HDU 5643 (神奇的打表 线段树)

King's Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 263    Accepted Submission(s): 154

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

2
2
3

 

Sample Output

2
2

Hint:
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.

 

题意：n个人排成一圈。从1开始，第i次次从前一次死掉的人往后数i个人并将它杀死，求最后的数字。

这个题目做过类似的有线段树的解法O(nlgn)但是会超时但是n比较小，于是就打个表。

看了题解的递推解法感觉自己好蠢。

打表代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
#define maxn 5111
#define lson tree[c].l,tree[c].mid,c<<1
#define rson tree[c].mid+1,tree[c].r,(c<<1)|1
#define pl c<<1
#define pr (c<<1)|1

int n;
struct node {
    int l, r, mid, sum, num;
}tree[maxn<<4];

void build_tree (int l, int r, int c) {
    tree[c].l = l, tree[c].r = r, tree[c].mid = (l+r)>>1;
    if (l == r) {
        tree[c].num = 1;
        tree[c].sum = 1;
        return ;
    }
    build_tree (lson);
    build_tree (rson);
    tree[c].sum = tree[pl].sum + tree[pr].sum;
}

int query (int l, int r, int c, int x, int y) {
    if (l == x && r == y) {
        return tree[c].sum;
    }
    if (tree[c].mid >= y) {
        return query (lson, x, y);
    }
    else if (tree[c].mid < x) {
        return query (rson, x, y);
    }
    else {
        return query (lson, x, tree[c].mid) + query (rson, tree[c].mid+1, y);
    }
}

int update (int l, int r, int c, int id) {//从前往后将第id个数杀死 返回这个数
    if (l == r) {
        tree[c].num = tree[c].sum = 0;
        return l;
    }
    int ans;
    if (tree[pl].sum >= id) {
        ans = update (lson, id);
    }
    else
        ans = update (rson, id-tree[pl].sum);
    tree[c].sum = tree[pl].sum + tree[pr].sum;
    return ans;
}

int update2 (int l, int r, int c, int id) {//从后往前
    if (l == r) {
        tree[c].num = tree[c].sum = 0;
        return l;
    }
    int ans;
    if (tree[pr].sum > id)
        ans = update2 (rson, id);
    else
        ans = update2 (lson, id-tree[pr].sum);
    tree[c].sum = tree[pl].sum + tree[pr].sum;
    return ans;
}

int main () {
    //freopen ("in.txt", "w", stdout);
    for (n = 1; n <= 5000; n++) {
        build_tree (1, n, 1);
        int pos = 1;
        for (int i = 1; ; i++) {
            if (pos > n)
                pos %= n;
            int id = i;
            int Max = tree[1].sum;
            if (Max == 1) {
                printf ("%d, ", update (1, n, 1, 1));
                if (n%10 == 0)
                    cout << endl;
                break;
            }
            if (id > Max)
                id %= Max;
            if (id == 0)
                id = Max;
            int a1 = query (1, n, 1, pos, n); 
            if (a1 < id) {
                id -= a1;
                pos = update (1, n, 1, id); 
                pos++;
            }
            else {
                pos = update2 (1, n, 1, a1-id); 
                pos++;
            }
        }
    }
    return 0;
}