UVA 1452 八 Jump

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原文链接：http://www.cnblogs.com/cyd308/p/4771582.html

本文介绍了一个用于生成Jump序列的算法，该序列通过特定规则从圆环排列的整数中提取。文章提供了一种高效的方法来确定序列的最后三个元素，并附带了完整的源代码。

Jump

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

Submit Status Practice UVA 1452

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 $\le$ 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].

$\epsfbox{p4727.eps}$

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 $\le$ n $\le$ 500, 000 and 2 $\le$ k $\le$ 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

 1 #include <stdio.h>
 2 
 3 int fun(int m,int k,int i){
 4  
 5     if(i==1)
 6         return (m+k-1)%m;
 7     else
 8         return (fun(m-1,k,i-1)+k)%m;
 9  
10 }
11 
12 int main()
13 {
14      int T;
15      scanf("%d",&T);
16      while(T--)
17      {
18          int n,k,a,b,c;
19          scanf("%d %d",&n,&k);
20          if(k%6==1)
21              a=1,b=2,c=3;
22          if(k%6==2)
23              a=2,b=1,c=3;
24          if(k%6==3)
25              a=3,b=1,c=2;
26          if(k%6==4)
27              a=1,b=3,c=2;
28          if(k%6==5)
29              a=2,b=3,c=1;
30          if(k%6==0)
31              a=3,b=2,c=1;
32          a--,b--,c--;
33          for(int i=4;i<=n;i++)
34          {
35              a=(a+k)%i;
36              b=(b+k)%i;
37              c=(c+k)%i;
38          }
39          a++,b++,c++;
40          printf("%d %d %d\n",a,b,c);
41         //printf("%d %d %d\n",fun(n,k,n-2)+1,fun(n,k,n-1)+1,fun(n,k,n)+1);
42     }
43     return 0;
44 }