noip模拟赛 对刚

3

分析：约瑟夫环问题，可以用链表模拟做，也可以套用公式,比较水.

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int n, t, l[100010], r[100010], st = 1;

int main()
{
    scanf("%d%d", &n, &t);
    for (int i = 1; i <= n; i++)
    {
        l[i] = i - 1;
        r[i] = i + 1;
        if (i == 1)
            l[i] = n;
        if (i == n)
            r[i] = 1;
    }
    for (int i = 1; i < n; i++)
    {
        for (int j = 1; j < t; j++)
            st = r[st];
        int t = st;
        st = r[st];
        l[r[t]] = l[t];
        r[l[t]] = r[t];
    }
    printf("%d\n", st);

    return 0;
}