1289: Zju1136 Multiple N的倍数
Time Limit: 1 Sec Memory Limit: 256 MBSubmit: 62 Solved: 46
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Description
a program that, given a natural number N between 0 and 4999 (inclusively), and M distinct decimal digits X1,X2..XM (at least one), finds the smallest strictly positive multiple of N that has no other digits besides X1,X2..XM (if such a multiple exists).
Input
On the first line - the number N
On the second line - the number M
On the following M lines - the digits X1,X2..XM.
On the second line - the number M
On the following M lines - the digits X1,X2..XM.
Output
For each data set, the program should write to standard output on a single line the multiple, if such a multiple exists, and 0 otherwise.
Sample Input
22
3
7
0
1
Sample Output
110
HINT
Source
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 using namespace std; 3 int num[10000],y[10000],pp[10000]; 4 bool o[5000]; 5 int f[10]; 6 int m,n,i,j,k,temp,head,tail; 7 8 void output(int head) 9 { 10 if (pp[head]!=0) 11 output (pp[head]); 12 cout<<num[head]; 13 } 14 void bfs() 15 { 16 head=0; 17 tail=0; 18 k=0; 19 do 20 { 21 for (i=1;i<=m;i++) 22 { 23 if (head==0 && f[i]==0) continue; 24 if (o[(y[head]*10+f[i])%n]) continue; 25 tail++; 26 y[tail]=(y[head]*10+f[i])%n; 27 pp[tail]=head; 28 num[tail]=f[i]; 29 o[y[tail]]=true; 30 if (y[tail]==0) 31 { 32 output(tail); 33 k=1; 34 break; 35 } 36 } 37 if (k==1) break; 38 head++; 39 } while(head<=tail); 40 } 41 int main() 42 { 43 cin>>n>>m; 44 for (i=1;i<=m;i++) 45 cin>>f[i]; 46 if (n==0) 47 { 48 cout<<0<<endl; 49 return 0; 50 } 51 for (i=1;i<=m-1;i++) 52 for (j=i+1;j<=m;j++) 53 if (f[j]<f[i]) 54 { 55 temp=f[j]; 56 f[j]=f[i]; 57 f[i]=temp; 58 } 59 bfs(); 60 if (k==0) 61 cout<<0<<endl; 62 cout<<endl; 63 return 0; 64 }