The Java implementation of the Joseph question (drawing on someone else's code + your own analysis of the understanding)

Http://www.blogjava.net/rorely/archive/2010/01/15/309732.html

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ImportJava.util.Scanner;/*** Use an array to implement the Joseph ring problem * The circle count is enclosed by a m-person. * Starting from the first person, count off from 1, check in n people out of the circle, * the remainder continues to count off from 1 until all the people are out of the loop. * For a given m and N, find out the order of all the rings.*/ Public classringyuesefu{ Public Static voidMain (string[] args) {System.out.println ("The procedure is described as follows:"); System.out.println ("The M-person is surrounded by a circle that is connected to the end." Starting from the first person, starting from 1, the number of people reporting n out of the circle, the rest of the people continue to count from 1, until all the people out of the circle so far. For a given M and N, find out the order in which everyone is out of the loop.); //Prompt to enter total number of peopleSystem.out.println ("Please enter the total number of people doing this game:"); Scanner SCA=NewScanner (system.in); intm=Sca.nextint (); //prompt to enter a value to be circledSystem.out.println ("Please enter a value to be out of the loop:"); intn=Sca.nextint (); System.out.println ("Output sequence number in the Order of the Rings:"); //create an array with M values        int[] A=New int[M]; //The initial length, after the ring one, the length is reduced by one        intlen=m; //Assigning a value to an array         for(inti=0;i<a.length;i++) A[i]=i+1; //I is the element below table, J represents the number of current to be reported        intI=0; intJ=1; //start the game, as long as someone in the circle to continue the game, judged as len>0, each out of a person len--until the end of 0 games. //but the length of the array does not change, the value of the array element is changed, the value of the loop is-1. All of the last array values are-1, which means everyone is out of the loop.         while(len>0){            //The game has two conditions, the first to ensure that the position of someone, that is, a[] is not 1, the reason is not 1 because the back to the position of the circle into 1//The reason for using a[i%m] is to prevent the array angle label out of bounds, such as M=10,i is from 0 to 9 this is a lap, and then i++ 10, 10%10=0,//The IF is not executed, but else I++,i is 11%10>0 to continue. Prevent a[10] Cross-border situations            if(a[i%m]>0){                if(j%n==0) {//find the person who wants to go out and reduce the number of people in the circle by oneSystem.out.print (a[i%m]+ ""); A[i%m]=-1; J= 1;//someone's out of the loop, starting from 1 .i++; Len--; }Else{//no one out of the circle, i++, Position backward Move, J + + slogan plus one, such as the report is 1, the next 2,3,4 until Ni++; J++; }            }Else{//when the position is 1, it jumps to the next i++, but J does not add one .i++;        }} System.out.println ();     System.out.println (i); //the end of the loop, the game is over    }}

The Java implementation of the Joseph question (drawing on someone else's code + your own analysis of the understanding)