ACdream 1020 - The Game about KILL（约瑟夫环）

最新推荐文章于 2024-04-07 10:16:59 发布

原创最新推荐文章于 2024-04-07 10:16:59 发布 · 321 阅读

0 ·

CC 4.0 BY-SA版权

杂同时被 2 个专栏收录

80 篇文章

订阅专栏

ACdream

10 篇文章

订阅专栏

本文探讨了约瑟夫环问题的一种特殊情形，即在指定条件下确定最后一个存活者的位置。通过寻找规律，给出了一种有效的算法实现，并附带源代码进行说明。

The Game about KILL

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

Submit Statistic Next Problem

Problem Description

Teacher HU and his 40 students were trapped by the brigands. To show their power, the head of the brigands want to select one people to kill.

Teacher HU and his 40 students will stand in a circle, and every second person would leave, and the last people in the circle would be killed.

For example, if there are 5 persons in the circle, counting proceeds as 2, 4, 1, 5 and person 3 will be killed.

To make his students alive, teacher HU calculated the position to be the last man standing, and sacrifice himself.

Now we consider a more common condition, if teacher HU has N - 1 students, which place should he stand to be the last person.

Input

There are multiple test cases.
Each test case only contains an integer N. (1 ≤ N ≤ 1,000,000,000)

Output

For each test case, output an integer indicating which place should teacher HU stand.

Sample Input

2
3

Sample Output

1
3

Source

dut200901102

Manager

nanae

Submit

题意：

就是约瑟夫环。

找个规律就行。。

#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <stdio.h>
using namespace std;
const int maxn = 100000+43;
const int inf = 0x3f3f3f3f;
#define LL long long

LL sum[maxn];

int main()
{
    LL n;
    int i;
    LL er=1;
    for(i=1;;i++){
        sum[i]=sum[i-1]+er;
        if(sum[i]>1000000000)
            break;
        er=2*er;
    }
    while(~scanf("%lld",&n)){
    	int now=lower_bound(sum,sum+1+i,n)-sum;
    //	printf("%d\n",now);
    	printf("%lld\n",-1+2*(n-sum[now-1]));
    }

    return 0;
}