hdu 3089 （快速约瑟夫环）

Josephus again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 741    Accepted Submission(s): 210

Problem Description

In our Jesephus game, we start with n people numbered 1 to n around a circle, and we eliminated every k remaining person until only one survives. For example, here's the starting configuration for n = 10, k = 2, The elimination order is 2, 4, 6, 8, 10, 3, 7, 1, 9. So 5 survives.The problem: determine the survivor's number , J(n, k).

 

Input

There are multiple cases, end with EOF
each case have two integer, n, k. (1<= n <= 10^12, 1 <= k <= 1000)

 

Output

each case a line J(n, k)

 

Sample Input

10 2
10 3

 

Sample Output

5
4

 

Source

2009 Multi-University Training Contest 18 - Host by ECNU

 

Recommend

lcy

 

---

 //约瑟夫环递推式 报到k出列 那么我们需要for(i:1~n) f[i]=(f[i-1]+k)%i。f[1]=0; 其中 编号是0~n-1
 //当f[i]+k>=i 就常规递推，否则快速跳跃，跳跃长度为(i-f[i])/(k-1)，利用i和k的增长速度差算长度。
 //因为编号是0~n-1，所以最后答案要+1和数组下标对应。
 #include<bits/stdc++.h>
 #define clr(x) memset(x,0,sizeof(x))
 #define clr_1(x) memset(x,-1,sizeof(x))
 #define mod 1000000007
 #define LL long long
 #define INF 0x3f3f3f3f
 #define mp make_pair
 #define pb push_back
 #define mp make_pair
 #define fi first
 #define se second
 using namespace std;
 const int N=5e5+;
 LL n,k,t;
 LL solve(LL n,LL k)
 {
     LL ans=,i;
     if(k==) return n;
     for(i=;i<=n;)
     {
         if(ans+k>=i)
             ans=(ans+k)%i,i++;
         else
         {
           t=min((i--ans)/(k-),n-(i-));
           i+=t-;
           ans+=k*t;
           ans%=i;
           i++;
         }
     }
     return ans%n+;
 }
 int main()
 {
     while(scanf("%lld%lld",&n,&k)!=EOF)
     {
         printf("%lld\n",solve(n,k));
     }
     return ;
 }