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

170

Sample Output

7

/*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;}