wustoj 1093 NEW RDSP MODE I

1093: NEW RDSP MODE I

Time Limit: 1 Sec   Memory Limit: 128 MB
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Problem 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 1 2
5 2 2

Sample Output

2 4
4 3

HINT

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.

Source

辽宁省赛2010

思路:

n、m都挺大，模拟或者对n个元素直接操作肯定会TLE。

这题的突破口在于：x的范围很小。

那么预处理每个位置的上一个位置，对前x个点处理就够了，从后往前推，每次找到它的前一个位置，当然位置也是周期性出现的，而且循环节不超过n，这样复杂度就降到O(x*n)了，可以过了。

ps:每个位置的前一个位置其实可以推公式的，我是算出来的，时间要的多一点。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
//#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1000005
#define MAXN 100005
#define mod 1000000007
#define INF 0x3f3f3f3f
using namespace std;

int n,m,x,k,ans,cnt;
int pre[maxn],tt[maxn],pos[maxn];

void presolve()
{
    int i,j,t,cur,now,st,step;
    memset(pre,-1,sizeof(pre));
    cnt=0;
    k=n>>1;
    for(i=0;i<n;i++)
    {
        if(pre[i]!=-1) continue ;
        st=now=i;
        step=0;
        cnt++;
        while(1)
        {
            pos[st]=cnt;
            cur=st;
            step++;
            if(st&1) st=st>>1;
            else st=(st>>1)+k;
            pre[st]=cur;
            if(st==now) break ;
        }
        tt[cnt]=step;
    }
}
void solve()
{
    int i,j,s,st;
    for(i=0;i<x;i++)
    {
        s=m%tt[pos[i]];
        st=i;
        for(j=0;j<s;j++)
        {
            st=pre[st];
        }
        if(i==0) printf("%d",st+1);
        else printf(" %d",st+1);
    }
    printf("\n");
}
int main()
{
    int i,j;
    while(~scanf("%d%d%d",&n,&m,&x))
    {
        presolve();
        solve();
    }
    return 0;
}