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 }
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 }
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 }
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。
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 }
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.
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).
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.
1
4
3
4
1 3 2 5
3223
4
1 5 2 3
3133
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 }