CQUPT WEEKLY TRAINING （2）DIV2 解题报告

先推荐一个很好用的网站 http://www.cplusplus.com/

所有C/C++的函数和文档都可以在上面查到。。

A (uva 11988 - Broken Keyboard)

'[' 表示碰到了 "home" ，']' 表示碰到了 "end"。用链表模拟，维护一个指针，当输入是'[' 时，指针移向表头，为 ']' 则移到表尾，否则直接插入。

PS: STL链表

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <stack>#include <cassert>#include <algorithm>#include <cmath>#include <set>#include <list>#include <map>#include <limits>using namespace std;#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MAX(a, b) ((a) > (b) ? (a) : (b))#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))#define MAXN 100000#define MAXM 999999#define MOD 1000000007#define PI 3.1415926535897932384626433832795typedef long long LL;typedef vector<int> veci;char buf[MAXN+5];int main() {    //freopen("input.in", "r", stdin);    while(scanf("%s",buf) != EOF) {        char *p = buf, c;        list<char> ans;        list<char>::iterator it = ans.begin();        while(*p != '\0') {            c = *p;            if (c == '[') {                it = ans.begin();            } else if (c == ']') {                it = ans.end();            } else {                ans.insert(it, c);            }            p++;        }        for(it = ans.begin();it != ans.end();++it)            putchar(*it);        putchar('\n');    }    return 0;}

B (11136 - Hoax or what)

我们需要一个能够快速插入元素，并且能够快速查询最大，最小值的数据结构。STL 已经为我们实现好了这样的容器，set 和 multiset，不同之处在于 set 里面的元素不可重，而 multiset 可重。set 和 multiset 都是基于 BST（Binary search tree）实现的，以后你可能需要实现自己的 BST 以满足特殊需求。。

PS:STL multiset

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <stack>#include <cassert>#include <algorithm>#include <cmath>#include <set>#include <list>#include <map>#include <limits>using namespace std;#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MAX(a, b) ((a) > (b) ? (a) : (b))#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))#define MAXN 100000#define MAXM 999999#define MOD 1000000007#define PI 3.1415926535897932384626433832795typedef long long LL;typedef vector<int> veci;char buf[MAXN+5];int main() {    //freopen("input.in", "r", stdin);    int N;    while(scanf("%d",&N) == 1 && N) {        multiset<int> myset;        LL ans = 0;        REP(i, 1, N) {            int num, t;            scanf("%d",&num);            REP(j, 1, num) {                scanf("%d",&t);                myset.insert(t);            }            int _min = *myset.begin();            int _max = *myset.rbegin();            myset.erase(myset.find(_min));            myset.erase(myset.find(_max));            ans += _max-_min;        }        printf("%lld\n",ans);    }    return 0;}

C (ZOJ 1089 Lotto)

因为数据非常小，所以有两种做法：

1）生成组合 （请自行 baidu =_=||）

2）暴力枚举

//生成组合#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <stack>#include <cassert>#include <algorithm>#include <cmath>#include <set>#include <list>#include <map>#include <limits>using namespace std;#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MAX(a, b) ((a) > (b) ? (a) : (b))#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))#define MAXN 100000#define MAXM 999999#define MOD 1000000007#define PI 3.1415926535897932384626433832795typedef long long LL;typedef vector<int> veci;int _map[15];int ans[15];int N;int upper(int x) {    return N - (6 - x);}void print() {    REP(i, 1, 5)        printf("%d ", _map[ans[i]]);    printf("%d\n",_map[ans[6]]);}int main() {    //freopen("input.in", "r", stdin);    int kase = 0;    while(scanf("%d",&N) == 1 && N) {         if (kase++ > 0)            putchar('\n');         REP(i, 1, N)            scanf("%d", &_map[i]);         REP(i, 1, 6)            ans[i] = i;         print();         while(true) {            int idx = 6;            while(idx >= 1 && ans[idx] >= upper(idx))                --idx;            if (idx < 1)                break;            ans[idx]++;            REP(i, idx+1, 6)                ans[i] = ans[i-1]+1;            print();         }    }    return 0;}

//枚举#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <stack>#include <cassert>#include <algorithm>#include <cmath>#include <set>#include <list>#include <map>#include <limits>using namespace std;#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MAX(a, b) ((a) > (b) ? (a) : (b))#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))#define MAXN 100000#define MAXM 999999#define MOD 1000000007#define PI 3.1415926535897932384626433832795typedef long long LL;typedef vector<int> veci;int _map[15];int ans[15];int N;int upper(int x) {    return N - (6 - x);}void print(int a, int b, int c, int d, int e, int f) {    printf("%d %d %d %d %d %d\n", _map[a], _map[b],           _map[c], _map[d], _map[e], _map[f]);}int main() {    //freopen("input.in", "r", stdin);    int kase = 0;    while(scanf("%d",&N) == 1 && N) {         if (kase++ > 0)            putchar('\n');         REP(i, 1, N)            scanf("%d", &_map[i]);         REP(i1, 1, N)            REP(i2, i1+1, N)                REP(i3, i2+1, N)                    REP(i4, i3+1, N)                        REP(i5, i4+1, N)                            REP(i6, i5+1, N) {                                print(i1, i2, i3, i4, i5, i6);                            }    }    return 0;}

D (Codeforces 272 Div2 A)

从小到大枚举 m， 判断。判断方法可以是：枚举 m 步里面的1，判断余下的能否被2整除。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <stack>#include <cassert>#include <algorithm>#include <cmath>#include <set>#include <list>#include <map>#include <limits>using namespace std;#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MAX(a, b) ((a) > (b) ? (a) : (b))#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))#define MAXN 100000#define MAXM 10000#define MOD 1000000007#define PI 3.1415926535897932384626433832795typedef long long LL;typedef vector<int> veci;typedef vector<pair<int, int> > vect;typedef pair<int, int> pairi;const double maxdouble = numeric_limits<double>::max();const double eps = 1e-10;const int INF = 0x7FFFFFFF;int main() {    int n, m;    cin >> n >> m;    bool ok = false;    int ans;    for (int i=1;m*i<=n && !ok;++i) {        int num = m*i;        for (int j=0;j<=num && !ok;++j)            if (j+2*(num-j) == n) {                ok = true;                ans = num;                break;            }    }    if (!ok)        cout << -1;    else        cout << ans;    return 0;}

E (Codeforces 269 Div2 A)

分情况判断下。。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <stack>#include <cassert>#include <algorithm>#include <cmath>#include <set>#include <limits>using namespace std;#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MAX(a, b) ((a) > (b) ? (a) : (b))#define F(i, n) for(int (i)=0;(i)<(n);++(i))#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))#define MAXN 100000#define MAXM 10000#define MOD 10000007#define PI 3.1415926535897932384626433832795typedef long long LL;const double maxdouble = numeric_limits<double>::max();const double eps = 1e-10;const int INF = 0x7FFFFFFF;int cnt[10];vector<int> buf;int main() {    //freopen("input.in", "r", stdin);    int tmp;    REP(i, 0, 5) {        cin >> tmp;        cnt[tmp]++;    }    bool ok = false;    REP(i, 0, 9)        if (cnt[i] > 0) {            if (cnt[i] >= 4) {                cnt[i] -= 4;                ok = true;            }            if (cnt[i] > 0) {                REP(j, 0, cnt[i]-1)                    buf.push_back(i);            }        }    if (!ok)        cout << "Alien\n";    else {        //cout << buf.size() << " " << buf[0] << " " << buf[1] << endl;        if (buf[0] != buf[1])            cout << "Bear\n";        else            cout << "Elephant\n";    }    return 0;

F (Codeforces 269 Div2 B)

找连续相等的部分。。然后。。这种和谐的数据，随便搞吧 = =||

因为是 DIV2 ，还是给出我丑陋的代码。。附加一份比较好的代码，对比看看吧~~

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <stack>#include <cassert>#include <algorithm>#include <cmath>#include <set>#include <limits>using namespace std;#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MAX(a, b) ((a) > (b) ? (a) : (b))#define F(i, n) for(int (i)=0;(i)<(n);++(i))#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))#define MAXN 100000#define MAXM 10000#define MOD 10000007#define PI 3.1415926535897932384626433832795typedef long long LL;const double maxdouble = numeric_limits<double>::max();const double eps = 1e-10;const int INF = 0x7FFFFFFF;int cnt[10];vector<int> buf;int main() {    //freopen("input.in", "r", stdin);    int tmp;    REP(i, 0, 5) {        cin >> tmp;        cnt[tmp]++;    }    bool ok = false;    REP(i, 0, 9)        if (cnt[i] > 0) {            if (cnt[i] >= 4) {                cnt[i] -= 4;                ok = true;            }            if (cnt[i] > 0) {                REP(j, 0, cnt[i]-1)                    buf.push_back(i);            }        }    if (!ok)        cout << "Alien\n";    else {        //cout << buf.size() << " " << buf[0] << " " << buf[1] << endl;        if (buf[0] != buf[1])            cout << "Bear\n";        else            cout << "Elephant\n";    }    return 0;}

牛犇的代码：

#include <iostream>#include <cstdio>#include <vector>#include <cmath>#include <algorithm>#include <iomanip>#include <string>#include <set>#include <stack>#define mp(x, y) make_pair(x, y)#define all(x)   x.begin(), x.end()#define eps 1e-6using namespace std;int main(){  //freopen("input.txt", "r", stdin);  int n; cin >> n;  vector<pair<int, int> > g(n), k(n), l(n);    for (int i = 0; i < n; ++i) {    cin >> g[i].first;    g[i].second = i + 1;  }  sort(all(g));  k = g;  l = g;  bool kk = false;  for (int i = 0; i < n - 1; ++i) {    if (g[i].first == g[i + 1].first && !kk) {      kk = true;      swap(k[i], k[i + 1]);      continue;    }    if (g[i].first == g[i + 1].first && kk)  {      swap(l[i], l[i + 1]);      cout << "YES" << endl;      for (int w = 0; w < n; ++w)        cout << g[w].second << " ";      cout << endl;      for (int w = 0; w < n; ++w)        cout << k[w].second << " ";      cout << endl;      for (int w = 0; w < n; ++w)        cout << l[w].second << " ";      cout << endl;      return 0;    }  }  cout << "NO" << endl;     return 0;}

By slowlight93