Codeforces Round #277(Div 2) A、B、C、D、E题解

2023-01-29 15:47:10

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A. Calculating Function

水题，判个奇偶即可

 #include <iostream>

 #include <sstream>

 #include <ios>

 #include <iomanip>

 #include <functional>

 #include <algorithm>

 #include <vector>

 #include <string>

 #include <list>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <set>

 #include <map>

 #include <cstdio>

 #include <cstdlib>

 #include <cmath>

 #include <cstring>

 #include <climits>

 #include <cctype>

 using namespace std;

 #define XINF INT_MAX

 #define INF 0x3FFFFFFF

 #define MP(X,Y) make_pair(X,Y)

 #define PB(X) push_back(X)

 #define REP(X,N) for(int X=0;X<N;X++)

 #define REP2(X,L,R) for(int X=L;X<=R;X++)

 #define DEP(X,R,L) for(int X=R;X>=L;X--)

 #define CLR(A,X) memset(A,X,sizeof(A))

 #define IT iterator

 typedef long long ll;

 typedef pair<int,int> PII;

 typedef vector<PII> VII;

 typedef vector<int> VI;

 int main()

 {

     ios::sync_with_stdio(false);

     ll n;

     while(cin>>n)

     {

         if(n&)cout<<-(n+)/<<endl;

         else cout<<n/<<endl;

     }

     return ;

 }

B. OR in Matrix

先把A全部初始化为1，再把B中为0的对应的行和列在A中设为0，最后再通过得到的A来求B，看是否一致

 #include <iostream>

 #include <sstream>

 #include <ios>

 #include <iomanip>

 #include <functional>

 #include <algorithm>

 #include <vector>

 #include <string>

 #include <list>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <set>

 #include <map>

 #include <cstdio>

 #include <cstdlib>

 #include <cmath>

 #include <cstring>

 #include <climits>

 #include <cctype>

 using namespace std;

 #define XINF INT_MAX

 #define INF 0x3FFFFFFF

 #define MP(X,Y) make_pair(X,Y)

 #define PB(X) push_back(X)

 #define REP(X,N) for(int X=0;X<N;X++)

 #define REP2(X,L,R) for(int X=L;X<=R;X++)

 #define DEP(X,R,L) for(int X=R;X>=L;X--)

 #define CLR(A,X) memset(A,X,sizeof(A))

 #define IT iterator

 typedef long long ll;

 typedef pair<int,int> PII;

 typedef vector<PII> VII;

 typedef vector<int> VI;

 int a[];

 vector<int>G[];

 int d,n;

 const ll MOD=;

 ll dp[];

 void dfs(int v,int u,int root,int x)

 {

     if((a[v]<x&&x-a[v]<=d)||(a[v]==x&&root>=v))

     {

         //cout<<v<<" ";

         dp[v]=;

         for(int i=;i<G[v].size();i++)

         {

             if(G[v][i]==u)continue;

             dfs(G[v][i],v,root,x);

             dp[v]*=(dp[G[v][i]]+);

             dp[v]%=MOD;

         }

     }

     return ;

 }

 int main()

 {

     ios::sync_with_stdio(false);

     //freopen("in.in","r",stdin);

     while(cin>>d>>n)

     {

         for(int i=;i<n;i++)

         {

             cin>>a[i];

             G[i].clear();

         }

         int u,v;

         for(int i=;i<n-;i++)

         {

             cin>>u>>v;

             u--,v--;

             G[u].PB(v);

             G[v].PB(u);

         }

         ll ans=;

         for(int i=;i<n;i++)

         {

             CLR(dp,);

             dfs(i,i,i,a[i]);

             ans+=dp[i];

             ans%=MOD;

         }

         cout<<ans<<endl;

     }

     return ;

 }

C. Palindrome Transformation

贪心，只需要看初始位置所在的一半字符串。取短的那一部分先改。

 #include <iostream>

 #include <sstream>

 #include <ios>

 #include <iomanip>

 #include <functional>

 #include <algorithm>

 #include <vector>

 #include <string>

 #include <list>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <set>

 #include <map>

 #include <cstdio>

 #include <cstdlib>

 #include <cmath>

 #include <cstring>

 #include <climits>

 #include <cctype>

 using namespace std;

 #define XINF INT_MAX

 #define INF 0x3FFFFFFF

 #define MP(X,Y) make_pair(X,Y)

 #define PB(X) push_back(X)

 #define REP(X,N) for(int X=0;X<N;X++)

 #define REP2(X,L,R) for(int X=L;X<=R;X++)

 #define DEP(X,R,L) for(int X=R;X>=L;X--)

 #define CLR(A,X) memset(A,X,sizeof(A))

 #define IT iterator

 typedef long long ll;

 typedef pair<int,int> PII;

 typedef vector<PII> VII;

 typedef vector<int> VI;

 string s;

 int main()

 {

     ios::sync_with_stdio(false);

     int n,p;

     cin>>n>>p>>s;

     int len=n;

     int ans=;

     int x=(n+)/;

     if(p<=x)

     {

         p--;

         int lx=,rx=;

         int i=;

         while(s[i]==s[len--i]&&i<p)i++;

         ans+=p-i;

         lx=p-i;

         int temp;

         while(i<=p)

         {

             if(s[i]!=s[len--i])

             {

                 temp=max(s[i],s[len--i])-min(s[i],s[len--i]);

                 ans+=min(temp,-temp);

             }

             i++;

         }

         i=x-;

         while(s[i]==s[len--i]&&i>p)i--;

         rx=i-p;

         ans+=i-p;

         while(i>p)

         {

             if(s[i]!=s[len--i])

             {

                 temp=max(s[i],s[len--i])-min(s[i],s[len--i]);

                 ans+=min(temp,-temp);

             }

             i--;

         }

         ans+=min(lx,rx);

     }

     else

     {

         p--;

         int lx=,rx=;

         int i=x;

         while(s[i]==s[len--i]&&i<p)i++;

         ans+=p-i;

         lx=p-i;

         int temp;

         while(i<=p)

         {

             if(s[i]!=s[len--i])

             {

                 temp=max(s[i],s[len--i])-min(s[i],s[len--i]);

                 ans+=min(temp,-temp);

             }

             i++;

         }

         i=len-;

         while(s[i]==s[len--i]&&i>p)i--;

         rx=i-p;

         ans+=i-p;

         while(i>p)

         {

             if(s[i]!=s[len--i])

             {

                 temp=max(s[i],s[len--i])-min(s[i],s[len--i]);

                 ans+=min(temp,-temp);

             }

             i--;

         }

         ans+=min(lx,rx);

     }

     cout<<ans<<endl;

     return ;

 }

D. Valid Sets

As you know, an undirected connected graph with n nodes and n - 1 edges is called a tree. You are given an integer d and a tree consisting of n nodes. Each node i has a value ai associated with it.

We call a set S of tree nodes valid if following conditions are satisfied:

S is non-empty.
S is connected. In other words, if nodes u and v are in S, then all nodes lying on the simple path between u and v should also be presented in S.
.

Your task is to count the number of valid sets. Since the result can be very large, you must print its remainder modulo 1000000007(109 + 7).

题目：要求所给的树中，满足任意两个节点的权值相差小于等于d的子树个数。

分析：树形dp。

为了方便计数，我采取以下方式：每次选取的子树都是保证根节点的权值为最大，当然如果单单是这样的话，权值均相等的子树是会被重复计数的，因此我在遇到权值相等的情况时，需要保证其结点的标号小于根节点。

 #include <iostream>

 #include <sstream>

 #include <ios>

 #include <iomanip>

 #include <functional>

 #include <algorithm>

 #include <vector>

 #include <string>

 #include <list>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <set>

 #include <map>

 #include <cstdio>

 #include <cstdlib>

 #include <cmath>

 #include <cstring>

 #include <climits>

 #include <cctype>

 using namespace std;

 #define XINF INT_MAX

 #define INF 0x3FFFFFFF

 #define MP(X,Y) make_pair(X,Y)

 #define PB(X) push_back(X)

 #define REP(X,N) for(int X=0;X<N;X++)

 #define REP2(X,L,R) for(int X=L;X<=R;X++)

 #define DEP(X,R,L) for(int X=R;X>=L;X--)

 #define CLR(A,X) memset(A,X,sizeof(A))

 #define IT iterator

 typedef long long ll;

 typedef pair<int,int> PII;

 typedef vector<PII> VII;

 typedef vector<int> VI;

 int a[];

 vector<int>G[];

 int d,n;

 const ll MOD=;

 ll dp[];

 void dfs(int v,int u,int root,int x)

 {

     if((a[v]<x&&x-a[v]<=d)||(a[v]==x&&root>=v))

     {

         //cout<<v<<" ";

         dp[v]=;

         for(int i=;i<G[v].size();i++)

         {

             if(G[v][i]==u)continue;

             dfs(G[v][i],v,root,x);

             dp[v]*=(dp[G[v][i]]+);

             dp[v]%=MOD;

         }

     }

     return ;

 }

 int main()

 {

     ios::sync_with_stdio(false);

     //freopen("in.in","r",stdin);

     while(cin>>d>>n)

     {

         for(int i=;i<n;i++)

         {

             cin>>a[i];

             G[i].clear();

         }

         int u,v;

         for(int i=;i<n-;i++)

         {

             cin>>u>>v;

             u--,v--;

             G[u].PB(v);

             G[v].PB(u);

         }

         ll ans=;

         for(int i=;i<n;i++)

         {

             CLR(dp,);

             dfs(i,i,i,a[i]);

             ans+=dp[i];

             ans%=MOD;

         }

         cout<<ans<<endl;

     }

     return ;

 }

E. LIS of Sequence

The next "Data Structures and Algorithms" lesson will be about Longest Increasing Subsequence (LIS for short) of a sequence. For better understanding, Nam decided to learn it a few days before the lesson.

Nam created a sequence a consisting of n (1 ≤ n ≤ 105) elements a1, a2, ..., an (1 ≤ ai ≤ 105). A subsequence ai1, ai2, ..., aik where 1 ≤ i1 < i2 < ... < ik ≤ n is called increasing if ai1 < ai2 < ai3 < ... < aik. An increasing subsequence is called longest if it has maximum length among all increasing subsequences.

Nam realizes that a sequence may have several longest increasing subsequences. Hence, he divides all indexes i (1 ≤ i ≤ n), into three groups:

group of all i such that ai belongs to no longest increasing subsequences.
group of all i such that ai belongs to at least one but not every longest increasing subsequence.
group of all i such that ai belongs to every longest increasing subsequence.

Since the number of longest increasing subsequences of a may be very large, categorizing process is very difficult. Your task is to help him finish this job.

Input

The first line contains the single integer n (1 ≤ n ≤ 105) denoting the number of elements of sequence a.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

Print a string consisting of n characters. i-th character should be '1', '2' or '3' depending on which group among listed above index ibelongs to.

Sample test(s)

input

1
4

output

input

4
1 3 2 5

output

input

4
1 5 2 3

output

Note

In the second sample, sequence a consists of 4 elements: {a1, a2, a3, a4} = {1, 3, 2, 5}. Sequence a has exactly 2 longest increasing subsequences of length 3, they are {a1, a2, a4} = {1, 3, 5} and {a1, a3, a4} = {1, 2, 5}.

In the third sample, sequence a consists of 4 elements: {a1, a2, a3, a4} = {1, 5, 2, 3}. Sequence a have exactly 1 longest increasing subsequence of length 3, that is {a1, a3, a4} = {1, 2, 3}.

题目：问所给的序列中，若ai不存在于任何的LIS中，则输出1，若存在于至少一个但不为全部LIS中，则输出2，若ai存在于任意一个LIS中，则输出3

分析：求出到该点（包括该点在内）的正向LIS和不包括该点的反向LIS的长度，通过此判断即可。

例如：序列 1 2 4 2 3 5 4

　　　l[i] 1 2 3 2 3 3 4　　表示正向的到该点为止包括该点在内的LIS的长度

　　 r[i] 3 2 1 2 1 0 0 表示的是反向的到该点为止不包括该点的最长下降子序列的长度

如果对于某一点，若l[i]+r[i]等于LIS的长度，则其必处于某一LIS之中，否则必然不在任意一个LIS之中，可判该点为1

接下来在由于之前我记录了正向的到该点为止包括该点在内的LIS的长度，即我知道每一个处于LIS中的位置，由此，我只需判LIS中该位置的点是否唯一，若唯一，则为3，否则为2；

 #include <iostream>

 #include <sstream>

 #include <ios>

 #include <iomanip>

 #include <functional>

 #include <algorithm>

 #include <vector>

 #include <string>

 #include <list>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <set>

 #include <map>

 #include <cstdio>

 #include <cstdlib>

 #include <cmath>

 #include <cstring>

 #include <climits>

 #include <cctype>

 using namespace std;

 #define XINF INT_MAX

 #define INF 0x3FFFFFFF

 #define MP(X,Y) make_pair(X,Y)

 #define PB(X) push_back(X)

 #define REP(X,N) for(int X=0;X<N;X++)

 #define REP2(X,L,R) for(int X=L;X<=R;X++)

 #define DEP(X,R,L) for(int X=R;X>=L;X--)

 #define CLR(A,X) memset(A,X,sizeof(A))

 #define IT iterator

 typedef long long ll;

 typedef pair<int,int> PII;

 typedef vector<PII> VII;

 typedef vector<int> VI;

 const int maxn=;

 int a[maxn],b[maxn],dp[maxn];

 int l[maxn],r[maxn];

 int ans[maxn];

 int deg[maxn];

 int main()

 {

     ios::sync_with_stdio(false);

     int n;

     while(cin>>n)

     {

         CLR(ans,);

         CLR(deg,);

         for(int i=;i<n;i++)

         {

             cin>>a[i];

             b[n--i]=-a[i];

         }

         fill(dp,dp+n,INF);

         for(int i=;i<n;i++)

         {

             int temp=lower_bound(dp,dp+n,a[i])-dp;

             l[i]=temp+;

             dp[temp]=a[i];

         }

         int len=lower_bound(dp,dp+n,INF)-dp;

         fill(dp,dp+n,INF);

         for(int i=;i<n;i++)

         {

             int temp=lower_bound(dp,dp+n,b[i])-dp;

             r[n--i]=temp;

             dp[temp]=b[i];

         }

         for(int i=;i<n;i++)

         {

             if(l[i]+r[i]==len)

             {

                 deg[l[i]]++;

             }

             else

             {

                 ans[i]=;

             }

         }

         for(int i=;i<n;i++)

         {

             if(ans[i])cout<<;

             else if(deg[l[i]]>)cout<<;

             else cout<<;

         }

         cout<<endl;

     }

     return ;

 }

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