"Joseph Ring variant" UVa 1394-and then there is one

First of all see this problem in the brain immediately jump out of the list. Then continue to look at home analysis said, the chain table method time complexity of O (N*k), will certainly tle, I realized that it is a simple mind t^t.

Look at home analysis not how to understand, later found that this article oneself understand easier (...) ) share a little bit of ~http://blog.csdn.net/chenguolinblog/article/details/8873444

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

The number 0~ (n-1) is meaningful because it is modulo n, so it is better to operate with 0-(N-1)
We know the first person (the number must be (m-1) mod n) after the dequeue, the remaining n-1 individuals make up a new Joseph ring (starting with the person numbered k=m mod n):

K k+1 k+2 n-2,n-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:
K--0
K+1-1
K+2-2
...
...
K-2-N-2
After the transformation has completely become the (n-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 above table to turn this x back is not exactly the solution of n personal situation?! The formula to change back is very simple, I believe everyone can push out: X ' = (x+k) mod n;

So as long as we repeat this process, we can get the first person's number, because this person's final number is 0 (he alone):0-> (0+k)%2-> ((0+k)%2+k)%3->
f[n]= (f[n-1]+k)%n,f[1]=0; F[i] Indicates that there is an I person when the last Victor number

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Note that this idea is applicable to the classical Joseph ring problem.

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The following will be deformed ...

Now the problem is starting with M, which is the first (m-1) number of people going out. And then just like the normal Joseph Ring. So as long as we f[n]= (f[n-1]+m)%n alone counted on the line. Other f[i]= (f[i-1]+k)%i; (1<i<n);

The code is as follows:

1#include <iostream>2#include <cstdio>3#include <cstdlib>4#include <cstring>5 using namespacestd;6 Const intMAXN =10005;7 intF[MAXN];8 intMain ()9 {Ten     intN, K, M, step, CNT; One      while(~SCANF ("%d%d%d", &n, &k, &m)) A     { -         if(!n &&!k &&!m) Break; -f[1] =0; the          for(inti =2; I < n; i++) -         { -F[i] = (f[i-1]+K)%i; -         } +         intAns = (f[n-1]+M)%N; -printf"%d\n", ans+1); +     } A     return 0; at}

"Joseph Ring variant" UVa 1394-and then there is one