Counting Game
There are n people standing in a line, playing a famous game called ``counting". When the game begins, the leftmost person says ``1" loudly, then the second person (people are numbered 1 to n from left to right) says ``2" loudly. This is followed by the 3rd person saying ``3" and the 4th person say ``4", and so on. When the n-th person (i.e. the rightmost person) said ``n" loudly, the next turn goes to his immediate left person (i.e. the (n - 1)-th person), who should say ``n + 1" loudly, then the (n - 2)-th person should say ``n + 2" loudly. After the leftmost person spoke again, the counting goes right again.
There is a catch, though (otherwise, the game would be very boring!): if a person should say a number who is a multiple of 7, or its decimal representation contains the digit 7, he should clap instead! The following tables shows us the counting process for n = 4 (`X‘ represents a clap). When the 3rd person claps for the 4th time, he‘s actually counting 35.
Person 1 2 3 4 3 2 1 2 3
Action 1 2 3 4 5 6 X 8 9
Person 4 3 2 1 2 3 4 3 2
Action 10 11 12 13 X 15 16 X 18
Person 1 2 3 4 3 2 1 2 3
Action 19 20 X 22 23 24 25 26 X
Person 4 3 2 1 2 3 4 3 2
Action X 29 30 31 32 33 34 X 36
Given n, m and k, your task is to find out, when the m-th person claps for the k-th time, what is the actual number being counted.
Input
There will be at most 10 test cases in the input. Each test case contains three integers n, m and k ( 2$ \le$n$ \le$100, 1$ \le$m$ \le$n, 1$ \le$k$ \le$100) in a single line. The last test case is followed by a line with n = m = k = 0, which should not be processed.
Output
For each line, print the actual number being counted, when the m-th person claps for the k-th time. If this can never happen, print ` -1‘.
Sample Input
4 3 1
4 3 2
4 3 3
4 3 4
0 0 0
Sample Output
17
21
27
35

就是不知道-1的情况，感觉一直报数下去肯定就能喊到啊。

程序如下，还是没过

#include<stdio.h>
main()
{int i;
int a,n,m,k,t;
while(scanf("%d %d %d",&n,&m,&k)!=EOF)
{i=0;
int x[108]={0};
if(n==0&m==0&k==0)
break;
while(1)
{
i++;
if(i%7==0||i%10==7||i/10%10==7||i/100%10==7)
{
if(i<n)
x[i]++;
else
{t=(i-n)%(2*n-2);
if(t<n)
x[n-t]++;
else
{
t=t-n+1;
x[t+1]++;
}

}
}if(x[m]==k)
break;
if(i>1000000)
break;
}if(i>1000000)
printf("-1\n");
else
printf("%d\n",i);
}
}