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。

View Code

 1 #include<iostream>

 2 #include<cstdio>

 3 using namespace std;

 4 typedef long long int ll;

 5 

 6 ll get_num(ll x)

 7 {

 8     if (x == 1)return 1;

 9     if (x&1)return (2*get_num(x/2)+1);

10     else return (2*get_num(x/2)-1);

11 }

12 

13 int main()

14 {

15     ll n,m;

16     while (scanf("%I64de%I64d",&n,&m) && !(n == 0 && m == 0)) {

17         for (int i(0); i<m; ++i) {

18             n *= 10;

19         }

20         printf("%I64d\n",get_num(n));

21     }

22     return 0;

23 }