Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1071 Accepted Submission(s): 273
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 0xAhex < 11dec.
You may assume that 1 <= N <= 104 and 2 <= K <= 10.
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 0xAhex < 11dec.
You may assume that 1 <= N <= 104 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).
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
解题：据说：
最多用两个数就可以表示任何数的倍数。
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。
然后暴力搜索即可
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 20010; 4 string ans,str; 5 int n,k,m,num[2],pre[maxn],b[maxn]; 6 bool bfs() { 7 queue<int>q; 8 memset(b,-1,sizeof b); 9 for(int i = 0; i < m; ++i) 10 if(num[i]) { 11 q.push(num[i]); 12 b[num[i]] = num[i]; 13 } 14 memset(pre,-1,sizeof pre); 15 while(!q.empty()) { 16 int u = q.front(); 17 q.pop(); 18 if(!u) return true; 19 for(int i = 0; i < m; ++i) { 20 int t = (u*k + num[i])%n; 21 if(b[t] == -1) { 22 b[t] = num[i]; 23 pre[t] = u; 24 q.push(t); 25 } 26 if(!t) return true; 27 } 28 } 29 return false; 30 } 31 void Path(int u) { 32 if(pre[u] != -1) Path(pre[u]); 33 str += (char)(b[u] + '0'); 34 } 35 bool cmp(string &a,string &b) { 36 if(b.size() == 0) return true; 37 if(a.size() > b.size()) return false; 38 if(a.size() < b.size()) return true; 39 return a < b; 40 } 41 int main() { 42 while(~scanf("%d%d",&n,&k)) { 43 if(n < k) { 44 printf("%d ",n); 45 continue; 46 } 47 bool flag = false; 48 ans = ""; 49 for(int i = m = 1; i < k; ++i) { 50 num[0] = i; 51 if(bfs()) { 52 str = ""; 53 Path(0); 54 flag = true; 55 if(cmp(str,ans)) ans = str; 56 } 57 } 58 if(!flag) { 59 m = 2; 60 for(int i = 1; i < k; ++i) { 61 num[1] = i; 62 for(int j = 0; j < i; ++j) { 63 num[0] = j; 64 if(bfs()) { 65 str = ""; 66 Path(0); 67 if(cmp(str,ans)) ans = str; 68 } 69 } 70 } 71 } 72 cout<<ans<<endl; 73 } 74 return 0; 75 }