ACM-simulation-nyoj 559-Report game-Hunan seventh provincial Competition

Nanyang OJ subject address: Portal

Game time limit: 1000 MS | memory limit: 65535 kb difficulty: 2

Description

N people stand in one row to play a report game. The numbers of all users are 1 to n from left to right. At the beginning of the game, the leftmost people reported 1, the people on the right reported 2, The people numbered 3 reported 3, and so on. After N is reported, the person on the left (I .e., the person on the rightmost) Reports n + 1, then the person numbered N-2 reported n + 2, and so on. When the leftmost person reports the number again, the report direction changes from left to right, and so on. To prevent games from getting bored, there is a special case when reporting: if the number should contain a number 7 or a multiple of 7, he should replace the number with a clap. The following table lists the numbers of N = 4 (X indicates a clap ). When a person with the number 3 clap his hands for 4th times, the number actually reaches 35.

Person	1	2	3	4	3	2	1	2	3
Reports	1	2	3	4	5	6	X	8	9
Person	4	3	2	1	2	3	4	3	2
Reports	10	11	12	13	X	15	16	X	18
Person	1	2	3	4	3	2	1	2	3
Reports	19	20	X	22	23	24	25	26	X
Person	4	3	2	1	2	3	4	3	2
Reports	X	29	30	31	32	33	34	X	36

Given n, m, and K, your task is to calculate the number of a person with the number of M to the number of K clap operations.

Input

The input contains no more than 10 groups of data. Each group of data occupies one row and contains three integers, n, m, and K (2 <= n <= 100, 1 <= m <= n, 1 <= k <= 100 ). The input end flag is n = m = k = 0.

Output

For each group of data, a row is output, that is, the integer to which a person numbered m actually counts when he clap his hands for the K time.

Sample Input

4 3 1 4 3 2 4 3 3 4 3 4 0 0 0 

Sample output

17 21 27 35 

The first approach is to directly simulate the process, first in order, and then in reverse order.

# Include <cstdio> # include <cstring>/** determines whether N is a multiple of 7 or contains 7 */bool judge (int n) {If (! (N % 7) return true; while (n) {If (N % 10 = 7) return true; N/= 10;} return false;} int main () {int n, m, K, I; while (scanf ("% d", & N, & M, & K) & N & M & K) {int Pos = 0, Count = 0, flag = 0; // sequential processing for (I = 1; I <= N; I ++) {pos ++; if (I = m) {If (Judge (POS) Count ++; // If (COUNT = k) {printf ("% d \ n", POS); break ;}} if (I = N) {I = n-1; // perform reverse processing for (; I> = 1; I --) {pos ++; if (I = m) {If (Judge (POS) Count ++; If (COUNT = k) {flag = 1; printf ("% d \ n", POS ); break ;}}if (FLAG) break; I = 1; // trace back to order }}} return 0 ;}

Reference blog: http://blog.csdn.net/whjkm/article/details/39830611

ACM-simulation-nyoj 559-Report game-Hunan seventh provincial Competition