POJ 1781

约瑟夫环的二进制解法。

当n=2*k,

序列1:1,2,3...2*k

序列2:1,3,5..2*k-1(k个)

序列3:1,2,3...k

所以有f(2*k) = 2*f(k) - 1。

当n=2*k-1,

序列1:1,2,3...2*k-1

序列2:3,5,7...2*k-1(k个)

序列3:1,2,3...k

所以有f(2*k+1) = 2*f(k) + 1。

#include<iostream>
#include<cstdio>
using namespace std;
typedef long long int ll;

ll get_num(ll x)
{
    if (x == 1)return 1;
    if (x&1)return (2*get_num(x/2)+1);
    else return (2*get_num(x/2)-1);
}

int main()
{
    ll n,m;
    while (scanf("%I64de%I64d",&n,&m) && !(n == 0 && m == 0)) {
        for (int i(0); i<m; ++i) {
            n *= 10;
        }
        printf("%I64d\n",get_num(n));
    }
    return 0;
}