Nbut 1225 NEW RDSP MODE I (law + fast Power)

• [1225] NEW RDSP MODE I
• Time limit: Ms Memory limit: 131072 K
• Problem description

• Little A has became fascinated and the game Dota recently, but he's not A good player. In all the modes, the RDSP mode was popular on online, in this mode, little A always loses games if he gets strange heroes, Because, the heroes is distributed randomly.

  Little A wants to win the game, so he cracks the code of the RDSP mode with his talent on programming. The following description is about the RDSP mode:

  There is N heroes in the game, and they all has a unique number between 1 and N. At the beginning of the game, all heroes'll be is sorted by the number in ascending order. So, all heroes form a sequence one.

  These heroes'll be operated by the following stages M times:

  1.Get out the heroes in odd position of sequence one to form a new sequence;

  2.Let The remaining heroes in even position to form a new sequence three;

  3.ADD the sequence from the back of sequence three to form a new sequence one.

  After M Times ' operation, the X heroes in the front of new sequence one would be chosen to be Little A ' s heroes. The problem for little A, the numbers of his heroes.

• Input
• There is several test cases.
  Each case contains three integers N (1<=n<1,000,000), M (1<=m<100,000,000), X (1<=x<=20).
  Proceed to the end of file.
• Output
• For each test case, output X integers indicate the number of heroes. There is a space between the numbers. The output of one test case occupied exactly one line.
• Sample input

• 5 1 25) 2 2

• Sample output

• 2 44 3

• Tips

• In case two:n=5,m=2,x=2,the initial sequence one is 1,2,3,4,5.after the first operation, the sequence Oneis 2,4,1,3,5. After the second operation, the sequence one is 4,3,2,1,5.so,output 4 3.

• Source

• Liaoning Province Race 2010

Test instructions: 1~n number of arrays A, an array of odd bits into an array B, an even number of bits to form an array c, and then connect B to C, to execute m times, the output of the last number of the first X-digits.

Analysis: This problem can be understood as a number in the final position after M-transformation, then the first from the two-time transformation of the relationship began to deduce

Suppose the position of this time is X, divided into even and odd two cases:

If x is an even number, then x's next coordinate will be X ' =X/2, which can be understood as half of the odd number in front is taken away, change the x=2*x '

If x is an odd number, then x's next coordinate will be X ' =N/2+X/2, which can be interpreted as having N/2 in front, and then x/2 an odd number in front of X, transforming X=2*x '-N

Can be found, whether it is odd or even, is x=2*x after the n modulo, then the M transformation can be regarded as 2 m power, here write a fast power on the line.

Also note that when N is an even number to +1, because N is even, the subscript may be 0, when n is an odd number, the nth position is unchanged in the transformation, so the situation and N-1.

#pragmaComment (linker, "/stack:1024000000,1024000000")#include<cstdio>#include<string>#include<iostream>#include<cstring>#include<cmath>#include<stack>#include<queue>#include<vector>#include<map>#include<stdlib.h>#include<algorithm>#defineLL __int64#defineFIN freopen ("In.txt", "R", stdin)using namespacestd; LL n,m,x; LL ans,tmp; ll Pow (ll x,ll n,ll MOD) {ll res=1;  while(n) {if(n&1) res= (x*res)%MOD; X= (x*x)%MOD; N>>=1; }    returnRes;}intMain () { while(SCANF ("%i64d%i64d%i64d", &n,&m,&x)! =EOF) {        if(N%2==0) n++; TMP=pow (2, M,n); Ans=tmp; printf ("%i64d", TMP);  for(intI=2; i<=x;i++) {ans+=tmp; Ans%=N; printf ("%i64d", ans); } printf ("\ n"); }    return 0;}

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Nbut 1225 NEW RDSP MODE I (law + fast Power)