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500pt:

A. Mashmokh and Lights

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mashmokh works in a factory. At the end of each day he must turn off all of the lights.

The lights on the factory are indexed from 1 to n. There are n buttons in Mashmokh's room indexed from 1 to n as well. If Mashmokh pushes button with index i, then each light with index not less than i that is still turned on turns off.

Mashmokh is not very clever. So instead of pushing the first button he pushes some of the buttons randomly each night. He pushed mdistinct buttons b1,?b2,?...,?bm (the buttons were pushed consecutively in the given order) this night. Now he wants to know for each light the index of the button that turned this light off. Please note that the index of button bi is actually bi, not i.

Please, help Mashmokh, print these indices.

Input

The first line of the input contains two space-separated integers n and m (1?≤?n,?m?≤?100), the number of the factory lights and the pushed buttons respectively. The next line contains m distinct space-separated integers b1,?b2,?...,?bm (1?≤?bi?≤?n).

It is guaranteed that all lights will be turned off after pushing all buttons.

Output

Output n space-separated integers where the i-th number is index of the button that turns the i-th light off.

Sample test(s)

input 

5 44 3 1 2

output 

1 1 3 4 4

input 

5 55 4 3 2 1

output 

1 2 3 4 5

Note

In the first sample, after pressing button number 4, lights 4 and 5 are turned off and lights 1, 2 and 3 are still on. Then after pressing button number 3, light number 3 is turned off as well. Pressing button number 1 turns off lights number 1 and 2 as well so pressing button number 2 in the end has no effect. Thus button number 4 turned lights 4 and 5 off, button number 3 turned light 3 off and button number 1 turned light 1 and 2 off.


分析：弄一个数组，开始初始化为-1，然后每次开关某light,就把比它大的还是-1的值设为该开关

代码：

#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;typedef long long ll;const int N=100010;int arr[N];int n;int main(){    while(cin>>n)    {        int m;        for(int i=1;i<=n;i++)            arr[i]=-1;        cin>>m;        for(int i=0;i<m;i++)        {            int b;            cin>>b;            for(int j=b;j<=n;j++)            {                if(arr[j]==-1)                    arr[j]=b;            }        }        for(int i=1;i<=n;i++)            cout<  
 1000pt:  
  
  
   
    B. Mashmokh and Tokens   
   
    
     time limit per test    
 1 second      
    
     memory limit per test    
 256 megabytes      
    
     input    
 standard input      
    
     output    
 standard output       
   
 Bimokh is Mashmokh's boss. For the following n days he decided to pay to his workers in a new way. At the beginning of each day he will give each worker a certain amount of tokens. Then at the end of each day each worker can give some of his tokens back to get a certain amount of money. The worker can save the rest of tokens but he can't use it in any other day to get more money. If a worker gives backw tokens then he'll get  dollars.
   
 Mashmokh likes the tokens however he likes money more. That's why he wants to save as many tokens as possible so that the amount of money he gets is maximal possible each day. He has n numbers x1,?x2,?...,?xn. Number xi is the number of tokens given to each worker on the i-th day. Help him calculate for each of n days the number of tokens he can save.
    
   
    Input   
   
 The first line of input contains three space-separated integers n,?a,?b (1?≤?n?≤?105; 1?≤?a,?b?≤?109). The second line of input containsn space-separated integers x1,?x2,?...,?xn (1?≤?xi?≤?109).
    
   
    Output   
   
 Output n space-separated integers. The i-th of them is the number of tokens Mashmokh can save on the i-th day.
    
   
    Sample test(s)   
   
    
     
      input     
     
5 1 412 6 11 9 1
        
     
      output     
     
0 2 3 1 1 
        
     
      input     
     
3 1 21 2 3
        
     
      output     
     
1 0 1 
        
     
      input     
     
1 1 11
        
     
      output     
          分析：数学题，不要理解错题意就行，反正就是不要多拿没用的来token来换钱就行  
  
代码：
  
  
#include int n, a, b, x[100000];int main(){	scanf("%d%d%d", &n, &a, &b);	for(int i = 0; i < n; i++)		scanf("%d", &x[i]);	for(int i = 0; i < n; i++)	{		long long tmp = (long long)x[i] * a / b;		printf("%d ", x[i] - (tmp * b % a ? tmp * b / a + 1 : tmp * b / a));	}	getchar(); getchar();	return 0;}
  
 1500pt:  
  
  
   
    C. Mashmokh and Numbers   
   
    
     time limit per test    
 1 second      
    
     memory limit per test    
 256 megabytes      
    
     input    
 standard input      
    
     output    
 standard output       
   
 It's holiday. Mashmokh and his boss, Bimokh, are playing a game invented by Mashmokh.
   
 In this game Mashmokh writes sequence of n distinct integers on the board. Then Bimokh makes several (possibly zero) moves. On the first move he removes the first and the second integer from from the board, on the second move he removes the first and the second integer of the remaining sequence from the board, and so on. Bimokh stops when the board contains less than two numbers. When Bimokh removes numbers x and y from the board, he gets gcd(x,?y) points. At the beginning of the game Bimokh has zero points.
   
 Mashmokh wants to win in the game. For this reason he wants his boss to get exactly k points in total. But the guy doesn't know how choose the initial sequence in the right way.
   
 Please, help him. Find n distinct integers a1,?a2,?...,?an such that his boss will score exactly k points. Also Mashmokh can't memorize too huge numbers. Therefore each of these integers must be at most 109.
    
   
    Input   
   
 The first line of input contains two space-separated integers n,?k (1?≤?n?≤?105; 0?≤?k?≤?108).
    
   
    Output   
   
 If such sequence doesn't exist output -1 otherwise output n distinct space-separated integers a1,?a2,?...,?an (1?≤?ai?≤?109).
    
   
    Sample test(s)   
   
    
     
      input     
     
5 2
        
     
      output     
     
1 2 3 4 5
        
     
      input     
     
5 3
        
     
      output     
     
2 4 3 7 1
        
     
      input     
     
7 2
        
     
      output     
     
-1
           
   
    Note   
   
 gcd(x,?y) is greatest common divisor of x and y.
    
  
  
分析：貌似乱搞搞过了。。。我是先平均每对分担到的k值，然后对于每个k,都用连续的两个值去乘，比如k=4,一共5个数的话，前两队每对得分为2，第一队的两个数为1*2和2*2，第二队的两个数为3*2和4*2，连续的两个数能保证gcd,针对平摊的最后一个k值要特殊处理一下
  
代码：
  
  
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;typedef long long ll;const int N=(int)1e9+1;map<int,bool> visited;int n,k;int main(){    cin>>n>>k;    if(n==1)    {        if(k==0)            cout<<1<<endl;        else            cout<<-1<<endl;        return 0;    }    vector ans;    int t = n/2;    if(t>k)    {        cout<<-1<<endl;        return 0;    }    int avg = k/t;    int last = k-avg*(t-1);    int cur = 1;    for(int i=1;i<t;i++)    {        int score = avg;        ll t1 = (ll)score*(ll)cur;        ll t2 = (ll)score*(ll)(cur+1);        if(t1>N||t2>N)        {            cout<<-1<<endl;            return 0;        }        if(!visited[t1]&&!visited[t2])        {            ans.push_back(t1);            ans.push_back(t2);            visited[t1]=true;            visited[t2]=true;            cur+=2;        }    }    for(int i=1;i<N;i++)    {        if(!visited[i*last]&&!visited[i*last+last])        {            ans.push_back(i*last);            ans.push_back(i*last+last);            visited[i*last+last]=true;            visited[i*last]=true;            break;        }    }    if(ans.size()<n)    {        for(int i=1;i<N;i++)        {            if(!visited[i])            {                ans.push_back(i);                break;            }        }    }    for(int i=0;i  
 2000pt:  
  
  
   
    D. Mashmokh and ACM   
   
    
     time limit per test    
 1 second      
    
     memory limit per test    
 256 megabytes      
    
     input    
 standard input      
    
     output    
 standard output       
   
 Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
   
 A sequence of l integers b1,?b2,?...,?bl (1?≤?b1?≤?b2?≤?...?≤?bl?≤?n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally  for all i (1?≤?i?≤?l?-?1).
   
 Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109?+?7).
    
   
    Input   
   
 The first line of input contains two space-separated integers n,?k (1?≤?n,?k?≤?2000).
    
   
    Output   
   
 Output a single integer ? the number of good sequences of length k modulo 1000000007 (109?+?7).
    
   
    Sample test(s)   
   
    
     
      input     
     
3 2
        
     
      output     
        
     
      input     
     
6 4
        
     
      output     
     
39
        
     
      input     
     
2 1
        
     
      output     
           
   
    Note   
   
 In the first sample the good sequences are: [1,?1],?[2,?2],?[3,?3],?[1,?2],?[1,?3].
   分析：早知道应该做这题的，连我都会的dp.....用dp[i][j]表示长度为i,最后一个元素为j的序列数  
  
代码：
  
  
#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;typedef long long ll;const int N=2010;const int MOD = (int)1e9+7;int dp[N][N];int n,k;int main(){    while(cin>>n>>k)    {		memset(dp,0,sizeof(dp));		for(int i=1;i<=n;i++)			dp[1][i]=1;		for(int length=1;length<k;length++)		{			for(int i=1;i<=n;i++)			{				for(int j=1;j*i<=n;j++)				{					dp[length+1][j*i]+=dp[length][i];					dp[length+1][j*i]%=MOD;				}			}		}		int ret = 0;		for(int i=1;i<=n;i++)		{			ret+=dp[k][i];			ret%=MOD;		}		cout<<ret<<endl;    }    return 0;}

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