UVA Josephus Problem 11089 （約瑟夫環） 數學好題

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 Type:  None

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 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 31st 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 kth 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

Sample Input	Output for Sample Input
6 2 1 2 2 3 1 3 2 3 3 4 6 #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 ++) //從零開始報數 { f[i] = (f[i-1] + k)%i; //數學推導當i個人數數時，赢家f[i]與f[i-1]有數學關系 } printf("Case %d: %d\n",T++,f[n]+1); } return 0; }	Case 1: 2 Case 2: 1 Case 3: 3 Case 4: 3 Case 5: 2 Case 6: 3