Algorithm for solving Joseph ring problem C language source code

Joseph algorithm: n Individuals in a circle, each has a different number, select a person as the starting point, and then clockwise from 1 to K number, every number to K of the people out of the circle, circle narrowing, and then continue from the next person from 1 to K number, repeat the above process. The original number of the person who asked for the last launch circle.

Idea: Follow the above algorithm to let people out of the circle, until there are n-1 individuals to roll out the circle, and then get the last one out of the circle of the number of people.

Program: The number of people sitting in a circle does not need to be ordered

#define N 100
int yuesefu1 (int data[],int sum,int k)
{
int i=0,j=0,count=0;
while (count<sum-1)
{
if (data[i]!=0)/* Current person in the circle * *
j + +;
if (j==k)/* If the person should exit the circle * *
{
Data[i]=0;/*0 said not in the circle.
count++;/* exit number plus 1*/
j=0;/* the number of new numbers * *
}
i++;/* judge the next person
if (i==sum)/* Surround in a circle * *
i=0;
}
for (i=0;i<sum;i++)
if (data[i]!=0)
Return data[i];/* returns the last person's number * *
}

void Main ()
{
int data[n];
int i,j,total,k;
printf ("\nplease input the number of every people.\n");
for (i=0;i<n;)/* For people in the circle to arrange the number * *
{
int input;
scanf ("%d", &input);
if (input==0)
Break;/*0 means end of input/
for (j=0;j<i;j++)/* Check number is duplicate * *
if (data[j]==input)
Break
if (j>=i&&input>0)/* No duplicates, record number, continue to enter * *
{
Data[i]=input;
i++;
}
Else
printf ("\ndata error.") Re-input: ");
}
Total=i;
printf ("\nyou have input:\n");
for (i=0;i<total;i++)
{
if (i%10==0)
printf ("\ n");
printf ("%4d", Data[i]);
}
printf ("\nplease input a number to count:");
scanf ("%d", &k);
printf ("\nthe last one ' s number is%d", YUESEFU1 (data,total,k));
}