/*读了老半天才把题读懂，读懂了题输出格式没注意，结果re了两次。

题意：先给一串数字S，然后每次给出对应相同数目的的一串数字Si，然后优先统计Si和S对应位相同并且相等的个数L，在统计不在对应位上但相等的的个数R.

当Si全0时，表示结束。

每个数只能用一次。

例如：有

S  1 3 5 5

S1 1 1 2 3

S2 4 3 3 5

Si . . . .

   0 0 0 0

对于S1:则第一步只有第一个对应相等，所以L = 1

就还剩下3 5 5

　　　　1 2 3，在统计R，下面只有3和上面的3相同，所以R = 1;

对于S2:L = 2,第二个位置和第四个位置对应相等。

就还剩下1 5

       4 3,所以R = 0;依次类推。

*/

#include <iostream>

#include <fstream>

#include <cstdio>

#include <cstdlib>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#define N 1005

using namespace std;

int n;

int a[N], b[N], c[N];

int main()

{

    int i, j, t = , l, r;

    while (cin>>n, n)

    {

        cout<<"Game "<<t++<<':'<<endl;

        for (i = ; i < n; i++)

            cin>>a[i];

        while (true)

        {

            for (i = ; i < n; i++)

                c[i] = a[i];

            l = r = ;

            for (i = ; i < n; i++)

            {

                cin>>b[i];

                if (b[i] == c[i])/**< 统计对应位相等l的个数 */

                {

                    l++;

                    c[i] = ;/**< 置相等位0,统计r时不影响 */

                    b[i] = -;

                }

            }

            if (!b[])

                break;

            for (i = ; i < n; i++)

            {

                if (b[i] == -)

                    continue;

                for (j = ; j < n; j++)

                {

                    if (b[i] == c[j])/**< 统计r的个数 */

                    {

                        r++;

                        c[j] = ;

                        break;

                    }

                }

            }

            cout<<"    "<<'('<<l<<','<<r<<')'<<endl;

        }

    }

    return ;

}

 /*题意:统计每个国家有多少个人被Giovanni喜欢。

 并按字典序输出。

 */

 #include <iostream>

 #include <string>

 #include <cstring>

 #include <cstdio>

 #include <cstdlib>

 #include <fstream>

 #include <climits>

 #define N 2005

 using namespace std;

 struct node

 {

     string c;

     int sum;

     node()

     {

         c.clear();

         sum = ;

     }

     bool operator >= (const node &b)

     {

         return c >= b.c;

     }

     friend istream& operator >> (istream &in, node &res)

     {

         in>>res.c;

         return in;

     }

     friend ostream &operator << (ostream &out, const node &res)

     {

         out<<res.c<<' '<<res.sum<<endl;

         return out;

     }

     bool operator == (const node &b)

     {

         return c == b.c;

     }

     bool operator != (const node &b)

     {

         return c != b.c;

     }

 };

 struct li

 {

     node s[N];

     int p;

     li()

     {

         p = ;

     }

     friend ostream &operator << (ostream &out, const li &res)

     {

         for (int i = ; i < res.p; i++)

             out<<res.s[i];

     }

     void insert(const node &res)

     {

         if (p)

         {

             int l = , r = p - , mid;

             while (l <= r)/**< 二分查找第一个比res大或等于的位置 */

             {

                 mid = (l + r) >> ;

                 if (s[mid] >= res)/**< 如果大于等于，在左区间位置 */

                     r = mid - ;

                 else/**< 在右区间 */

                     l = mid + ;

             }

             if (s[l] != res)/**< 不存在，插入 */

             {

                 for (r = p++; r > l; r--)

                     s[r] = s[r - ];

                 s[l] = res;

             }

             s[l].sum++;

             return;

         }

         s[p] = res;

         s[p++].sum++;

     }

 };

 li a;

 int main()

 {

     int n, i;

     cin>>n;

     string tt;

     node t;

     for (i = ; i < n; i++)

     {

         cin>>t;

         getline(cin, tt);

         a.insert(t);

     }

     cout<<a;

     return ;

 }

/**< 题意:给一组数据无序，查询给定数在这组数出现的第一个位置(在有序中) */

#include <iostream>

#include <string>

#include <cstring>

#include <cstdio>

#include <cstdlib>

#include <fstream>

#include <climits>

#define N 10005

using namespace std;

int n, q;

int sum[N], con[N];

int main()

{

    int t = , i, m;

    while (cin>>n>>q, n || q)

    {

        memset(con, , sizeof(con));

        cout<<"CASE# "<<t++<<':'<<endl;

        for (i = ; i < n; i++)

        {

            cin>>m;

            con[m]++;/**< 计数排序,每个数不超过10000 */

        }

        sum[] = ;

        sum[] = con[];

        for (i = ; i < N; i++)

            sum[i] = sum[i - ] + con[i - ];

        for (i = ; i < q; i++)

        {

            cin>>m;

            if (con[m])

                cout<<m<<" found at "<<sum[m] + <<endl;

            else

                cout<<m<<" not found"<<endl;

        }

    }

    return ;

}

/**< 题意:(又是读了半天也没读懂，好纠结的英语)

在三维空间给一些点，计算每个点与其他点的距离，但只取最短距离，并统计最短距离在[0-10)的个数

比如某个点的最短距离在[1,2)之间，则对1这个位置的个数加1 */

#include <iostream>

#include <string>

#include <cstring>

#include <cstdio>

#include <cstdlib>

#include <fstream>

#include <climits>

#include <cmath>

#include <algorithm>

#include<iomanip>

#define N 5005

using namespace std;

struct node

{

    int x, y, z;

    /*

    bool operator > (const node &b)

    {

        if (x == b.x)

        {

            if (y == b.y)

                return z > b.z;

            return y > b.y;

        }

        return x > b.x;

    }*/

}a[N];

int con[];

int dis(const node &a, const node &b)

{

    int x = a.x - b.x, y = a.y - b.y, z = a.z - b.z;

    return int(sqrt(x * x + y * y + z * z));

}

int cmp(const void *a, const void *b)/**<  */

{

    return ((node *)a)->x > ((node *)b)->x ?  : -;

}

int main()

{

    int n = , i, j, tm, t;

    while (cin>>a[n].x>>a[n].y>>a[n].z, (a[n].x || a[n].y || a[n].z))

    {

        n++;

    }

    //qsort(a, n, sizeof(a[0]), cmp);

    memset(con, , sizeof(con));

    for (i = ; i < n; i++)

    {

        tm = ;

        for (j = ; j < n; j++)

        {

            if (i != j)

            {

                t = dis(a[i], a[j]);

                if (t < tm)

                    tm = t;

            }

        }

        con[tm]++;

    }

    for (i = ; i < ; i++)

        cout<<setw()<<con[i];

    cout<<endl;

    return ;

}

/**<

题目啰嗦了半天，就是火车进行从小到大重排序(冒泡排序)交换的次数

 */

#include <iostream>

#include <algorithm>

using namespace std;

#define N 55

int a[N];

int main()

{

    int n, l, i, j, s;

    cin>>n;

    while (n--)

    {

        cin>>l;

        for (i = ; i < l; i++)

            cin>>a[i];

        s = ;

        for (i = ; i < l; i++)

            for (j = l - ; j > i; j--)

            {

                if (a[j] < a[j - ])

                {

                    swap(a[j], a[j - ]);

                    s++;

                }

            }

        cout<<"Optimal train swapping takes "<<s<<" swaps."<<endl;

    }

    return ;

}

/**<  翻转从小到大排序*/

#include <iostream>

#include <algorithm>

#include <cstdio>

using namespace std;

#define N 35

int a[N];

int main()

{

    int i, j, n = , p, mm, t;

    char ch = ;

    while (ch != EOF && cin>>a[n])

    {

        ch = cin.get();

        if (ch == ' ')

        {

            n++;

            continue;

        }

        t = ++n;

        cout<<a[];

        for (i = ; i < n; i++)

            cout<<' '<<a[i];

        cout<<endl;

        for (i =  ; i < n; n--)

        {

            mm = a[i];

            p = ;

            for (j = ; j < n; j++)

            {

                if (mm < a[j])

                {

                    mm = a[j];

                    p = j;

                }

            }

            if (p != n - )

            {

                if (p)

                {

                    reverse(a, a + p + );

                    reverse(a, a + n);

                    cout<<t - p<<' '<<t - n + <<' ';

                }

                else

                {

                    reverse(a, a + n);

                    cout<<t - n + <<' ';

                }

            }

        }

        cout<<<<endl;

    }

    return ;

}

/**<

给一些单词，去掉相同字母且个数一样不区分大小写，再给剩下的单词进行排序输出

 */

#include <iostream>

#include <fstream>

#include <string>

#include <cstring>

#include <cstdio>

#include <cstdlib>

#include <algorithm>

#define N 1005

using namespace std;

struct node

{

    string str;

    bool flag;

    int tag[];

    bool operator == (const node &b)

    {

        for (int i = ; i < ; i++)

            if (tag[i] != b.tag[i])

                return false;

        return true;

    }

    bool operator > (const node &b)

    {

        return str > b.str;

    }

    node(const string &s)

    {

        str = s;

        flag = false;

        memset(tag, , sizeof(tag));

        for (int i = ; s[i]; i++)

            tag[(s[i] | ) - 'a']++;

    }

    node()

    {

        flag = false;

        memset(tag, , sizeof(tag));

    }

}s[N];

int p = ;

void insert(const string &v)

{

    if (!p)

    {

        s[p++] = v;

        return;

    }

    for (int i = ; i < p; i++)

        if (s[i] == v)

        {

            s[i].flag = true;

            return;

        }

    int l = , r = p - , mid;

    while (l <= r)

    {

        mid = (l + r) >> ;

        if (s[mid] > v)

            r = mid - ;

        else

            l = mid + ;

    }

    for (r = p++; r > l; r--)

        s[r] = s[r - ];

    s[l] = v;

}

string t;

int main()

{

    int i;

    char ch;

    cin.get(ch);

    while (ch != '#')

    {

        t.clear();

        while (ch == ' ' || ch == '\n')

            cin.get(ch);

        if (ch == '#')

            break;

        while (true)

        {

            t += ch;

            cin.get(ch);

            if (ch == ' ' || ch == '\n')

                break;

        }

        insert(t);

    }

    for (i = ; i < p; i++)

        if (!s[i].flag)

            cout<<s[i].str<<endl;

    return ;

}

/**<

对输入的字符串进行排序，按格式输出，最后一列的宽度为所有字符串最长宽度，其余列的宽度为最长宽度加2.

要求输出所用的行最少

 */

#include <iostream>

#include <fstream>

#include <string>

#include <cstring>

#include <cstdio>

#include <cstdlib>

#include <algorithm>

#include <iomanip>

using namespace std;

#define N 105

string s[N];

int main()

{

    int n, i, mm, len, j, t, k, kk;

    while (cin>>n)

    {

        mm = ;

        for (i = ; i < n; i++)

        {

            cin>>s[i];

            len = s[i].length();

            if (mm < len)

                mm = len;

        }

        sort(s, s + n);

        len = ( - mm) / (mm + ) + ;

        for (i = ; i < ; i++)

            cout<<'-';

        cout<<endl;

        cout<<left;

        t = n / len;/**< 要输出t行len列 */

        if (n % len)

            t++;

        for (i = ; i < t; i++)

        {

            kk = mm + ;

            for (j = i, k = ; j < n; j += t, k++)

            {

                if (k == len)/**< 最后一列输出宽度为mm */

                    kk = mm;

                cout<<setw(kk)<<s[j];

            }

            cout<<endl;

        }

    }

    return ;

}

#include <iostream>

#include <fstream>

#include <string>

#include <cstring>

#include <cstdio>

#include <cstdlib>

#include <algorithm>

#include <iomanip>

#include <set>

using namespace std;

struct snode

{

    string str;

    int x, y;/**< 字符串位置 */

    bool operator < (const snode &b)const

    {

        if (str == b.str)

        {

            if (x == b.x)

                return y < b.y;

            return x < b.x;

        }

        return str < b.str;

    }

    snode()

    {

        str.clear();

        x = y = ;

    }

    snode(string v, int xx, int yy)

    {

        str = v;

        x = xx;

        y = yy;

    }

    friend ostream & operator << (ostream &out, const snode &v)

    {

        out<<v.str<<' '<<v.x<<' '<<v.y;

        return out;

    }

};

set<string> ks;

multiset<snode> res;

string ts[];

int main()

{

    int i, j, p;

    string tmp;

    ks.clear();

    res.clear();

    while (cin>>tmp, tmp != "::")

    {

        ks.insert(tmp);

    }

    i = ;

    cin.get();

    while (getline(cin, ts[i]))

    {

        j = ;

        while (true)

        {

            p = j;

            tmp.clear();

            while (ts[i][j] && ts[i][j] != ' ')

            {

                ts[i][j] |= ;

                tmp += ts[i][j++];

            }

            if (ks.find(tmp) == ks.end())/**< 不存在 */

                res.insert(snode(tmp, i, p));

            if (!ts[i][j])

                break;

            j++;

        }

        i++;

    }

    for (multiset<snode>::iterator it = res.begin(); it != res.end(); it++)

    {

        for (i = ; i < it->y; i++)

            cout<<ts[it->x][i];

        for (; ts[it->x][i] != ' ' && ts[it->x][i]; i++)/**< 大写 */

            cout<<char(ts[it->x][i] & );

        for (; ts[it->x][i]; i++)

            cout<<ts[it->x][i];

        //cout<<*it;

        cout<<endl;

    }

    return ;

}

#include <iostream>

#include <cstdio>

#include <fstream>

#include <cstring>

#include <string>

#include <cctype>

#include <iomanip>

#include <cstdlib>

#include <algorithm>

#include <set>

#include <map>

using namespace std;

struct snode

{

    snode()

    {

        score = in = out = win = con = tie = lose = ;

        name.clear();

        _name.clear();

    }

    snode(string v)

    {

        score = in = out = win = con = tie = lose = ;

        name = v;

        _name.clear();

        for (int i = ; v[i]; i++)

        {

            if (isalnum(v[i]))

                _name += (v[i] | );

            else

                _name += v[i];

        }

    }

    string name;

    string _name;

    int score;/**< 总得分 */

    int in;/**< 进球 */

    int out;/**< 对方进球 */

    int win;/**< 赢得场数 */

    int con;/**< 参加比赛次数 */

    int tie;/**< 平局次数 */

    int lose;/**< 输的次数 */

    bool operator < (const snode &b)const

    {

        if (score == b.score)

        {

            if (win == b.win)

            {

                int ta = in - out, tb = b.in - b.out;

                if (ta == tb)

                {

                    if (in == b.in)

                    {

                        if (con == b.con)

                        {

                            return _name < b._name;

                        }

                        return con < b.con;

                    }

                    return in > b.in;

                }

                return ta > tb;

            }

            return win > b.win;

        }

        return score > b.score;

    }

    friend istream &operator >> (istream &in, snode &v)

    {

        string tmp;

        getline(in, tmp);

        v = tmp;

        return in;

    }

    friend ostream &operator << (ostream &out, const snode &v)

    {

        cout<<v.name<<' '<<v.score<<"p, "<<v.con<<"g ("<<v.win<<'-'<<v.tie<<'-'<<v.lose<<"), ";

        cout<<v.in - v.out<<"gd ("<<v.in<<'-'<<v.out<<')';

        return out;

    }

    bool operator == (const snode &v)

    {

        return name == v.name;

    }

}a[];

int n, t, g;

int find_str(string v)

{

    for (int i = ; i < t; i++)

        if (a[i] == v)

            return i;

    return -;

}

void fun(string v)

{

    int i = -;

    while (v[++i] != '#');

    string s1(v.begin(), v.begin() + i);

    /**< 数字处 */

    int t1 = v[++i] - '';

    if (isdigit(v[++i]))/**< 第二个是不是数字 */

        t1 =  * t1 + (v[i++] - '');

    int t2 = v[++i] - '';

    if (isdigit(v[++i]))/**< 第二个是不是数字 */

        t2 =  * t2 + (v[i++] - '');

    string s2(v.begin() + i + , v.end());

    int p1 = find_str(s1), p2 = find_str(s2);

    //cout<<p1<<p2<<endl;

    if (t1 > t2)/**< 1赢 */

    {

        a[p1].win++;

        a[p2].lose++;

        a[p1].score += ;

    }

    else

        if (t1 == t2)/**< 平 */

        {

            a[p1].tie++;

            a[p2].tie++;

            a[p1].score++;

            a[p2].score++;

        }

        else/**< 2赢 */

        {

            a[p2].win++;

            a[p1].lose++;

            a[p2].score += ;

        }

        a[p1].con++;

        a[p2].con++;

        a[p1].in += t1;

        a[p2].in += t2;

        a[p1].out += t2;

        a[p2].out += t1;

}

int main()

{

    int i, j;

    cin>>n;

    cin.get();

    string st;

    while (n--)

    {

        getline(cin, st);

        cout<<st<<endl;

        cin>>t;

        cin.get();

        for (i = ; i < t; i++)

            cin>>a[i];

        cin>>g;

        cin.get();

        for (i = ; i < g; i++)

        {

            getline(cin, st);

            fun(st);

        }

        sort(a, a + t);

        for (i = ; i < t; i++)

        {

            cout<<i + <<") "<<a[i]<<endl;

        }

        if (n)

            cout<<endl;

    }

    return ;

}

#include <iostream>

#include <cstdio>

#include <fstream>

#include <cstring>

#include <string>

#include <cctype>

#include <iomanip>

#include <cstdlib>

#include <algorithm>

#include <set>

#include <map>

using namespace std;

int ls[] =

{

    , , , , , , , , , ,

    , , , , , , , , , ,

    , , , , ,

};

char get(char ch)

{

    return ls[ch - 'A'] + '';

}

map<string, int> mp;

int main()

{

    int t, n, i, tag;

    cin>>t;

    char ch;

    string tmp;

    while (t--)

    {

        mp.clear();

        cin>>n;

        cin.get();

        for (i = ; i < n; i++)

        {

            tmp.clear();

            while (ch = cin.get(), ch != '\n')/**< 对输入进行处理换成全数字的字符串 */

            {

                if (isdigit(ch))

                {

                    tmp += ch;

                    continue;

                }

                if (isalpha(ch))

                {

                    tmp += get(ch);

                    continue;

                }

            }

            mp[tmp]++;/**< 对应字符串个数加1 */

        }

        tag = true;

        for (map<string, int>::iterator it = mp.begin(); it != mp.end(); ++it)

        {

            if (it->second > )

            {

                tmp = it->first;

                tmp.insert(tmp.begin() + , '-');

                cout<<tmp<<' '<<it->second<<endl;

                tag = false;

            }

        }

        if (tag)

            cout<<"No duplicates."<<endl;

        if (t)

            cout<<endl;

    }

    return ;

}

#include <iostream>

#include <fstream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <string>

#include <iomanip>

#include <cctype>

#include <climits>

#include <algorithm>

#include <map>

#include <set>

using namespace std;

map<char, int> mp1, mp2;

char s1[] =

{

    'J', 'S', 'B', 'K', 'T', 'C', 'L', 'D', 'M', 'V',

    'N', 'W', 'F', 'X', 'G', 'P', 'Y', 'H', 'Q', 'Z',

    'R'

};

char s2[] =

{

    'A' , 'U', 'E', 'O', 'I'

};

int n1, n2, p1, p2;

void fun1()

{

    while (n1 > )

    {

        p1++;

        if (n1 >= )

            mp1[s1[p1]] = ;

        else

            mp1[s1[p1]] = n1;

        n1 -= ;

    }

}

void fun2()

{

    while (n2 > )

    {

        p2++;

        if (n2 >= )

            mp2[s2[p2]] = ;

        else

            mp2[s2[p2]] = n2;

        n2 -= ;

    }

}

void print()

{

    int i;

    string s1, s2;

    s1.clear();

    s2.clear();

    for (map<char, int>::iterator it = mp1.begin(); it != mp1.end(); ++it)

    {

        for (i = it->second; i > ; i--)

            s1 += it->first;

    }

    for (map<char, int>::iterator it = mp2.begin(); it != mp2.end(); ++it)

    {

        for (i = it->second; i > ; i--)

            s2 += it->first;

    }

    i = ;

    while (s1[i] && s2[i])

    {

        cout<<s2[i]<<s1[i];

        i++;

    }

    if (s2[i])

        cout<<s2[i];

    cout<<endl;

}

int main()

{

    int m, n, t = ;

    cin>>m;

    while (m--)

    {

        cout<<"Case "<<t++<<": ";

        cin>>n;

        p1 = p2 = -;

        n1 = n / ;

        n2 = n1 + (n & );

        mp1.clear();

        mp2.clear();

        fun1();

        fun2();

        print();

    }

    return ;

}