Bnuoj Musical Chairs Joseph ring non-recursive

/* Problem Description: N Person (number 0~ (N1-1)), starting from 0 count, reporting (M-1) exit, the remainder continues to count off from 0. The winner's number.


We know that the first person (the number must be m%n-1) after the dequeue, the rest of the n1-1 individuals formed a new Joseph Ring (starting with the person numbered k=m%n1):
K k+1 k+2 ... n1-2, n1-1, 0, 1, 2, ... k-2
And at the beginning of the K is reported 0.


Now let's do a conversion of their numbers:
n1=n+1;


K--0-->0
K+1-1-->1
K+2-2-->2
...
...
K-2-N1-2-->n-1
K-1-N1-1-->n




(3)
(1) (2) +k
K--0--and K
K+1-1-->k+1
K+2-2-->k+2
...                                                     ...
...                                                     ...
N1-2-> (N1-2-k)--(N1-2-K) +k
N1-1-> (N1-1-k)--(N1-1-K) +k
0---(N1-1-K) + (1)--(N1-1-K) + (1) +k
1---(N1-1-K) + (2)--(N1-1-K) + (2) +k
...
...
K-2---n1-1-k+ (K-1) (based on the corresponding relationship above his value is n1-2=n-1)--(N1-1-k) + (k-1) +k
K-1---n1-1-k+ (k) (based on the corresponding relationship above his value is N1-1=n)--(N1-1-k) + (k) +k
N1=n+1 into (3) after (1) ≡ (3) (mod n1) that is right plus K and left is about N1 is the same remainder
K--0
K+1-1
K+2-2
...
...
K-2-N-1
After the transformation has become completely (n1-1) personal count of the sub-problem, if we know the solution of this sub-problem: for example, X is the final winner, then
According to the table above, the x is changed back to not just the solution of the n personal situation?!! The formula that changes back is very simple, I believe everyone can push out:
X ' = (x+k)%n1,k=m%n1, because K is the remainder, so smaller than n1, x ' = (x%n1+k)%n1= (x%n1+m%n1)%n1= (x+m)%n i.e. x ' = (x+m)%n1, that is, the solution of the current n number is X,
N1=n+1,m is fixed, can be obtained through the formula, how to Know (n1-1) the problem of personal count off? Yes, as long as you Know (n1-2) the individual's solution. (n1-2) A personal solution? Of course it is.
First Ask (n1-3) The situation----This is obviously a backward problem! Well, the idea came out, the following to write a recursive formula: make F[i] means I play games
The number of the final winner of the report M exit, the final result is naturally f[n]
Recursive formulas
f[1]=0;
f[i]= (f[i-1]+m)%i; (i>1)

*/

#include <stdio.h>
int main ()
{
int n,m,i,s;
while (scanf ("%d%d", &n,&m)!=eof&& (n+m)!=0)
{
s=0;
for (i=2;i<=n;i++)
S= (s+m)%i;
printf ("%d%d%d\n", n,m,s+1);
}
return 0;
}

Bnuoj Musical Chairs Joseph ring non-recursive