Light oj 1179 - Josephus Problem 【思维】

 

1179 - Josephus Problem

PDF (English)

Statistics

Forum

Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
6 2 1 2 2 3 1 3 2 3 3 4 6	Case 1: 2 Case 2: 1 Case 3: 3 Case 4: 3 Case 5: 2 Case 6: 3

 

---

PROBLEM SETTER: JANE ALAM JAN

 

 

题意：
给你n个人，从1到n进行编号，每k个人出对，求最后一个剩下的人的编号（约瑟夫环问题）。
思路：
对于n个人，当第一个人（编号为k%n-1)）出去之后，剩下的一个序列为k%n，k%n+1，k%n+2，……，0，1，……，k%n-2。就成了一个新的序列，然后发现对于新的环与老的环之间存在f(n)=(f(n-1)+k)%n。进行递推就可以求出来。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
typedef long long LL;
//typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double Pi = acos(-1.0);
const int MOD = 1e9+5;
const int MAXN = 1e6 + 5;
int  main() {
    int k, n, t;
    int Kcase = 0;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &k);
        int ans = 0;
        for (int i = 2; i <= n; i++) {
            ans = (ans + k)%i;
        }
        printf("Case %d: %d\n", ++Kcase, ans + 1);
    }
    return 0;
}