浙南联合训练赛20180214

In the first sample remainders modulo 1 and 4 coincide.

只是试了一下暴力就过了，可能大数据很难造吧，或者不存在

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    ll n,k;
    cin>>n>>k;
    for(ll i=1;i<=k;i++)
        if(n%i!=i-1)
        {
            puts("No\n");
            return 0;
        }
    puts("Yes\n");
    return 0;
}

E - And Now, a Remainder from Our Sponsor

 HDU - 1930

InputThe input consists of multiple test cases. The first line of input consists of a single positive integer n indicating the number of test cases. The next 2n lines of the input consist of the test cases. The first line of each test case contains a positive integer (< 50) giving the number of groups in the encoded message. The second line of each test case consists of the four keys followed by the encoded message. 
Each message group is separated with a space. 
OutputFor each test case write the decoded message. You should not print any trailing blanks.Sample Input

2
6
34 81 65 43 1062038 1043103 1473907 22794503 15135731 16114011
3
20 31 53 39 5184133 14080210 7090922

Sample Output

THE CAT IN THE HAT
THE END

这个题目是数学题啦

给n串密码，再给4个被取余的数，要求恢复成原来的一串数组，所以两位两位分，运用扩欧或者中国剩余定理即可解决

还有玄学代码使用272727这个数字的，应该是数据不强，这个题目是这次最难得题目

#include<bits/stdc++.h>
using namespace std;
const int N=55;
int a[5],TT[N][5],out[N];
char num[N],m[5];
void exgcd(int a,int b,int &d,int &x,int &y)
{
    if(!b)d=a,x=1,y=0;
    else exgcd(b,a%b,d,y,x),y-=x*(a/b);
}
int main()
{
    ios::sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        for(int i=1; i<=4; i++)cin>>a[i];
        for(int i=1,x; i<=n; i++)
        {
            cin>>x;
            for(int j=4; j>0; j--)TT[i][j]=x%100,x/=100;
        }
        for(int i=1; i<=n; i++)
        {
            int c=a[1],t=TT[i][1],d,z,x,y;
            for(int j=2; j<5; j++)
            {
                exgcd(c,a[j],d,x,y);
                z=a[j]/d;
                x=((TT[i][j]-t)/d*x%z+z)%z;
                t+=c*x;
                c=c/d*a[j];
            }
            out[i]=t;
        }
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<4; j++)
                m[j]=out[i]%100,out[i]/=100;
            for(int j=3; j>0; j--)
            {
                if(m[j]==27)
                {
                    if(i!=n)printf(" ");
                    else break;
                }
                else printf("%c",m[j]+‘A‘-1);
            }
        }
        putchar(10);
    }
    return 0;
}

F - Connected Components?

 CodeForces - 920E 

You are given an undirected graph consisting of n vertices and  edges. Instead of giving you the edges that exist in the graph, we give you m unordered pairs (x,?y) such that there is no edge between x and y, and if some pair of vertices is not listed in the input, then there is an edge between these vertices.

You have to find the number of connected components in the graph and the size of each component. A connected component is a set of vertices X such that for every two vertices from this set there exists at least one path in the graph connecting these vertices, but adding any other vertex to X violates this rule.

Input

Output

Example

Input

5 5
1 2
3 4
3 2
4 2
2 5

Output

2
1 4 

---

图论啦，bfs出序列就可以了

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
set<int>M[N+N];
queue<int>Q;
int q[N],yy[N],n,m,k;
void bfs()
{
    while(!Q.empty())
    {
        int tot=1,u=Q.front();
        Q.pop();
        q[0]=u;
        for(int i=0; i<tot; i++)
        {
            int now=n-tot;
            while(now>0&&!Q.empty())
            {
                int v=Q.front();
                Q.pop();
                if(!M[q[i]].count(v)) q[tot++]=v;
                else Q.push(v);
                --now;
            }
        }
        yy[k++]=tot;
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin>>n>>m;
    for(int i=0,u,v; i<m; i++)
        cin>>u>>v,M[u].insert(v),M[v].insert(u);
    for(int i=1; i<=n; i++) Q.push(i);
    bfs();
    sort(yy,yy+k);
    printf("%d\n",k);
    for(int i=0; i<k; i++)printf("%d ",yy[i]);
    return 0;
}

G - MADMAX

 CodeForces - 917B 

As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actual game just to prove that she‘s not the best at games. The game is played on a directed acyclic graph (a DAG) with nvertices and m edges. There‘s a character written on each edge, a lowercase English letter.

Max and Lucas are playing the game. Max goes first, then Lucas, then Max again and so on. Each player has a marble, initially located at some vertex. Each player in his/her turn should move his/her marble along some edge (a player can move the marble from vertex v to vertex u if there‘s an outgoing edge from v to u). If the player moves his/her marble from vertex v to vertex u, the "character" of that round is the character written on the edge from v to u. There‘s one additional rule; the ASCII code of character of round i should be greater than or equal to the ASCII code of character of round i?-?1 (for i?>?1). The rounds are numbered for both players together, i. e. Max goes in odd numbers, Lucas goes in even numbers. The player that can‘t make a move loses the game. The marbles may be at the same vertex at the same time.

Since the game could take a while and Lucas and Max have to focus on finding Dart, they don‘t have time to play. So they asked you, if they both play optimally, who wins the game?

You have to determine the winner of the game for all initial positions of the marbles.

Input

The first line of input contains two integers n and m (2?≤?n?≤?100, ).

The next m lines contain the edges. Each line contains two integers v, u and a lowercase English letter c, meaning there‘s an edge from v to u written c on it (1?≤?v,?u?≤?n, v?≠?u). There‘s at most one edge between any pair of vertices. It is guaranteed that the graph is acyclic.

Output

Print n lines, a string of length n in each one. The j-th character in i-th line should be ‘A‘ if Max will win the game in case her marble is initially at vertex iand Lucas‘s marble is initially at vertex j, and ‘B‘ otherwise.

Example

Input

4 4
1 2 b
1 3 a
2 4 c
3 4 b

Output

BAAA
ABAA
BBBA
BBBB

Input

5 8
5 3 h
1 2 c
3 1 c
3 2 r
5 1 r
4 3 z
5 4 r
5 2 h

Output

BABBB
BBBBB
AABBB
AAABA
AAAAB

Note

Here‘s the graph in the first sample test case:

Here‘s the graph in the second sample test case:

 

因为dag环，所以可以记忆化搜索

#include<bits/stdc++.h>
using namespace std;
const int N=105;
int SG[105][105][26];
int head[5005],nxt[5005],to[5005],W[5005];
int tot;
int DFS(int i,int j,int k)
{
    if(SG[i][j][k])return SG[i][j][k];
    for(int p=head[i]; p; p=nxt[p])
        if(W[p]>=k&&!DFS(j,to[p],W[p]))SG[i][j][k]=1;
    return SG[i][j][k];
}
void add(int u,int v,int w)
{
    nxt[++tot]=head[u],head[u]=tot,to[tot]=v,W[tot]=w;
}
int main()
{
    int n,m;
    string s;
    cin>>n>>m;
    for(int i=1,u,v; i<=m; i++)
        cin>>u>>v>>s,add(u,v,s[0]-‘a‘);
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
            if(DFS(i,j,0))cout<<‘A‘;
            else cout <<‘B‘;
        cout<<"\n";
    }
    return 0;
}

H - Color a Tree

 HDU - 1055 

Bob is very interested in the data structure of a tree. A tree is a directed graph in which a special node is singled out, called the "root" of the tree, and there is a unique path from the root to each of the other nodes. 

Bob intends to color all the nodes of a tree with a pen. A tree has N nodes, these nodes are numbered 1, 2, ..., N. Suppose coloring a node takes 1 unit of time, and after finishing coloring one node, he is allowed to color another. Additionally, he is allowed to color a node only when its father node has been colored. Obviously, Bob is only allowed to color the root in the first try. 

Each node has a “coloring cost factor”, Ci. The coloring cost of each node depends both on Ci and the time at which Bob finishes the coloring of this node. At the beginning, the time is set to 0. If the finishing time of coloring node i is Fi, then the coloring cost of node i is Ci * Fi. 

For example, a tree with five nodes is shown in Figure-1. The coloring cost factors of each node are 1, 2, 1, 2 and 4. Bob can color the tree in the order 1, 3, 5, 2, 4, with the minimum total coloring cost of 33. 

 

Given a tree and the coloring cost factor of each node, please help Bob to find the minimum possible total coloring cost for coloring all the nodes. 

InputThe input consists of several test cases. The first line of each case contains two integers N and R (1 <= N <= 1000, 1 <= R <= N), where N is the number of nodes in the tree and R is the node number of the root node. The second line contains N integers, the i-th of which is Ci (1 <= Ci <= 500), the coloring cost factor of node i. Each of the next N-1 lines contains two space-separated node numbers V1 and V2, which are the endpoints of an edge in the tree, denoting that V1 is the father node of V2. No edge will be listed twice, and all edges will be listed. 

A test case of N = 0 and R = 0 indicates the end of input, and should not be processed. 
OutputFor each test case, output a line containing the minimum total coloring cost required for Bob to color all the nodes. 
Sample Input

5 1
1 2 1 2 4
1 2
1 3
2 4
3 5
0 0

Sample Output

33

---

树染色经典题目，可以并查集维护的

#include<bits/stdc++.h>
using namespace std;
const int N=1005;
int a[N],f[N],fa[N],num[N],sum[N],n,r;
int find(int u)
{
    return u==fa[u]?u:(fa[u]=find(fa[u]));
}
int main()
{
    ios::sync_with_stdio(false);
    while(cin>>n>>r)
    {
        if(!n&&!r)break;
        f[r]=0,memset(sum,0,sizeof sum);
        for(int i=0; i<=n; i++)
            fa[i]=i,num[i]=1;
        for(int i=1; i<=n; i++)cin>>a[i];
        for(int i=1,u,v; i<n; i++)
            cin>>u>>v,f[v]=u;
        for(int i=0; i<n; i++)
        {
            int u=-1;
            for(int j=1; j<=n; j++)
            if(fa[j]==j&&(u==-1||a[u]*num[j]<a[j]*num[u]))u=j;
            int v=find(f[u]);
            sum[v]+=num[v]*a[u]+sum[u],num[v]+=num[u],a[v]+=a[u],fa[u]=v;
        }
        cout<<sum[0]<<"\n";
    }
    return 0;
}