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



分析：報數問題，凡是7的倍數或者含7的數就不喊出來。n我的，當第m我的喊這類數p次後輸出他們一共喊了多少次。
     n+n-2表明循環報數是一個循環有這麼屢次  如12345432 少了一個5和一個1  換爲n也是如此比較

代碼:

#include<stdio.h>
#include<string.h>
#define N 1000001
int pu(int x)
{
    if(x%7==0)
      return 1;
    while(x)
    {
        if(x%10==7)
          return 1;
        x/=10;
    }
    return 0;
}
int main()
{
    int n,m,k,i,j,sum,count[10001];
    while(~scanf("%d %d %d",&n,&m,&k))
    {
        sum=0;
        if(n==0&&m==0&&k==0)
           break;
        for(i=1;i<=n;i++)
           count[i]=i;
        j=n-1;
        for(i=n+1;i<n+n-2;i++)
        {
            count[i]=j;
            j--;
        }
        for(i=1;i<N;i++)
        {
            if(pu(i)==1)
            {
                if(i%(n+n-2)==0)
                {
                    if(count[n+n-2]==m)
                       sum++; 
                }
                else
                {
                    if(count[i%(n+n-2)]==m)
                       sum++;
                }
            }
            if(sum==k)
            {
                printf("%d\n",i);
                break;
            }
        }
        if(i==N)
           printf("-1\n");
    }
    return 0;
}

可參考的代碼：

#include<set>
#include<map>
#include<stack>
#include<queue>
#include<ctime>
#include<cmath>
#include<vector>
#include<string>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n, m, t;
bool seven(int k){
    if (!(k % 7))return 1;
    while (k){
        if ((k % 10) == 7)return 1;
        k /= 10;
    }
    return 0;
}
void solve(){
    int x = m;
    int cnt=0;
    int step = 0;
    int k1 = 2 * (n - m);//若是向右走回到本身 步數是2*(n-m)
    int k2 = 2 * (m - 1);//向左走回到本身是2*（m-1)
    if (seven(x))cnt++;
    if (t == cnt){ printf("%d\n", x); return;}
    while (x <= 100000000){
        step++;
        if (step % 2 && k1 == 0)continue;
        if (!(step % 2) && k2 == 0)continue;
        if (step % 2)x += k1;
        else x += k2;
        if (seven(x))cnt++;
        if (cnt == t){ printf("%d\n", x); return; }
    }
}
int main(){
    while (1){
        scanf("%d%d%d", &n, &m, &t);
        if (!n)break;
        solve();
    }
    return 0;
}