Power Crisis

最新推荐文章于 2020-10-01 19:45:02 发布

雁阵惊寒声

最新推荐文章于 2020-10-01 19:45:02 发布

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本文链接：https://blog.youkuaiyun.com/u010760034/article/details/16281781

Power Crisis

Time Limit : 3000/1000ms (Java/Other) Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 2 Accepted Submission(s) : 2

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Problem Description

During the power crisis in New Zealand this winter (caused by a shortage of rain and hence low levels in the hydro dams), a contingency scheme was developed to turn off the power to areas of the country in a systematic,
totally fair, manner. The country was divided up into N regions (Auckland was region number 1, and Wellington number 13). A number, m, would be picked `at random', and the power would first be turned off in region 1
(clearly the fairest starting point) and then in every m'th region after that, wrapping around to 1 after N, and ignoring regions already turned off. For example, if N = 17 and m = 5, power would be turned off to the
regions in the order:1,6,11,16,5,12,2,9,17,10,4,15,14,3,8,13,7. The problem is that it is clearly fairest to turn off Wellington last (after all, that is where the Electricity headquarters are), so for a given
N, the `random' number m needs to be carefully chosen so that region 13 is the last region selected.
Write a program that will read in the number of regions and then determine
the smallest number m that will ensure that Wellington (region 13) can function while the rest of the country is blacked out.

Input

Input will consist of a series of lines, each line containing the number of regions (N) with . The file will be terminated by a line consisting of a single 0.

Output

Output will consist of a series of lines, one for each line of the input. Each line will consist of the number m according to the above scheme.

Sample Input

17
0

Sample Output

/*约瑟夫环问题 去掉首个就是查找的11*/

#include <iostream>
#include <stdio.h>
using namespace std;
int n;
int work(int m)
{
    int a=0;
    for(int j=2;j<n;j++)
        a=(a+m)%j;
        return a;
}
int main ()
{ int t,i;
    while(scanf("%d",&n)!=EOF&&n)
    {
        for( i=0;(t=work(i))!=11;i++);
           printf("%d\n",i);
    }
    return 0;
}