HDU 1443 Joseph 约瑟夫环 暴力

题意：有k个好人在队伍前面，k个坏人在队伍后面。每次去掉一个在第m个位置的人，问使所有坏人都比好人先去掉的最小的m是多少？

思路：因为k比较小，直接模拟这个过程。要注意一个情况，每次找第mm个数的要找在现在这种情况的第m0m_0个数。假设已经找到cntcnt个数，则m0=(m−1)%(2∗k−cnt)+1m_0 = (m-1) \%(2 * k - cnt) + 1。mm和m0m_0是等价的，m0m_0比mm在很多情况下小很多。用mm可能就超时了。

/*********************************************
Problem : HDU 1443
Author  : NMfloat
InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++)
#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --)
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next)
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m,k;

int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,-1,1};

int vis[30];
int ans[14];

void init() {
rep(i,1,13) {
rep(j,i+1,INF) {
cls(vis,0);
int pos = (j - 1 ) % (2 * i) + 1;
if(pos <= i) continue;
vis[pos] = 1;
int cnt = 1;//被删除的个数;
int flag = 0;
while(cnt < i) {
int limit = (j - 1) % ( 2 * i - cnt ) + 1;
rep(ia,1,limit) {
++pos;
pos = (pos - 1) % (2 * i) + 1;
while(vis[pos]) {
++ pos;
pos = (pos - 1) % (2 * i) + 1;
}
}
if(pos <= i) {flag = 1 ; break;}
vis[pos] = 1;
cnt ++;
}
if(flag) continue ;
else {
ans[i] = j;
//printf("%d %d\n",i,ans[i]);
break;
}
}
}
}

void input() {

}

void solve() {
printf("%d\n",ans
);
}

int main(void) {
//freopen("a.in","r",stdin);
init();
while(scanf("%d",&n),n) {
input();
solve();
}
return 0;
}