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最新推荐文章于 2021-05-06 13:09:14 发布

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最新推荐文章于 2021-05-06 13:09:14 发布

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文章标签： java

原文链接：http://www.cnblogs.com/devtang/archive/2012/08/21/2649928.html

约瑟夫环问题

序列1:0,1,2...m-2,m...n-2,n-1(n-1个)

序列2:m,m+1,m+2...n-1,0,1..m-2(n-1个)

序列3:0,1,2...n-2,n-1(n-1个)

序列4:k,k+1,k+2...k-2(n-2个)

先由f[1] = 0;

f[i] = (f[i-1]+k)%i递推到n-1,然后f[n] = (f[n-1] + m)%n+1即可

View Code

 1 #include<iostream>
 2 #include<cstdio>
 3 #define MAXN 10004
 4 using namespace std;
 5 int f[MAXN];
 6 
 7 
 8 int main()
 9 {
10     int n,m,k;
11     while (scanf("%d%d%d",&n,&k,&m) != EOF) {
12         if (n == 0 && k == 0 && m == 0)break;
13         f[1] = 0;
14         for (int i(2); i<n; ++i) {
15             f[i] = (f[i-1] + k)%i;
16         }
17         f[n] = (f[n-1] + m)%n;
18         cout<<f[n]+1<<endl;
19     }
20     return 0;
21 }