hdu 5643 King's Game (dp约瑟夫环)

http://acm.hdu.edu.cn/showproblem.php?pid=5643

King's Game

Problem Description

In order to remember history, King plans to play losephus problem in the parade gap.He calls n(1≤n≤5000) soldiers, counterclockwise in a circle, in label 1,2,3...n.

The first round, the first person with label 1 counts off, and the man who report number 1 is out.

The second round, the next person of the person who is out in the last round counts off, and the man who report number 2 is out.

The third round, the next person of the person who is out in the last round counts off, and the person who report number 3 is out.



The N - 1 round, the next person of the person who is out in the last round counts off, and the person who report number n−1 is out.

And the last man is survivor. Do you know the label of the survivor?

 

Input

The first line contains a number T(0<T≤5000), the number of the testcases.

For each test case, there are only one line, containing one integer n, representing the number of players.

 

Output

Output exactly T lines. For each test case, print the label of the survivor.

 

Sample Input

223

 

Sample Output

22Hint:For test case #1：the man who report number $1$ is the man with label $1$, so the man with label $2$ is survivor.For test case #1：the man who report number $1$ is the man with label $1$, so the man with label 1 is out. Again the the man with label 2 counts $1$,  the man with label $3$ counts $2$, so the man who report number $2$ is the man with label $3$. At last the man with label $2$ is survivor.

 

问题描述：n个人（编号1~n)，从1开始报数，报到m的退出，剩下的人继续从1开始报数。求胜利者的编号。数学解法复杂度：O(n)。

无论是用链表实现还是用数组实现都有一个共同点：要模拟整个游戏过程，不仅程序写起来比 较烦，而且时间复杂度高达O(nm)，当n，m非常大(例如上百万，上千万)的时候，几乎是没有办法在短时间内出结果的。我们注意到原问题仅仅是要求出最 后的胜利者的序号，而不是要读者模拟整个过程。因此如果要追求效率，就要打破常规，实施一点数学策略。

为了讨论方便，先把问题稍微改变一下，并不影响原意：

问题描述：n个人（编号0~(n-1))，从0开始报数，报到(m-1)的退出，剩下的人继续从0开始报数。求胜利者的编号。

5个人

n个人的标号：1 2 3 4 5

对应的下标：  0 1 2 3 4

i=1 第一次报数：  x 0 1 2 3

i=2 第二次报数：  x 2 x 0 1

i=3 第三次报数：  x x x 0 1

i=4 第四次报数：  x x x 0 x

上图就是模拟的过程，那么最后胜利的人num在前一次（i）报数过程中的下标为 （num + i）% （n-i）;

#include <iostream>#include <cstdio>#include <algorithm>#include <queue>#include <cmath>#include <cstring>using namespace std;#define N 5010#define MOD 1000000007#define INF 0x3f3f3f3f#define met(a, b) memset(a, b, sizeof (a))typedef long long LL;int dp[N][2];int main (){    int t, n;    scanf ("%d", &t);    for (int i=2; i<N; i++)    {        for (int j=i-2; j>=0; j--)        {            dp[i][j%2] = (dp[i][(j%2)^1] + j+1) % (i-j);        }    }    while (t--)    {        scanf ("%d", &n);        printf ("%d\n", dp[n][0]+1);    }    return 0;}