Counting Game (Uva)

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 (

2n100,

1mn,

1k100) 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

---

The Seventh Hunan Collegiate Programming Contest
Problemsetter: Rujia Liu, Special Thanks: Yiming Li & Jane Alam Jan

 

 

#include<stdio.h>
#include<string.h>
bool aa[1000000];
bool solve(int n)
{
    if(n%7==0)  return true;
    int t=n;
    while(t)
    {
        if(t%10==7)  return true;
        t/=10;
    }    
    return false;
}  
void calc()
{
    for(int i=1;i<1000000;i++)
    {
        if(solve(i))  aa[i]=true;
        else  aa[i]=false;
    }    
    
}      
int main()
{
    int n,m,k;
    int t;
    int cnt;
    calc();
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        if(n==0&&m==0&&k==0) break;
        t=m;
        cnt=0;
        while(1)
        {
            if(aa[t])cnt++;
            if(cnt==k)
            {
                printf("%d\n",t);
                break;
            }  
            if(n==m)
            {
                t+=2*(m-1); 
                continue;
            }  
            if(m==1)
            {
                t+=2*(n-m);
                continue;
            }      
            t+=2*(n-m);
            if(aa[t])cnt++;
            if(cnt==k)
            {
                printf("%d\n",t);
                break;
            }  
               
            t+=2*(m-1);  
        }    
    } 
    return 0;   
}