题目描述
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.
输入
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.
输出
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.
样例输入
<span style="color:#333333">5 1 2
5 2 2</span>样例输出
<span style="color:#333333">2 4
4 3</span>提示
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 One is 2,4,1,3,5. After the second operation, the sequence One is 4,3,2,1,5.So,output 4 3.
题目意思是：输入n，m，x，刚开始有一个1……n的排列，然后定义了一种操作，是将数组中的奇数位的数字选出来，按照顺序放到数组最后面，偶数位按照顺序放到奇数位的后面，进行m次这样的操作之后，输出前x个数字。
分析：找到循环节T，利用T去约m，然后再将很小的m拿去模拟，输出前x个.
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
const int mx = 1e6 + 10;
const int inf = 0x3f3f3f3f;
int num[mx];
int n,m,x;
int find_t() { //找循环周期T
int cnt=0,cur=1;
do{
if(cur*2<=n)
cur*=2;
else
cur=(cur-n/2)*2-1;
cnt++;
} while(cur!=1);
return cnt;
}
int main() {
while(~scanf("%d%d%d",&n,&m,&x)) {
for(int i=1; i<=n; i++)
num[i]=i;
int T=find_t();
m%=T;
for(int i=1; i<=x; i++) {
if(i!=1)
printf(" ");
for(int j=1; j<=m; j++) {
if(num[i]*2<=n)
num[i]*=2;
else
num[i]=(num[i]-n/2)*2-1;
}
printf("%d",num[i]);
}
printf("\n");
}
return 0;
}