UVA Josephus Problem 11089 （约瑟夫环） 数学好题

Josephus Problem

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on 

LightOJ. Original ID:  1179

64-bit integer IO format: 

%lld      Java class name:  Main

Prev 

Submit  Status  Statistics  Discuss

Next

None

•  None
• Graph Theory
•     2-SAT 
•     Articulation/Bridge/Biconnected Component
•      Cycles/Topological Sorting/Strongly Connected Component
•      Shortest Path
•         Bellman Ford
•          Dijkstra/Floyd Warshall
•     Euler Trail/Circuit 
•     Heavy-Light Decomposition
•      Minimum Spanning Tree
•     Stable Marriage Problem 
•     Trees
•     Directed Minimum Spanning Tree 
•     Flow/Matching
•         Graph Matching 
•             Bipartite Matching
•              Hopcroft–Karp Bipartite Matching
•             Weighted Bipartite Matching/Hungarian Algorithm 
•         Flow
•             Max Flow/Min Cut 
•             Min Cost Max Flow
•  DFS-like
•     Backtracking with Pruning/Branch and Bound 
•     Basic Recursion
•     IDA* Search 
•     Parsing/Grammar
•     Breadth First Search/Depth First Search 
•     Advanced Search Techniques
•          Binary Search/Bisection
•         Ternary Search 
• Geometry
•     Basic Geometry 
•     Computational Geometry
•      Convex Hull
•     Pick's Theorem
•  Game Theory
•     Green Hackenbush/Colon Principle/Fusion Principle 
•     Nim
•     Sprague-Grundy Number 
• Matrix
•     Gaussian Elimination 
•     Matrix Exponentiation
•  Data Structures
•     Basic Data Structures
•      Binary Indexed Tree
•     Binary Search Tree
•      Hashing
•     Orthogonal Range Search
•      Range Minimum Query/Lowest Common Ancestor
•     Segment Tree/Interval Tree 
•     Trie Tree
•     Sorting 
•     Disjoint Set
• String
•      Aho Corasick
•     Knuth-Morris-Pratt
•      Suffix Array/Suffix Tree
• Math
•      Basic Math
•     Big Integer Arithmetic
•      Number Theory
•         Chinese Remainder Theorem
•          Extended Euclid
•         Inclusion/Exclusion
•          Modular Arithmetic
•     Combinatorics
•          Group Theory/Burnside's lemma
•         Counting 
•     Probability/Expected Value
•  Others
•     Tricky
•     Hardest 
•     Unusual
•     Brute Force 
•     Implementation
•     Constructive Algorithms 
•     Two Pointer
•     Bitmask 
•     Beginner
•     Discrete Logarithm/Shank's Baby-step Giant-step Algorithm 
•     Greedy
•     Divide and Conquer 
• Dynamic Programming
• Tag it!

The historian Flavius Josephus relates how, in the Romano-Jewish conflict of 67 A.D., the Romans took the town of Jotapata which he was commanding. Escaping, Josephus found himself trapped in a cave with 40 companions. The Romans discovered his whereabouts and invited him to surrender, but his companions refused to allow him to do so. He therefore suggested that they kill each other, one by one, the order to be decided by lot. Tradition has it that the means for affecting the lot was to stand in a circle, and, beginning at some point, count round, every third person being killed in turn. The sole survivor of this process was Josephus, who then surrendered to the Romans. Which begs the question: had Josephus previously practiced quietly with 41 stones in a dark corner, or had he calculated mathematically that he should adopt the 31^st position in order to survive?

Now you are in a similar situation. There are n persons standing in a circle. The persons are numbered from 1 to n circularly. For example, 1 and n are adjacent and 1 and 2 are also. The count starts from the first person. Each time you count up to k and the k^th person is killed and removed from the circle. Then the count starts from the next person. Finally one person remains. Given n and k you have to find the position of the last person who remains alive.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains two positive integers n (1 ≤ n ≤ 10⁵) and k (1 ≤ k < 2³¹).

Output

For each case, print the case number and the position of the last remaining person.

Sample Input

#include<stdio.h>
#include<string.h>
using namespace std;
int f[100010];
int main()
{
    int t,T=1;
	int n,k,i;
	scanf("%d",&t);
	while(t--)   
 	{
	 	scanf("%d%d",&n,&k);
 	    f[1] = 0;               
 		for(i = 2; i <= n; i ++)    //从零开始报数
 		{
 			f[i] = (f[i-1] + k)%i;  //数学推导当i个人数数时，赢家f[i]与f[i-1]有数学关系
 		}
 		printf("Case %d: %d\n",T++,f[n]+1);
 	}
 	return 0;
}