Codeforces Round #277 (Div. 2) 题解

Codeforces Round #277 (Div. 2)

A. Calculating Function

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

For a positive integer n let's define a function f:

f(n) =  - 1 + 2 - 3 + .. + ( - 1)nn

Your task is to calculate f(n) for a given integer n.

Input

The single line contains the positive integer n (1 ≤ n ≤ 1015).

Output

Print f(n) in a single line.

Sample test(s)

input

4

output

2

input

5

output

-3

Note

f(4) =  - 1 + 2 - 3 + 4 = 2

f(5) =  - 1 + 2 - 3 + 4 - 5 =  - 3

简单公式:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long int LL;

LL n;

int main()
{
    cin>>n;
    if(n%2==0)
    {
        LL t=n/2;
        cout<<t<<endl;
    }
    else
    {
        LL t=(n-1)/2;
        cout<<t-n<<endl;
    }
    return 0;
}

B. OR in Matrix

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's define logical OR as an operation on two logical values (i. e. values that belong to the set {0, 1})
that is equal to 1 if either or both of the logical values is set to 1,
otherwise it is 0. We can define logical OR of three or more
logical values in the same manner:

 where  is
equal to 1 if some ai = 1,
otherwise it is equal to 0.

Nam has a matrix A consisting of m rows and n columns.
The rows are numbered from 1 to m, columns are numbered from 1 to n.
Element at row i(1 ≤ i ≤ m) and column j (1 ≤ j ≤ n)
is denoted as Aij.
All elements of A are either 0 or 1. From matrix A,
Nam creates another matrix B of the same size using formula:

.

(Bij is OR of
all elements in row i and column j of matrix A)

Nam gives you matrix B and challenges you to guess matrix A.
Although Nam is smart, he could probably make a mistake while calculating matrix B, since size of A can
be large.

Input

The first line contains two integer m and n (1 ≤ m, n ≤ 100),
number of rows and number of columns of matrices respectively.

The next m lines each contain n integers separated
by spaces describing rows of matrix B (each element of B is
either 0 or 1).

Output

In the first line, print "NO" if Nam has made a mistake when calculating B,
otherwise print "YES". If the first line is "YES", then also
print m rows consisting of n integers representing
matrix A that can produce given matrix B. If there
are several solutions print any one.

Sample test(s)

input

2 2
1 0
0 0

output

NO

input

2 3
1 1 1
1 1 1

output

YES
1 1 1
1 1 1

input

2 3
0 1 0
1 1 1

output

YES
0 0 0
0 1 0

水题:将A里全部可能是1的点加上就能够了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int n,m;

int B[220][220];
int A[220][220];
bool vis[220][220];

bool check(int x,int y)
{
    bool flag=true;
    for(int i=0;i<m&&flag;i++)
    {
        if(vis[x][i]==false&&B[x][i]==0) flag=false;
    }
    for(int i=0;i<n&&flag;i++)
    {
        if(vis[i][y]==false&&B[i][y]==0) flag=false;
    }
    return flag;
}

void CL(int x,int y)
{
    for(int i=0;i<m;i++)
        vis[x][i]=true;
    for(int i=0;i<n;i++)
        vis[i][y]=true;
}

int main()
{
    cin>>n>>m;
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
            cin>>B[i][j];
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(check(i,j))///ONE
            {
                A[i][j]=1;
                CL(i,j);
            }
        }
    }
    bool flag=true;
    for(int i=0;i<n&&flag;i++)
        for(int j=0;j<m&&flag;j++)
            if(B[i][j]==1&&vis[i][j]==0) flag=false;

    if(flag)
    {
        puts("YES");
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
                cout<<A[i][j]<<" ";
            cout<<endl;
        }
    }
    else puts("NO");
    return 0;
}

C. Palindrome Transformation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more
beautiful, that is to make it be a palindrome by using 4 arrow keys: left, right, up, down.

There is a cursor pointing at some symbol of the string. Suppose that cursor is at position i (1 ≤ i ≤ n,
the string uses 1-based indexing) now. Left and right arrow keys are used to move cursor around the string. The string is cyclic, that means that when Nam presses left arrow key, the cursor will move to position i - 1 if i > 1 or
to the end of the string (i. e. position n) otherwise. The same holds when he presses the right arrow key (if i = n,
the cursor appears at the beginning of the string).

When Nam presses up arrow key, the letter which the text cursor is pointing to will change to the next letter in English alphabet (assuming that alphabet is also cyclic, i. e. after 'z'
follows 'a'). The same holds when he presses the down arrow key.

Initially, the text cursor is at position p.

Because Nam has a lot homework to do, he wants to complete this as fast as possible. Can you help him by calculating the minimum number of arrow keys presses to make the string to be a palindrome?

Input

The first line contains two space-separated integers n (1 ≤ n ≤ 105)
and p (1 ≤ p ≤ n), the length of Nam's
string and the initial position of the text cursor.

The next line contains n lowercase characters of Nam's string.

Output

Print the minimum number of presses needed to change string into a palindrome.

Sample test(s)

input

8 3
aeabcaez

output

6

Note

A string is a palindrome if it reads the same forward or reversed.

In the sample test, initial Nam's string is:  (cursor
position is shown bold).

In optimal solution, Nam may do 6 following steps:

The result, ,
is now a palindrome.

由于没有删除操作,最后的回文串是什么样已经是确定的了, A-->A'  或 A'--->A 或到A和A'的中间值上下移动的次数是一样的,所以没有必要跨越中点

仅仅要计算出在一边的左右移动次数就能够了....

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int maxn=200100;
const int INF=0x3f3f3f3f;

char str[maxn];
char rstr[maxn];
int n,p;

int change[maxn][2];

void getC()
{
    for(int i=0;i<n;i++)
    {
        /// 0: A --> A'
        change[i][0]=max(str[i],rstr[i])-min(str[i],rstr[i]);
        change[i][0]=min(change[i][0],min(str[i],rstr[i])+26-max(str[i],rstr[i]));
        /// 1: A --> m <-- A'
        change[i][1]=change[i][0];
    }
}

int main()
{
    cin>>n>>p;
    p--;
    cin>>str;
    int tt=n/2;
    if(n%2==0) tt--;
    if(p>tt)
    {
        reverse(str,str+n);
        p=n-1-p;
    }
    memcpy(rstr,str,sizeof(str));
    reverse(rstr,rstr+n);
    getC();

    int temp=0;
    ///改变字符的花费
    for(int i=0;i<=tt;i++)
    {
        temp+=change[i][0];
    }
    ///移动的花费
    ///须要改变的左右边界

    int R=-1;

    for(int i=tt;i>=0;i--)
    {
        if(change[i][0]!=0)
        {
            R=i; break;
        }
    }
    int L=-1;
    for(int i=0;i<=tt;i++)
    {
        if(change[i][0]!=0)
        {
            L=i; break;
        }
    }

    if(L==-1||R==-1)
    {
        puts("0");
    }
    else if(L==R)
    {
        cout<<abs(p-L)+temp<<endl;
    }
    else
    {
        /// L <----> R
        if(p>=L&&p<=R)
        {
            int t=min(abs(R-p),abs(L-p));
            cout<<R-L+t+temp<<endl;
        }
        else if(p<L)
        {
            cout<<abs(R-p)+temp<<endl;
        }
        else if(p>R)
        {
            cout<<abs(L-p)+temp<<endl;
        }
    }
    return 0;
}

D. Valid Sets

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 nnodes.
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).

Input

The first line contains two space-separated integers d (0 ≤ d ≤ 2000)
and n (1 ≤ n ≤ 2000).

The second line contains n space-separated positive integers a1, a2, ..., an(1 ≤ ai ≤ 2000).

Then the next n - 1 line each contain pair of integers u and v (1 ≤ u, v ≤ n)
denoting that there is an edge between u and v.
It is guaranteed that these edges form a tree.

Output

Print the number of valid sets modulo 1000000007.

Sample test(s)

input

1 4
2 1 3 2
1 2
1 3
3 4

output

8

input

0 3
1 2 3
1 2
2 3

output

3

input

4 8
7 8 7 5 4 6 4 10
1 6
1 2
5 8
1 3
3 5
6 7
3 4

output

41

Note

In the first sample, there are exactly 8 valid sets: {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {3, 4} and {1, 3, 4}.
Set {1, 2, 3, 4} is not valid, because the third condition isn't satisfied. Set {1, 4} satisfies
the third condition, but conflicts with the second condition.

树型DP,从每个节点走仅仅扩展和根节点 root  权值 在 root<=w[v]<=root+D 之内的点, DP[u]= 全部子节点(DP[v]+1)相乘

假设扩展到某个节点 w[v]==w[root]   则标记一下,不要反复走

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>

using namespace std;

typedef long long int LL;

const int maxn=2222;
const LL mod=1000000007;

int n,d,root;
LL w[maxn];
vector<int> g[maxn];
bool vis[maxn][maxn];

LL dp[maxn];

LL dfs(int u,int fa)
{
    dp[u]=1;
    for(int i=0,sz=g[u].size();i<sz;i++)
    {
        int v=g[u][i];
        if(v==fa) continue;
        if(!((w[root]<=w[v])&&(w[v]<=w[root]+d))) continue;
        if(vis[root][v]) continue;
        if(w[root]==w[v]) vis[root][v]=vis[v][root]=true;
        int temp=dfs(v,u);
        dp[u]=(dp[u]+temp*dp[u])%mod;
    }
    return dp[u];
}

int main()
{
    cin>>d>>n;
    for(int i=1; i<=n; i++)
        cin>>w[i];
    for(int i=0; i<n-1; i++)
    {
        int a,b;
        cin>>a>>b;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    LL sum=0;
    for(int i=1; i<=n; i++)
    {
        root=i;
        sum=(sum+dfs(i,i))%mod;
    }
    cout<<sum<<endl;
    return 0;
}

E. LIS of Sequence

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 i belongs 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}.

Solution 2:

// Some notation is re-defined.

• Let F1i be the
  length of LIS ending exactly at ai of
  sequence {a1, a2, ..., ai}.

• Let F2i be the
  length of LIS beginning exactly at ai of
  sequence {ai, ai + 1, ..., an}.

• l = length of LIS of {a1, a2, ..., an} = max{F1i} = max{F2j}.

• Let Fi be the
  length of LIS of sequence {a1, a2, ..., ai - 1, ai + 1, ..., an}
  (i.e the length of LIS of initial sequence a after removing element ai).

• Index i must in group:

  1) if F1i + F2i - 1 < l,
  otherwise:

  2) if Fi = l

  3) if Fi = l - 1

• How to caculate Fi?
  We have: Fi = max{F1u + F2v} among 1 ≤ u < i < v ≤ n such
  that au < av.
  From this formula, we can use Segment tree to calculate Fi.
  Due to limitation of my English, it is really hard to write exactly how. I will post my code soon.

正反求两遍LIS,比較一下就可以.....

假设F1[i]+F2[j]-1==LIS 要用map记录下有没有同样的F1[i],F2[i]   有输出2 没有输出3

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <set>
#include <map>

using namespace std;

const int maxn=100100;

int n,a[maxn],b[maxn];
int f1[maxn],f2[maxn];
int v1[maxn],n1,v2[maxn],n2;
set<int> st;
map<pair<int,int>,int> mp;

int r[maxn],rn;
int ans[maxn];

int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",a+i);
        r[rn++]=a[i];
    }

    sort(r,r+rn);
    rn=unique(r,r+rn)-r;
    ///.....rhash.....
    for(int i=0;i<n;i++)
    {
        int id=lower_bound(r,r+rn,a[i])-r;
        id=rn-1-id;
        b[n-1-i]=r[id];
    }
    int LIS=1;
    for(int i=0;i<n;i++)
    {
        if(i==0)
        {
            v1[n1++]=a[i];
            v2[n2++]=b[i];
            f1[0]=f2[0]=1;
        }
        else
        {
            int p1=lower_bound(v1,v1+n1,a[i])-v1;
            v1[p1]=a[i];
            if(p1==n1) n1++;
            f1[i]=p1+1;
            LIS=max(LIS,f1[i]);

            int p2=lower_bound(v2,v2+n2,b[i])-v2;
            v2[p2]=b[i];
            if(p2==n2) n2++;
            f2[i]=p2+1;
        }
    }
    for(int i=0;i<n;i++)
    {
        int x=i,y=n-1-i;
        if(f1[x]+f2[y]-1<LIS)
            ans[i]=1;
        else if(f1[x]+f2[y]-1==LIS)
        {
            ans[i]=4;
            mp[make_pair(f1[x],f2[y])]++;
        }
    }

    for(int i=0;i<n;i++)
    {
        if(ans[i]==4)
        {
            int x=i,y=n-1-i;
            if(mp[make_pair(f1[x],f2[y])]==1) ans[i]=3;
            else ans[i]=2;
        }
        printf("%d",ans[i]);
        if(i==n-1) putchar('\n');
    }
    return 0;
}