Jump（递推）

4727 - Jump

Asia - Seoul - 2009/2010

Integers 1, 2, 3,..., n are placed on a circle in the increasing order as in the following figure. We want to

construct a sequence from these numbers on a circle. Starting with the number 1, we continually go round by

picking out each k-th number and send to a sequence queue until all numbers on the circle are exhausted. This

linearly arranged numbers in the queue are called Jump(n, k) sequence where 1 n, k.

Let us compute Jump(10, 2) sequence. The first 5 picked numbers are 2, 4, 6, 8, 10 as shown in the following

figure. And 3, 7, 1, 9 and 5 will follow. So we get Jump(10, 2) = [2,4,6,8,10,3,7,1,9,5]. In a similar way, we

can get easily Jump(13, 3) = [3,6,9,12,2,7,11,4,10,5,1,8,13], Jump(13, 10) = [10,7,5,4,6,9,13,8,3,12,1,11,2]

and Jump(10, 19) = [9,10,3,8,1,6,4,5,7,2].

Jump(10,2) = [2,4,6,8,10,3,7,1,9,5]

You write a program to print out the last three numbers of Jump(n, k) for n, k given. For example suppose that

n = 10, k = 2, then you should print 1, 9 and 5 on the output file. Note that Jump(1, k) = [1].

 

Input

Your program is to read the input from standard input. The input consists of T test cases. The number of test

cases T is given in the first line of the input. Each test case starts with a line containing two integers n and k,

where 5 n 500, 000 and 2 k 500, 000.

 

Output

Your program is to write to standard output. Print the last three numbers of Jump(n, k) in the order of the last

third, second and the last first. The following shows sample input and output for three test cases.

 

Sample Input

3

10 2

13 10

30000 54321

 

Sample Output

1 9 5

1 11 2

10775 17638 23432

 

         题意：

         约瑟夫环问题。给出 N 和 M，输出最后输出的三个数是什么。

 

         思路：

         递推。a，b，c 保存最后三个的数是什么。

 

         AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int main() {
        int t;
        scanf("%d", &t);

        while (t--) {
                int n, m;
                int a, b, c;
                scanf("%d%d", &n, &m);

                if (m % 6 == 1) a = 1, b = 2, c = 3;
                if (m % 6 == 2) a = 2, b = 1, c = 3;
                if (m % 6 == 3) a = 3, b = 1, c = 2;
                if (m % 6 == 4) a = 1, b = 3, c = 2;
                if (m % 6 == 5) a = 2, b = 3, c = 1;
                if (m % 6 == 0) a = 3, b = 2, c = 1;

                for (int i = 4; i <= n; ++i) {
                        a = !((a + m) % i) ? i : (a + m) % i;
                        b = !((b + m) % i) ? i : (b + m) % i;
                        c = !((c + m) % i) ? i : (c + m) % i;
                }

                printf("%d %d %d\n", a, b, c);
        }

        return 0;
}