Volume 1. Sorting/Searching(uva)

340 - Master-Mind Hints

/*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 ;
}

10420 - List of Conquests

 /*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 ;
 }

10474 - Where is the Marble?

/**< 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 ;
}

152 - Tree's a Crowd

/**< 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 ;
}

299 - Train Swapping

/**<
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 ;
}

120 - Stacks of Flapjacks

/**<  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 ;
}

156 - Ananagrams

/**<
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 ;
}

400 - Unix ls

/**<
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 ;
}

123 - Searching Quickly

#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 ;
}

10194 - Football (aka Soccer)

#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 ;
}

755 - 487--3279

#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 ;
}

10785 - The Mad Numerologist

#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 ;
}