HDOJ 4259 Double Dealing

HDOJ 4259 Double Dealing

Find the cyclic section of each digit and find the lcm

Double Dealing Time Limit: 50000/20000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 1806 Accepted Submission (s): 622


Problem DescriptionTake a deck NUnique cards. Deal the entire deck out KPlayers in the usual way: the top card to player 1, the next to player 2, KTh to player K, K+ 1st to player 1, and so on. Then pick up the cards-place player 1's cards on top, then player 2, and so on, so that player K'S cards are on the bottom. Each player's cards are in reverse order-the last card that they were dealt is on the top, and the first on the bottom.
How many times, including the first, must this process be repeated before the deck is back in its original order?
InputThere will be multiple test cases in the input. Each case will consist of a single line with two integers, NAnd K(1 ≤ N≤ 800, 1 ≤ K≤ 800). The input will end with a line with two 0 s.

OutputFor each test case in the input, print a single integer, indicating the number of deals required to return the deck to its original order. output each integer on its own line, with no extra spaces, and no blank lines between answers. all possible inputs yield answers which will fit in a signed 64-bit integer.
Sample Input

1 310 352 40 0

Sample Output

1413

SourceThe University of Chicago Invitational Programming Contest 2012

#include 
 
  #include 
  
   #include 
   
    #include #include 
    
     using namespace std;int n,m;typedef long long int LL;int next[880],to[880];bool vis[880];LL gcd(LL a,LL b){    if(b==0) return a;    return gcd(b,a%b);}LL lcm(LL a,LL b){    return a/gcd(a,b)*b;}int get_int(){    char ch;    int ret=0;    while(ch=getchar())    {        if(ch>='0'&&ch<='9')        {            ret=ret*10+ch-'0';        }        else break;    }    return ret;}int main(){    while(true)    {        n=get_int();m=get_int();        if(n==0&&m==0) break;        if(n<=m)        {            puts("1"); continue;        }        ///mo ni yi chi        for(int i=1;i<=n;i++)            next[i]=i;        for(int i=1;i<=m;i++)        {            to[i]=n/m;            if(i<=n%m) to[i]++;            to[i]+=to[i-1];        }        for(int i=1;i<=n;i++)        {            next[i]=to[(i-1)%m+1]--;        }        LL ans=1;        memset(vis,false,sizeof(vis));        for(int i=1;i<=n;i++)        {            if(vis[i]) continue;            int t=next[i];            LL temp=1;            while(t!=i)            {                vis[t]=true;                t=next[t];                temp++;            }            ans=lcm(ans,temp);        }        printf("%I64d\n",ans);    }    return 0;}