card

Time limit: 1 Sec Memory Limit: 128 MB

Title Description

You have a stack of numbered 1 to n card.
You have an operation, these cards can be rearranged, as follows:
1. the card into a front half and a second half.
2. sequentially depicting a card from the latter half, the front half portion, into a new sequence.
For example, the card sequence (1,2,3,4,5,6) for the result of the operation (4,1,5,2,6,3).
Now you have an initially (1,2,3, ⋯, n) of the sequence of cards, you need to determine the reference card on the x-th position after operation m times.

Entry

The first line contains three non-negative integer n, m, x.

Export

Output line a number that represents the answer.

Sample input

6 2 3

Sample Output

prompt

For 60% of the data, m≤10 ^{. 7} .
To 100% of the data, 0 ≦ n, m, x≤10 ^{. 9} .
Graded data, to ensure that n is even.

answer:

Suppose the original position x (also values), the current position p;

Because once transformed, or changed positions 2x, or becomes 2x- (n + 1), it is the equation (2 ^ m) * x + (n + 1) * y = p;

Since we only required the x, so the equation can be written as ((2 ^ m)% (n + 1)) * x + (n + 1) * y = p;

Simple to understand what extend_gcd during the execution of the function you can get the above conclusions.

AC Code:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 ll qpow(ll a,ll b,ll mod)
 5 {
 6     ll res=1;
 7     while(b){
 8         if(b&1) res=res*a%mod;
 9         a=a*a%mod;
10         b>>=1;
11     }
12     return res;
13 }
14 ll extend_gcd(ll a,ll b,ll &x,ll &y)
15 {
16     if(b==0){
17         x=1;y=0;
18         return a;
19     }
20     ll r=extend_gcd(b,a%b,y,x);
21     y-=x*(a/b);
22     return r;
23 }
24 int main()
25 {
26     ll n,m,p,a,b,x,y;
27     scanf("%lld %lld %lld",&n,&m,&p);
28     a=qpow(2,m,n+1);b=n+1;
29     ll gcd=extend_gcd(a,b,x,y);
30     b/=gcd;p/=gcd;
31     x=x%b*p%b;
32     if(x<0) x+=b;
33     printf("%lld\n",x);
34     return 0;
35 }

View Code