UVA 133 The Dole Queue

题目链接：https://vjudge.net/problem/UVA-133

题目翻译摘自《算法禁赛入门经典》

题目大意

　　N 个人站成一圈，逆时针编号为 1 ~ N，有两个官员，A 从 1 开始逆时针数，B 从 N 开始顺时针数，在每一轮中，官员 A 数 k 个就停下来，官员 B 数 m 个就停下来（注意有可能两个官员停在同一个人身上）。接下来被官员选中的人（1个或者两个）离开队伍。

　　输入 N，k，m 输出每轮里被选中的人的编号，如果有两个人，先输出被A选中的。
　　PS：输出的每个数应当恰好占3列。

分析

　　只要是线性表都能做这道题，不过我私以为静态链表比较美观。

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< int, PII > PIPII;
typedef pair< string, int > PSI;
typedef pair< int, PSI > PIPSI;
typedef set< int > SI;
typedef set< PII > SPII;
typedef vector< int > VI;
typedef vector< VI > VVI;
typedef vector< PII > VPII;
typedef map< int, int > MII;
typedef map< int, PII > MIPII;
typedef map< PII, int > MPIII;
typedef map< string, int > MSI;
typedef multimap< int, int > MMII;
//typedef unordered_map< int, int > uMII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-6;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 1e4 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

struct StaticList{
    int sl[21], nt[21], pv[21];
    int *next = nt, *prev = pv;
    int sz;
    
    void init(int x) {
        sz = x;
        For(i, 0, sz) {
            if(i) sl[i] = i;
            prev[i] = (sz + i) % (sz + 1);
            next[i] = (i + 1) % (sz + 1);
        }
    }
    
    void erase(int x) {
        next[prev[x]] = next[x];
        prev[next[x]] = prev[x];
    }
    
    bool empty() {
        return next[0] == 0;
    }
    
    void print() {
        cout << "SL:  ";
        For(i, 0, sz) cout << " " << sl[i];
        cout << endl;
        
        cout << "prev:";
        For(i, 0, sz) cout << " " << prev[i];
        cout << endl;
        
        cout << "next:";
        For(i, 0, sz) cout << " " << next[i];
        cout << endl;
        
    }
}SL;

int N, k, m, A, B;

int main(){
    //freopen("MyOutput.txt","w",stdout);
    //freopen("input.txt","r",stdin);
    //INIT();
    while(cin >> N >> k >> m) {
        if(!N) break;
        SL.init(N);
        A = N;
        B = 1;
        
        while(!SL.empty()) {
            Rep(i, k) {
                A = SL.next[A];
                if(A == 0) A = SL.next[A];
            }
            Rep(i, m) {
                B = SL.prev[B];
                if(B == 0) B = SL.prev[B];
            }
            if(A == B) {
                --N;
                printf("%3d%c", A, ",
"[N == 0]);
            }
            else {
                N -= 2;
                printf("%3d%3d%c", A, B, ",
"[N == 0]);
            }
            
            SL.erase(A);
            A = SL.prev[A];
            SL.erase(B);
            B = SL.next[B];
        }
    }
    return 0;
}