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

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A. Calculating Function

水题，判个奇偶即可

 1 #include <iostream>
 2 #include <sstream>
 3 #include <ios>
 4 #include <iomanip>
 5 #include <functional>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <string>
 9 #include <list>
10 #include <queue>
11 #include <deque>
12 #include <stack>
13 #include <set>
14 #include <map>
15 #include <cstdio>
16 #include <cstdlib>
17 #include <cmath>
18 #include <cstring>
19 #include <climits>
20 #include <cctype>
21 using namespace std;
22 #define XINF INT_MAX
23 #define INF 0x3FFFFFFF
24 #define MP(X,Y) make_pair(X,Y)
25 #define PB(X) push_back(X)
26 #define REP(X,N) for(int X=0;X<N;X++)
27 #define REP2(X,L,R) for(int X=L;X<=R;X++)
28 #define DEP(X,R,L) for(int X=R;X>=L;X--)
29 #define CLR(A,X) memset(A,X,sizeof(A))
30 #define IT iterator
31 typedef long long ll;
32 typedef pair<int,int> PII;
33 typedef vector<PII> VII;
34 typedef vector<int> VI;
35 
36 int main()
37 {
38     ios::sync_with_stdio(false);
39     ll n;
40     while(cin>>n)
41     {
42         if(n&1)cout<<-(n+1)/2<<endl;
43         else cout<<n/2<<endl;
44     }
45     return 0;
46 }
View Code

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

 1 #include <iostream>
 2 #include <sstream>
 3 #include <ios>
 4 #include <iomanip>
 5 #include <functional>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <string>
 9 #include <list>
10 #include <queue>
11 #include <deque>
12 #include <stack>
13 #include <set>
14 #include <map>
15 #include <cstdio>
16 #include <cstdlib>
17 #include <cmath>
18 #include <cstring>
19 #include <climits>
20 #include <cctype>
21 using namespace std;
22 #define XINF INT_MAX
23 #define INF 0x3FFFFFFF
24 #define MP(X,Y) make_pair(X,Y)
25 #define PB(X) push_back(X)
26 #define REP(X,N) for(int X=0;X<N;X++)
27 #define REP2(X,L,R) for(int X=L;X<=R;X++)
28 #define DEP(X,R,L) for(int X=R;X>=L;X--)
29 #define CLR(A,X) memset(A,X,sizeof(A))
30 #define IT iterator
31 typedef long long ll;
32 typedef pair<int,int> PII;
33 typedef vector<PII> VII;
34 typedef vector<int> VI;
35 int a[2010];
36 vector<int>G[2010];
37 int d,n;
38 const ll MOD=1000000007;
39 ll dp[2010];
40 void dfs(int v,int u,int root,int x)
41 {
42     if((a[v]<x&&x-a[v]<=d)||(a[v]==x&&root>=v))
43     {
44         //cout<<v<<" ";
45         dp[v]=1;
46         for(int i=0;i<G[v].size();i++)
47         {
48             if(G[v][i]==u)continue;
49             dfs(G[v][i],v,root,x);
50             dp[v]*=(dp[G[v][i]]+1);
51             dp[v]%=MOD;
52         }
53 
54     }
55     return ;
56 }
57 int main()
58 {
59     ios::sync_with_stdio(false);
60     //freopen("in.in","r",stdin);
61     while(cin>>d>>n)
62     {
63         for(int i=0;i<n;i++)
64         {
65             cin>>a[i];
66             G[i].clear();
67         }
68         int u,v;
69         for(int i=0;i<n-1;i++)
70         {
71             cin>>u>>v;
72             u--,v--;
73             G[u].PB(v);
74             G[v].PB(u);
75         }
76         ll ans=0;
77         for(int i=0;i<n;i++)
78         {
79             CLR(dp,0);
80             dfs(i,i,i,a[i]);
81             ans+=dp[i];
82             ans%=MOD;
83         }
84         cout<<ans<<endl;
85 
86     }
87     return 0;
88 }
View Code

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。

  1 #include <iostream>
  2 #include <sstream>
  3 #include <ios>
  4 #include <iomanip>
  5 #include <functional>
  6 #include <algorithm>
  7 #include <vector>
  8 #include <string>
  9 #include <list>
 10 #include <queue>
 11 #include <deque>
 12 #include <stack>
 13 #include <set>
 14 #include <map>
 15 #include <cstdio>
 16 #include <cstdlib>
 17 #include <cmath>
 18 #include <cstring>
 19 #include <climits>
 20 #include <cctype>
 21 using namespace std;
 22 #define XINF INT_MAX
 23 #define INF 0x3FFFFFFF
 24 #define MP(X,Y) make_pair(X,Y)
 25 #define PB(X) push_back(X)
 26 #define REP(X,N) for(int X=0;X<N;X++)
 27 #define REP2(X,L,R) for(int X=L;X<=R;X++)
 28 #define DEP(X,R,L) for(int X=R;X>=L;X--)
 29 #define CLR(A,X) memset(A,X,sizeof(A))
 30 #define IT iterator
 31 typedef long long ll;
 32 typedef pair<int,int> PII;
 33 typedef vector<PII> VII;
 34 typedef vector<int> VI;
 35 string s;
 36 int main()
 37 {
 38     ios::sync_with_stdio(false);
 39     int n,p;
 40     cin>>n>>p>>s;
 41     int len=n;
 42     int ans=0;
 43     int x=(n+1)/2;
 44     if(p<=x)
 45     {
 46         p--;
 47         int lx=0,rx=0;
 48         int i=0;
 49         while(s[i]==s[len-1-i]&&i<p)i++;
 50         ans+=p-i;
 51         lx=p-i;
 52         int temp;
 53         while(i<=p)
 54         {
 55             if(s[i]!=s[len-1-i])
 56             {
 57                 temp=max(s[i],s[len-1-i])-min(s[i],s[len-1-i]);
 58                 ans+=min(temp,26-temp);
 59             }
 60             i++;
 61         }
 62         i=x-1;
 63         while(s[i]==s[len-1-i]&&i>p)i--;
 64         rx=i-p;
 65         ans+=i-p;
 66         while(i>p)
 67         {
 68             if(s[i]!=s[len-1-i])
 69             {
 70                 temp=max(s[i],s[len-1-i])-min(s[i],s[len-1-i]);
 71                 ans+=min(temp,26-temp);
 72             }
 73             i--;
 74         }
 75         ans+=min(lx,rx);
 76 
 77     }
 78     else
 79     {
 80         p--;
 81         int lx=0,rx=0;
 82         int i=x;
 83         while(s[i]==s[len-1-i]&&i<p)i++;
 84         ans+=p-i;
 85         lx=p-i;
 86         int temp;
 87         while(i<=p)
 88         {
 89             if(s[i]!=s[len-1-i])
 90             {
 91                 temp=max(s[i],s[len-1-i])-min(s[i],s[len-1-i]);
 92                 ans+=min(temp,26-temp);
 93             }
 94             i++;
 95         }
 96         i=len-1;
 97         while(s[i]==s[len-1-i]&&i>p)i--;
 98         rx=i-p;
 99         ans+=i-p;
100         while(i>p)
101         {
102             if(s[i]!=s[len-1-i])
103             {
104                 temp=max(s[i],s[len-1-i])-min(s[i],s[len-1-i]);
105                 ans+=min(temp,26-temp);
106             }
107             i--;
108         }
109         ans+=min(lx,rx);
110     }
111     cout<<ans<<endl;
112     return 0;
113 }
View Code

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:

  1. S is non-empty.
  2. 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.
  3. .

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。

 1 #include <iostream>
 2 #include <sstream>
 3 #include <ios>
 4 #include <iomanip>
 5 #include <functional>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <string>
 9 #include <list>
10 #include <queue>
11 #include <deque>
12 #include <stack>
13 #include <set>
14 #include <map>
15 #include <cstdio>
16 #include <cstdlib>
17 #include <cmath>
18 #include <cstring>
19 #include <climits>
20 #include <cctype>
21 using namespace std;
22 #define XINF INT_MAX
23 #define INF 0x3FFFFFFF
24 #define MP(X,Y) make_pair(X,Y)
25 #define PB(X) push_back(X)
26 #define REP(X,N) for(int X=0;X<N;X++)
27 #define REP2(X,L,R) for(int X=L;X<=R;X++)
28 #define DEP(X,R,L) for(int X=R;X>=L;X--)
29 #define CLR(A,X) memset(A,X,sizeof(A))
30 #define IT iterator
31 typedef long long ll;
32 typedef pair<int,int> PII;
33 typedef vector<PII> VII;
34 typedef vector<int> VI;
35 int a[2010];
36 vector<int>G[2010];
37 int d,n;
38 const ll MOD=1000000007;
39 ll dp[2010];
40 void dfs(int v,int u,int root,int x)
41 {
42     if((a[v]<x&&x-a[v]<=d)||(a[v]==x&&root>=v))
43     {
44         //cout<<v<<" ";
45         dp[v]=1;
46         for(int i=0;i<G[v].size();i++)
47         {
48             if(G[v][i]==u)continue;
49             dfs(G[v][i],v,root,x);
50             dp[v]*=(dp[G[v][i]]+1);
51             dp[v]%=MOD;
52         }
53 
54     }
55     return ;
56 }
57 int main()
58 {
59     ios::sync_with_stdio(false);
60     //freopen("in.in","r",stdin);
61     while(cin>>d>>n)
62     {
63         for(int i=0;i<n;i++)
64         {
65             cin>>a[i];
66             G[i].clear();
67         }
68         int u,v;
69         for(int i=0;i<n-1;i++)
70         {
71             cin>>u>>v;
72             u--,v--;
73             G[u].PB(v);
74             G[v].PB(u);
75         }
76         ll ans=0;
77         for(int i=0;i<n;i++)
78         {
79             CLR(dp,0);
80             dfs(i,i,i,a[i]);
81             ans+=dp[i];
82             ans%=MOD;
83         }
84         cout<<ans<<endl;
85 
86     }
87     return 0;
88 }
View Code

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:

  1. group of all i such that ai belongs to no longest increasing subsequences.
  2. group of all i such that ai belongs to at least one but not every longest increasing subsequence.
  3. 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
3
input
4
1 3 2 5
output
3223
input
4
1 5 2 3
output
3133
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；

 1 #include <iostream>
 2 #include <sstream>
 3 #include <ios>
 4 #include <iomanip>
 5 #include <functional>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <string>
 9 #include <list>
10 #include <queue>
11 #include <deque>
12 #include <stack>
13 #include <set>
14 #include <map>
15 #include <cstdio>
16 #include <cstdlib>
17 #include <cmath>
18 #include <cstring>
19 #include <climits>
20 #include <cctype>
21 using namespace std;
22 #define XINF INT_MAX
23 #define INF 0x3FFFFFFF
24 #define MP(X,Y) make_pair(X,Y)
25 #define PB(X) push_back(X)
26 #define REP(X,N) for(int X=0;X<N;X++)
27 #define REP2(X,L,R) for(int X=L;X<=R;X++)
28 #define DEP(X,R,L) for(int X=R;X>=L;X--)
29 #define CLR(A,X) memset(A,X,sizeof(A))
30 #define IT iterator
31 typedef long long ll;
32 typedef pair<int,int> PII;
33 typedef vector<PII> VII;
34 typedef vector<int> VI;
35 const int maxn=100010;
36 int a[maxn],b[maxn],dp[maxn];
37 int l[maxn],r[maxn];
38 int ans[maxn];
39 int deg[maxn];
40 int main()
41 {
42     ios::sync_with_stdio(false);
43     int n;
44     while(cin>>n)
45     {
46         CLR(ans,0);
47         CLR(deg,0);
48         for(int i=0;i<n;i++)
49         {
50             cin>>a[i];
51             b[n-1-i]=-a[i];
52         }
53         fill(dp,dp+n,INF);
54         for(int i=0;i<n;i++)
55         {
56             int temp=lower_bound(dp,dp+n,a[i])-dp;
57             l[i]=temp+1;
58             dp[temp]=a[i];
59         }
60         int len=lower_bound(dp,dp+n,INF)-dp;
61         fill(dp,dp+n,INF);
62         for(int i=0;i<n;i++)
63         {
64             int temp=lower_bound(dp,dp+n,b[i])-dp;
65             r[n-1-i]=temp;
66             dp[temp]=b[i];
67         }
68         for(int i=0;i<n;i++)
69         {
70             if(l[i]+r[i]==len)
71             {
72                 deg[l[i]]++;
73             }
74             else
75             {
76                 ans[i]=1;
77             }
78         }
79         for(int i=0;i<n;i++)
80         {
81             if(ans[i])cout<<1;
82             else if(deg[l[i]]>1)cout<<2;
83             else cout<<3;
84         }
85         cout<<endl;
86 
87     }
88     return 0;
89 }
View Code
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原文地址：https://www.cnblogs.com/fraud/p/4100847.html