Uva 12289 - Counting Game

你为何这么叼！

Counting Game 

Time limit: 1.000 seconds

 

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 <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n,m,k;
int ok(int x){
    if(x%7==0)return 1;
    while(x){
        if((x%10)==7)return 1;
        x/=10;
    }
    return 0;
}
int main(){
    while(scanf("%d%d%d",&n,&m,&k)&&(n+m+k)){
        int k1=0,i=0,j;
        for(j=m;i<k;k1++){
            if(ok(j)){i++;if(i>=k)break;}
            if(((k1&1)||n==m)&&m!=1)j=j+2*m-2;
            else j=j+2*n-2*m;
        }
        printf("%d\n",j);
    }
    return 0;
}