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

 

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//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+10;
LL n,k,t;
LL solve(LL n,LL k)
{
    LL ans=0,i;
    if(k==1) return n;
    for(i=2;i<=n;)
    {
        if(ans+k>=i)
            ans=(ans+k)%i,i++;
        else
        {
          t=min((i-1-ans)/(k-1),n-(i-1));
          i+=t-1;
          ans+=k*t;
          ans%=i;
          i++;
        }
    }
    return ans%n+1;
}
int main()
{
    while(scanf("%lld%lld",&n,&k)!=EOF)
    {
        printf("%lld\n",solve(n,k));
    }
    return 0;
}