Brush over a NOIP2013 circle game

There are n small partners (numbered from 0 to n-1) to sit around and play games. The n positions are numbered in a clockwise direction, from 0 to n-1. Initially, the No. 0 partner at position No. 0, the 1th small partner in the 1th position, ..., and so on.
The rules of the game are as follows: each round of the No. 0 position of the small partner clockwise to the location of position m, the 1th position of the small partner to the location of position m+1, ..., and so on, the location of the small partner in position n−m to the No. 0 position, the n-m+1 position of the small partners to the 1th position, ... The small partner in position n-1 clockwise to position m-1.
Now, altogether has carried on the 10^k round, ask the X small partner finally reached the number of the first place.

Input:

Enter a total of 1 rows, containing 4 integers n, m, K, X, separated by a space between each of the two integers.

Output:

Outputs a total of 1 rows, containing 1 integers representing the position number of the X-10^k after the wheel.

Input Example:

10 3 4 5

Output Example:

5

Data range: 0<n<10^6,0<m<n,0<x<=n,0<k<10^9

Were you extremely surprised when you saw 0<k<10^9? What the! 10 of 100 million times!!!! Then you can only use the fast power ...

1#include <iostream>2 #definell Long Long3 using namespacestd;4 ll N,m,x,k;5ll Fast_mi (Long LongXLong LongY//Fast Power6 {7ll temp=1;8      while(y)9     {Ten         if(y&1) temp=temp*x%N; Onex=x*x%N; Ay>>=1; -     } -     returntemp%N; the } - intMain () - { -scanf"%lld%lld%lld%lld",&n,&m,&k,&x); +ll Ans=fast_mi (Ten, k)%N; -ans=ans*m%N; +ans= (ans+x)%N; Aprintf"%lld", ans); at     //System ("Pause>nul"); -     return 0; -}

C + + answer

Brush over a NOIP2013 circle game