POJ 3517

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

#include<iostream>
#include<cstdio>
#define MAXN 10004
using namespace std;
int f[MAXN];


int main()
{
    int n,m,k;
    while (scanf("%d%d%d",&n,&k,&m) != EOF) {
        if (n == 0 && k == 0 && m == 0)break;
        f[1] = 0;
        for (int i(2); i<n; ++i) {
            f[i] = (f[i-1] + k)%i;
        }
        f[n] = (f[n-1] + m)%n;
        cout<<f[n]+1<<endl;
    }
    return 0;
}