hdu 4294 Multiple

Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 534    Accepted Submission(s): 138


Problem Description
  Given a positive integer N, you’re to solve the following problem:
  Find a positive multiple of N, says M, that contains minimal number of different digits in base-K notation. If there’re several solutions, you should output the numerical smallest one. By saying numerical smallest one, we compar their numerical value, so 0xA hex < 11 dec.
  You may assume that 1 <= N <= 10 4 and 2 <= K <= 10.
 

Input
  There’re several (less than 50) test cases, one case per line.
  For each test case, there is a line with two integers separated by a single space, N and K.
  Please process until EOF (End Of File).
 

Output
  For each test case, you should print a single integer one line, representing M in base-K notation,the answer.
 

Sample Input
  
10 8 2 3 7 5
 

Sample Output
  
2222 2 111111
 

Source
 


题目:在K进制下,用最少的不同的数,表示成n的倍数

首先要得到一个结论,就是最多用两个数就行了。

证明 :对于一个数字a,可以构造出的数字有a,aa,aaa,aaaa,aaaaa,……

每一个数对于n都有一个余数,余数最多有n个,根据鸽巢原理,前n+1个数中,必然有两个余数相等

那么二者之差,必定为n的倍数,形式为a……a0……0。

有这个结论,就简单了,先枚举一个数,然后枚举两个数,BFS即可


代码:

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N=10002;
int n,c,k,l;
int pre[N],res[N],step[N],num[2];
char ans[N],re[N];
bool bfs()
{
	queue<int> Q;
	int i,p,q;
	memset(step,0,sizeof(step));
	for(i=0;i<c;i++)
	{
		int y=num[i]%n;
		if(!num[i]||step[y]) continue;
		step[y]=1;
		pre[y]=-1;
		res[y]=num[i];
		Q.push(y);
	}
	while(!Q.empty())
	{
		p=Q.front();Q.pop();
		if(p==0) return true;
		if(l&&step[p]>l) return false;
		for(i=0;i<c;i++)
		{
			q=(p*k+num[i])%n;
			if(!step[q])
			{
				step[q]=step[p]+1;
				pre[q]=p;
				res[q]=num[i];
				Q.push(q);
			}
		}
	}
	return false;
}
int l1;
void solve(int i)
{
	if(pre[i]!=-1) solve(pre[i]);
	re[l1++]=res[i]+'0';
}
bool judge()
{
	int i;
	if(!l||l1<l) return true;
	if(l1>l) return false;
	for(i=0;i<l;i++)
	{
		if(re[i]<ans[i]) return true;
		if(re[i]>ans[i]) return false;
	}
	return false;
}
int main()
{
	int i,j;
	while(~scanf("%d%d",&n,&k))
	{
		l=0;
		bool f=false;
		for(i=1;i<k;i++)
		{
			num[0]=i;c=1;
			if(bfs())
			{
				f=true;
				l1=0;
				solve(0);
				if(judge()) {l=l1;memcpy(ans,re,sizeof(ans));}
			}
		}
		if(!f)
		{
			for(i=0;i<k;i++)for(j=i+1;j<k;j++)
			{
				num[0]=i;num[1]=j;c=2;
				if(bfs())
				{
					l1=0;
					solve(0);
					if(judge()) {l=l1;memcpy(ans,re,sizeof(ans));}
				}
			}
		}
		for(i=0;i<l;i++)putchar(ans[i]);
		puts("");
	}
	return 0;
}



### HDU 1466 Problem Description and Solution The problem **HDU 1466** involves calculating the expected number of steps to reach a certain state under specific conditions. The key elements include: #### Problem Statement Given an interactive scenario where operations are performed on numbers modulo \(998244353\), one must determine the expected number of steps required to achieve a particular outcome. For this type of problem, dynamic programming (DP) is often employed as it allows breaking down complex problems into simpler subproblems that can be solved iteratively or recursively with memoization techniques[^1]. In more detail, consider the constraints provided by similar problems such as those found in references like HDU 6327 which deals with random sequences using DP within given bounds \((1 \leq T \leq 10, 4 \leq n \leq 100)\)[^2]. These types of constraints suggest iterative approaches over small ranges might work efficiently here too. Additionally, when dealing with large inputs up to \(2 \times 10^7\) as seen in reference materials related to counting algorithms [^4], efficient data structures and optimization strategies become crucial for performance reasons. However, directly applying these methods requires understanding how they fit specifically into solving the expectation value calculation involved in HDU 1466. For instance, if each step has multiple outcomes weighted differently based on probabilities, then summing products of probability times cost across all possible states until convergence gives us our answer. To implement this approach effectively: ```python MOD = 998244353 def solve_expectation(n): dp = [0] * (n + 1) # Base case initialization depending upon problem specifics for i in range(1, n + 1): total_prob = 0 # Calculate transition probabilities from previous states for j in transitions_from(i): # Placeholder function representing valid moves prob = calculate_probability(j) next_state_cost = get_next_state_cost(j) dp[i] += prob * (next_state_cost + dp[j]) % MOD total_prob += prob dp[i] %= MOD # Normalize current state's expectation due to accumulated probability mass if total_prob != 0: dp[i] *= pow(total_prob, MOD - 2, MOD) dp[i] %= MOD return dp[n] # Example usage would depend heavily on exact rules governing transitions between states. ``` This code snippet outlines a generic framework tailored towards computing expectations via dynamic programming while adhering strictly to modular arithmetic requirements specified by the contest question format.
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