# LA 3882 and then there is one

Source: Internet
Author: User

Problem-Solving ideas: analysis to a long time, too lazy analysis, affixed to the analysis of a Daniel, the code is my own writing.

n the number of rows in a circle, the first time to delete m, after each k number deleted once, the last one to be deleted.

If the problem is to simulate the whole process with a linked list or an array, the time complexity will be up to O (NK), and the n<=10000,k<=10000 will be directly tle.

So is there any other way? The answer is yes.

We ignore the M first, analyze the number of each k deleted once, that is the classic Joseph problem.

Then, each number (1~n) is numbered sequentially 0~n-1

The number of the first deleted number is x, then x= K%n-1 (note is number, really deleted number is number + 1)

Then the remaining number of n-1 can form a new Joseph ring.

What is the number now? Obviously: (Make x+1=y, that is, y= k%n)

Y, y+1, y+2 ... n-1, 0, 1 ... y-2

Putting y in the first order is the next time you start counting from it.

Re-start number of k numbers.

You said re-, uh. Then you can renumber it like this:

Y-0

Y+1->1

Y+2->2

...

...

Y-2-N-2

Now it becomes the n-1 number (numbered from 0~n-2) of the Joseph Question!

Assuming that Z is the number left by the last n-1 number, then Z ' is the number left by the n person, then the z ' = (z+y)% n is clearly

How do you know the n-1 of a solution? Just give it a hand, you know n-2.

So, there are:

ans [1]=0;

Ans [n] = (ans[n-1]+k)%n;

(one might ask: Is it not z ' = (z+y)% n?) Now how did you become K? Because y= k%n, modulo operation)

Then, the answer is +1 (number and number)

So what about the first time this question is M?

Also very simple, we move k every time, there are n number, then the answer is Ans[n]

But the first move is M, so the trailing movement has a constant gap (K-M)

So the answer is: (ans[n]– (k–m))% n (note may be less than 0 and final answer +1)

`1#include <cstdio>2 intMain ()3 {4     intN, K, M, a[10005];5      while(~SCANF ("%d%d%d", &n, &k, &m) && (n | | m | |k))6     {7a[1] =0;8          for(inti =2; I <= N; i++) A[i] = (a[i-1]+K)%i;9         intA = (M-k +1+ a[n])%N;Ten         if(A <=0) A + = n;//note may be less than 0 Oneprintf"%d\n", a); A     } -     return 0; -}`
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LA 3882 and then there is one

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