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1060: NEW RDSP MODE I
1060: NEW RDSP MODE I
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 33 Solved: 9
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Description
Little A has became fascinated with the game Dota recently, but he is not a good player. In all the modes, the rdsp Mode is popular on online, in this mode, little A always loses games if he gets strange heroes, because, the heroes are 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 are N heroes in the game, and they all have a unique number between 1 and N. At the beginning of game, all heroes will be sorted by the number in ascending order. So, all heroes form a sequence One.
These heroes will be operated by the following stages M times:
1.Get out the heroes in odd position of sequence One to form a new sequence Two;
2.Let the remaining heroes in even position to form a new sequence Three;
3.Add the sequence Two to 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 will be chosen to be Little A's heroes.
The problem for you is to tell little A the numbers of his heroes.
Input
There are 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 two numbers.
The output of one test case occupied exactly one line.
Sample Input
5 2 2
Sample Output
4 3
#include <iostream>#include <vector>using
namespace std;
#define MAX_LEN 26int
main()
{ long
v[MAX_LEN+1];
long
hero, half;
long
long opNum;
int
x, i, loop;
int
remainder;
while
(cin >> hero >> opNum >> x)
{
for
(i = 1; i <= x; i++)
{
v[i] = i;
}
half = hero / 2;
loop = 0;
for
(i = 1; i <= opNum; i++)
{
if
(v[1] <= half)
{
v[1] *= 2;
}
else
{
if
(hero % 2 == 0)//偶数
v[1] = v[1] * 2 - hero - 1;
else
v[1] = v[1] * 2 - hero;
}
if
(1 == v[1])
{
loop = i;
break;
}
}
if
(loop != 0)
{
remainder = opNum % loop;
}
else
remainder = opNum;
for
(i = 1; i <= x; i++)
{
v[i] = i;
}
for
(i = 1; i <= remainder; i++)
{
for
(int
j = 1; j <= x; j++)
{
if
(v[j] <= half)
v[j] *= 2;
else
{
if
(hero % 2 == 0)
v[j] = v[j] * 2 - hero - 1;
else
v[j] = v[j] * 2 - hero;
}
}
}
for
(i = 1; i <= x; i++)
{
if
(i != 1)
cout <<
" ";
cout << v[i];
}
cout << endl;
}
return
0;
}/************************************************************** Problem: 1060
User: 1006440533
Language: C++
Result: Accepted
Time:124 ms
Memory:1272 kb
****************************************************************/最后更新：2017-04-02 15:14:59