HDU5860-Death Sequence

Death Sequence

                                                                     Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                                 Total Submission(s): 848    Accepted Submission(s): 368

Problem Description

You may heard of the Joseph Problem, the story comes from a Jewish historian living in 1st century. He and his 40 comrade soldiers were trapped in a cave, the exit of which was blocked by Romans. They chose suicide over capture and decided that they would form a circle and start killing themselves using a step of three. Josephus states that by luck or maybe by the hand of God, he and another man remained the last and gave up to the Romans.

Now the problem is much easier: we have N men stand in a line and labeled from 1 to N, for each round, we choose the first man, the k+1-th one, the 2*k+1-th one and so on, until the end of the line. These poor guys will be kicked out of the line and we will execute them immediately (may be head chop, or just shoot them, whatever), and then we start the next round with the remaining guys. The little difference between the Romans and us is, in our version of story, NO ONE SURVIVES. Your goal is to find out the death sequence of the man.

For example, we have N = 7 prisoners, and we decided to kill every k=2 people in the line. At the beginning, the line looks like this:

1 2 3 4 5 6 7

after the first round, 1 3 5 7 will be executed, we have

2 4 6

and then, we will kill 2 6 in the second round. At last 4 will be executed. So, you need to output 1 3 5 7 2 6 4. Easy, right?

But the output maybe too large, we will give you Q queries, each one contains a number m, you need to tell me the m-th number in the death sequence.

 

Input

Multiple cases. The first line contains a number T, means the number of test case. For every case, there will be three integers N (1<=N<=3000000), K(1<=K), and Q(1<=Q<=1000000), which indicate the number of prisoners, the step length of killing, and the number of query. Next Q lines, each line contains one number m(1<=m<=n).

 

Output

For each query m, output the m-th number in the death sequence.

 

Sample Input

  

   1
7 2 7
1
2
3
4
5
6
7
  

 

Sample Output

  

   1
3
5
7
2
6
4
  

 

Author

BUPT

 

Source

2016 Multi-University Training Contest 10

 

Recommend

wange2014

 

题意：n个人排成一行，从第一个人开始，每k个人死一个，剩下的人重新排成一行再报数。一共q个询问，每次询问第qi个死的人是谁

解题思路： 线段树维护区间还有多少人即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n, k, q;
int x[3000000 << 2];
int ans[3000000],a,ans1;

void build(int k,int l, int r)
{
	if (l == r) { x[k] = 1; return; }
	int mid = (l + r) >> 1;
	build(k << 1, l, mid);
	build(k << 1 | 1, mid + 1, r);
	x[k] = x[k << 1] + x[k << 1 | 1];
}

void query(int k, int l, int r,int sum)
{
	if (l == r) { x[k] = 0; ans1 = l; return; }
	int mid = (l + r) >> 1;
	if (x[k << 1] >= sum) query(k << 1, l, mid, sum);
	else query(k << 1 | 1, mid + 1, r, sum - x[k << 1]);
	x[k] = x[k << 1] + x[k << 1 | 1];
}

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		memset(x, 0, sizeof x);
		scanf("%d %d %d", &n, &k, &q);
		build(1,1, n);
		int nn = n,cnt = 1;
		while (nn)
		{
			int kk = (nn - 1) / k;
			for (int i = 0; i <= kk; i++)
			{
				int sum = 1 + i*k - i;
				query(1,1, n, sum);
				ans[cnt++] = ans1;
				nn--;
			}
		}
		for (int i = 1; i <= q; i++)
		{
			scanf("%d", &a);
			printf("%d\n", ans[a]);
		}
	}
	return 0;
}