0; int known=0,unknown=0,len=0;
		for (i=1;i<=n&&flag;++i)
		{
			if (s[i]=='1')
			{
				if (sign) flag=0; known+=unknown; unknown=0;
			}
			if (s[i]=='0')
			{
				if (!sign)
				{
					if (!known&&unknown>1) sign=1,--unknown; else
					if (known>=1&&unknown>=1) sign=1,--unknown; else flag=0;
				}
			}
			if (s[i]=='+') sign?++len:++unknown;
			if (s[i]=='-')
			{
				if (len>0) --len; else
				if (sign) sign=0; else
				if (unknown>0) --unknown; else
				if (known>0) --known; else flag=0;
			}
		}
		puts(flag?"YES":"NO");
	}
	return 0;
}

5|0D. Sorting By Multiplication


傻逼题，比C简单到不知道哪里去了

考虑如果不能乘负数就是把序列分成m段，每段内的元素都是单调递增的，最后答案就是m−1，要使段数最小就直接贪心地划分即可

如果能乘负数也很简单，就是确定一个分界点后，后面部分划成单调递增的段，前面部分划成单调递减的段即可

#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_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 __int128 i128;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=200005;
int t,n,a[N],suf[N],pre[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]);
		for (suf[n]=1,i=n-1;i>=1;--i) suf[i]=suf[i+1]+(a[i]>=a[i+1]);
		for (pre[1]=1,i=2;i<=n;++i) pre[i]=pre[i-1]+(a[i-1]<=a[i]);
		int ans=min(suf[1]-1,pre[n]); for (i=2;i<=n;++i)
		ans=min(ans,pre[i-1]+suf[i]-1); printf("%d\n",ans);
	}
	return 0;
}

6|0E. Non-Intersecting Subpermutations


ORZ祁神，比赛时候教了我怎么做，但我因为一个铸币下标写错了没调出来

考虑DP，设fi,j表示做了前i个位置，从位置i往前的j个数互不相同的方案数，转移的话有以下情况：

  • fi,j×(k−j)→fi+1,j+1
  • fi,j→fi+1,s，其中s∈[1,j]

然后要计算答案的话由于题目要求区间不能相交，但我们贪心地每次一旦出现j=k的情况就直接计算它的贡献，fi,j×kn−i→ans，然后注意转移变成k→fi+1,1即可

复杂度O(nk)

#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_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 __int128 i128;
typedef pair <int,int> pi;
typedef vector <int> VI;
typedef array <int,3> tri;
const int N=4005,mod=998244353;
int t,n,k,f[N][N],g[N][N],pw[N],ans;
inline void inc(int& x,CI y)
{
	if ((x+=y)>=mod) x-=mod;
}
int main()
{
	//freopen("CODE.in","r",stdin); freopen("CODE.out","w",stdout);
	RI i,j; for (scanf("%d%d",&n,&k),pw[0]=i=1;i<=n;++i) pw[i]=1LL*pw[i-1]*k%mod;
	for (f[1][1]=k,i=1;i<=n;++i)
	{
		for (j=k-1;j>=1;--j) inc(g[i][j],g[i][j+1]);
		for (j=1;j<=k;++j) inc(f[i][j],g[i][j]);
		for (j=1;j<=k;++j)
		{
			if (j==k)
			{
				inc(ans,1LL*f[i][j]*pw[n-i]%mod);
				inc(f[i+1][1],1LL*f[i][j]*k%mod); continue;
			}
			inc(f[i+1][j+1],1LL*f[i][j]*(k-j)%mod);
			inc(g[i+1][j],f[i][j]);
		}
	}
	return printf("%d",ans),0;
}

7|0Postscript


so weak……


__EOF__

本文作者：hl666
本文链接：https://www.cnblogs.com/cjjsb/p/17674143.html
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