Josephus formula

Josephus: numbering starts at 0, the first people out there who are (k-1)% n, renumbered people out there who's next numbered 0, and so on, the last one out of a certain number of people to 0, f [1] = 0; when the first person out, the remaining n - 1 numbers out individual f [9] = (k - 1)% (n - 1), the queue restore the original number (f [n - 1] + k)% (n - 1 + 1);

 

1. Numbering starts at 0
2. Each person went out a renumbering
3. Restore the original arrangement of the formula: f [x] = (f [x] + k)% (x + 1) (+ k Sagamihara continue until the original number of +1 the number of + 1 = n);

 

The initial value was

n individuals, the number of number k

The first out: left people n; Initial value: f [n] = (k - 1)% n;

The second out: remaining n - 1 person; Initial value: f [n - 1] = (k - 1)% (n - 1);

Third out: remaining n - 2 persons; Initial value: f [n - 2] = (k - 1)% (n - 2);

。。。

 

Restore original arrangement number

With the initial value, then you can restore the number 

First out: n left people without reduction

The second out: remaining n - 1 person, f [n - 1] = (f [n - 1] + k)% n;

Third out: remaining n - 2 persons, f [n - 2] = (f [n - 2] + k)% (n - 1), f [n - 2] = (f [n - 2] + k)% n;

。。。

Is to keep the number of mold + k + 1; until the person to n

 

Code

#include <cstdio>
 #define N 100001 int n-, K;
 int F [N]; int main () { 
    Scanf ( " % D% D " , & n-, & K);
     for ( int I = n-; I> = . 1 ; - I) { 
        F [I] = (K - . 1 )% I;
         for ( int J = I + . 1 ; J <= n-; ++ J)      
            F [I] = (F [I] + K )% J; 
        the printf ( " % D " , F [I] + . 1 ); // output number +1, as numbered from 0




    }
    return 0;
}