This is a explanation of this problem from USACO's training website. I have converted it to markdown. Please do not just copy code; you will not learn anything; at least type it out and understand so you can do it yourself in the future!

Runaround numbers are integers with unique digits, none of which is zero (e.g., 81362) that also have an interesting property, exemplified by this demonstration:

• If you start at the left digit (8 in our number) and count that number of digits to the right (wrapping back to the first digit when no digits on the right are available), you’ll end up at a new digit (a number which does not end up at a new digit is not a Runaround Number). Consider: 8 1 3 6 2 which cycles through eight digits: 1 3 6 2 8 1 3 6 so the next digit is 6.
• Repeat this cycle (this time for the six counts designed by the ‘6’) and you should end on a new digit: 2 8 1 3 6 2, namely 2.
• Repeat again (two digits this time): 8 1
• Continue again (one digit this time): 3
• One more time: 6 2 8 and you have ended up back where you started, after touching each digit once. If you don’t end up back where you started after touching each digit once, your number is not a Runaround number.

Given a number M (that has anywhere from 1 through 9 digits), find and print the next runaround number higher than M, which will always fit into an unsigned long integer for the given test data.

### INPUT FORMAT

A single line with a single integer, M

### SAMPLE INPUT (file runround.in)

``````81361
``````

### OUTPUT FORMAT

A single line containing the next runaround number higher than the input value, M.

### SAMPLE OUTPUT (file runround.out)

``````81362
``````

Java

C++

Pascal

## ANALYSIS

Russ Cox

The trick to this problem is noticing that since runaround numbers must have unique digits, they must be at most 9 digits long. There are only 9! = 362880 nine-digit numbers with unique digits, so there are fewer than 9*362880 numbers with up to 9 unique digits. Further, they are easy to generate, so we generate all of them in increasing order, test to see if they are runaround, and then take the first one bigger than our input.

``````#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

int m;
FILE *fout;

/* check if s is a runaround number;  mark where we've been by writing 'X' */
int
isrunaround(char *s)
{
int oi, i, j, len;
char t;

strcpy(t, s);
len = strlen(t);

i=0;
while(t[i] != 'X') {
oi = i;
i = (i + t[i]-'0') % len;
t[oi] = 'X';
}

/* if the string is all X's and we ended at 0, it's a runaround */
if(i != 0)
return 0;

for(j=0; j<len; j++)
if(t[j] != 'X')
return 0;

return 1;
}

/*
* create an md-digit number in the string s.
* the used array keeps track of which digits are already taken.
* s already has nd digits.
*/
void
permutation(char *s, int *used, int nd, int md)
{
int i;

if(nd == md) {
s[nd] = '\0';
if(atoi(s) > m && isrunaround(s)) {
fprintf(fout, "%s\n", s);
exit(0);
}
return;
}

for(i=1; i<=9; i++) {
if(!used[i]) {
s[nd] = i+'0';
used[i] = 1;
permutation(s, used, nd+1, md);
used[i] = 0;
}
}
}
``````