A

无trick水题。。。

  1.  /*
  2.   * Author:  Plumrain
  3.   * Created Time:  2013-12-24 22:26
  4.   * File Name: B.cpp
  5.   */
  6.  #include <iostream>
  7.  #include <cstdio>
  8.  #include <cstring>
  9.  #include <string>
  10.  #include <cmath>
  11.  #include <algorithm>
  12.  #include <vector>
  13.  #include <cstdlib>
  14.  #include <sstream>
  15.  #include <fstream>
  16.  #include <list>
  17.  #include <deque>
  18.  #include <queue>
  19.  #include <stack>
  20.  #include <map>
  21.  #include <set>
  22.  #include <bitset>
  23.  #include <cctype>
  24.  #include <ctime>
  25.  #include <utility>
  26.  
  27.  using namespace std;
  28.  
  29.  #define clr0(x) memset(x, 0, sizeof(x))
  30.  #define clr1(x) memset(x, -1, sizeof(x))
  31.  #define pb push_back
  32.  #define sz(v) ((int)(v).size())
  33.  #define all(t) t.begin(),t.end()
  34.  #define INF 999999999999999999
  35.  #define zero(x) (((x)>0?(x):-(x))<eps)
  36.  #define out(x) cout<<#x<<":"<<(x)<<endl
  37.  #define tst(a) cout<<a<<" "
  38.  #define tst1(a) cout<<#a<<endl
  39.  #define CINBEQUICKER std::ios::sync_with_stdio(false)
  40.  
  41.  const double eps = 1e-;
  42.  const double PI = atan(1.0)*;
  43.  const int inf =  / ;
  44.  
  45.  typedef vector<int> vi;
  46.  typedef vector<string> vs;
  47.  typedef vector<double> vd;
  48.  typedef pair<int, int> pii;
  49.  typedef long long int64;
  50.  
  51.  inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;}
  52.  
  53.  int ru[], chu[];
  54.  
  55.  int main()
  56.  {
  57.  //    freopen("a.in","r",stdin);
  58.  //    freopen("a.out","w",stdout);
  59.  //    std::ios::sync_with_stdio(false);
  60.      int n, m;
  61.      while (scanf ("%d%d", &n , &m) != EOF){
  62.          int t1, t2, w;
  63.          clr0 (chu); clr0 (ru);
  64.          for (int i = ; i < m; ++ i){
  65.              scanf ("%d%d%d", &t1, &t2, &w);
  66.              chu[--t1] += w;
  67.              ru[--t2] += w;
  68.          }
  69.          int ans = ;
  70.          for (int i = ; i < n; ++ i)
  71.              ans += abs(chu[i] - ru[i]);
  72.          printf ("%d\n", ans / );
  73.      }
  74.      return ;
  75.  }

B

YY题。。。结论直接看代码就好了。。。比赛的时候YY出了结论算样例算错了，然后就去想别的方法了。。。。。。

  1.  /*
  2.   * Author:  Plumrain
  3.   * Created Time:  2013-12-24 22:26
  4.   * File Name: B.cpp
  5.   */
  6.  #include <iostream>
  7.  #include <cstdio>
  8.  #include <cstring>
  9.  #include <string>
  10.  #include <cmath>
  11.  #include <algorithm>
  12.  #include <vector>
  13.  #include <cstdlib>
  14.  #include <sstream>
  15.  #include <fstream>
  16.  #include <list>
  17.  #include <deque>
  18.  #include <queue>
  19.  #include <stack>
  20.  #include <map>
  21.  #include <set>
  22.  #include <bitset>
  23.  #include <cctype>
  24.  #include <ctime>
  25.  #include <utility>
  26.  
  27.  using namespace std;
  28.  
  29.  #define clr0(x) memset(x, 0, sizeof(x))
  30.  #define clr1(x) memset(x, -1, sizeof(x))
  31.  #define pb push_back
  32.  #define sz(v) ((int)(v).size())
  33.  #define all(t) t.begin(),t.end()
  34.  #define INF 999999999999999999
  35.  #define zero(x) (((x)>0?(x):-(x))<eps)
  36.  #define out(x) cout<<#x<<":"<<(x)<<endl
  37.  #define tst(a) cout<<a<<" "
  38.  #define tst1(a) cout<<#a<<endl
  39.  #define CINBEQUICKER std::ios::sync_with_stdio(false)
  40.  
  41.  const double eps = 1e-;
  42.  const double PI = atan(1.0)*;
  43.  const int inf =  / ;
  44.  
  45.  typedef vector<int> vi;
  46.  typedef vector<string> vs;
  47.  typedef vector<double> vd;
  48.  typedef pair<int, int> pii;
  49.  typedef long long int64;
  50.  
  51.  inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;}
  52.  
  53.  int ru[], chu[];
  54.  
  55.  int main()
  56.  {
  57.  //    freopen("a.in","r",stdin);
  58.  //    freopen("a.out","w",stdout);
  59.  //    std::ios::sync_with_stdio(false);
  60.      int n, m;
  61.      while (scanf ("%d%d", &n , &m) != EOF){
  62.          int t1, t2, w;
  63.          clr0 (chu); clr0 (ru);
  64.          for (int i = ; i < m; ++ i){
  65.              scanf ("%d%d%d", &t1, &t2, &w);
  66.              chu[--t1] += w;
  67.              ru[--t2] += w;
  68.          }
  69.          int ans = ;
  70.          for (int i = ; i < n; ++ i)
  71.              ans += abs(chu[i] - ru[i]);
  72.          printf ("%d\n", ans / );
  73.      }
  74.      return ;
  75.  }

C

题意：给一个很大的数m，这个数的各个位上的数字中一定含有至少1个1,6,8,9。你可以重新组织所有数字的顺序，使得重新排列之后的数能被7整除。10^4 <= m <= 10^(10^6)。

　　　注意，排列之后的数不能含有前导0。

解法：把1,6,8,9四个数字放在最后面，根据前面的数*10000除以7的余数，来决定1,6,8,9四个数字的排列顺序即可。至于有没有前导0，特殊处理一下即可。

tag:math, think

  1.  /*
  2.   * Author:  Plumrain
  3.   * Created Time:  2013-12-25 14:09
  4.   * File Name: C.cpp
  5.   */
  6.  #include <iostream>
  7.  #include <cstdio>
  8.  #include <cstring>
  9.  #include <string>
  10.  #include <cmath>
  11.  #include <algorithm>
  12.  #include <vector>
  13.  #include <cstdlib>
  14.  #include <sstream>
  15.  #include <fstream>
  16.  #include <list>
  17.  #include <deque>
  18.  #include <queue>
  19.  #include <stack>
  20.  #include <map>
  21.  #include <set>
  22.  #include <bitset>
  23.  #include <cctype>
  24.  #include <ctime>
  25.  #include <utility>
  26.  
  27.  using namespace std;
  28.  
  29.  #define clr0(x) memset(x, 0, sizeof(x))
  30.  #define clr1(x) memset(x, -1, sizeof(x))
  31.  #define pb push_back
  32.  #define sz(v) ((int)(v).size())
  33.  #define all(t) t.begin(),t.end()
  34.  #define INF 999999999999999999
  35.  #define zero(x) (((x)>0?(x):-(x))<eps)
  36.  #define out(x) cout<<#x<<":"<<(x)<<endl
  37.  #define tst(a) cout<<a<<" "
  38.  #define tst1(a) cout<<#a<<endl
  39.  #define CINBEQUICKER std::ios::sync_with_stdio(false)
  40.  
  41.  const double eps = 1e-;
  42.  const double PI = atan(1.0)*;
  43.  const int inf =  / ;
  44.  
  45.  typedef vector<int> vi;
  46.  typedef vector<string> vs;
  47.  typedef vector<double> vd;
  48.  typedef pair<int, int> pii;
  49.  typedef long long int64;
  50.  
  51.  inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;}
  52.  
  53.  bool del[];
  54.  string temp = "";
  55.  map<int, string> mp;
  56.  
  57.  void gao(string s)
  58.  {
  59.      stringstream stm(s);
  60.      int num; stm >> num;
  61.      int yu = num % ;
  62.      if (!mp.count(yu)) mp[yu] = s;
  63.  }
  64.  
  65.  int main()
  66.  {
  67.  //    freopen("a.in","r",stdin);
  68.  //    freopen("a.out","w",stdout);
  69.  //    std::ios::sync_with_stdio(false);
  70.      mp.clear();
  71.      string tt = "";
  72.      gao(tt);
  73.      while (next_permutation(tt.begin(), tt.end())) gao(tt);
  74.  
  75.      string s;
  76.      while (cin >> s){
  77.          clr0 (del);
  78.          string ss; ss.clear();
  79.          int n = sz(s), cnt = ;
  80.          for (int i = ; i < n; ++ i){
  81.              if (s[i] == ''){
  82.                  ++ cnt; continue;
  83.              }
  84.              bool ok = ;
  85.              for (int j = ; j < ; ++ j) if (s[i] == temp[j] && !del[j]){
  86.                  del[j] = ; ok = ;
  87.              }
  88.              if (!ok) ss.pb (s[i]);
  89.          }
  90.  
  91.          int len = sz(ss);
  92.          if (!len){
  93.              ss = mp[];
  94.              for (int i = ; i < cnt; ++ i) ss.pb ('');
  95.              cout << ss << endl;
  96.              continue;
  97.          }
  98.          for (int i = ; i < cnt; ++ i) ss.pb ('');
  99.          len = sz(ss);
  100.          int flag = ;
  101.          for (int i = ; i < len; ++ i)
  102.              flag = (flag* + ss[i] - '')  % ;
  103.          flag = flag *  % ;
  104.          ss += mp[( - flag) % ];
  105.          cout << ss << endl;
  106.      }
  107.      return ;
  108.  }

D

题意：有一个0，1矩阵(最大5000*5000)，你可以无限次数地交换任意两行的位置。求交换之后，单个只含有1的矩形的面积最大，并返回这个面积值。

解法：枚举矩形的左下角是从哪一列开始，统计每一行从这一列开始连续的1有多少个记录在num数组里，然后从大到小遍历num数组，并更新面积值即可。

tag:dp, think, good

  1.  /*
  2.   * Author:  Plumrain
  3.   * Created Time:  2013-12-26 12:49
  4.   * File Name: D.cpp
  5.   */
  6.  #include <iostream>
  7.  #include <cstdio>
  8.  #include <cstring>
  9.  #include <string>
  10.  #include <cmath>
  11.  #include <algorithm>
  12.  #include <vector>
  13.  #include <cstdlib>
  14.  #include <sstream>
  15.  #include <fstream>
  16.  #include <list>
  17.  #include <deque>
  18.  #include <queue>
  19.  #include <stack>
  20.  #include <map>
  21.  #include <set>
  22.  #include <bitset>
  23.  #include <cctype>
  24.  #include <ctime>
  25.  #include <utility>
  26.  
  27.  using namespace std;
  28.  
  29.  #define clr0(x) memset(x, 0, sizeof(x))
  30.  #define clr1(x) memset(x, -1, sizeof(x))
  31.  #define pb push_back
  32.  #define sz(v) ((int)(v).size())
  33.  #define all(t) t.begin(),t.end()
  34.  #define INF 999999999999999999
  35.  #define zero(x) (((x)>0?(x):-(x))<eps)
  36.  #define out(x) cout<<#x<<":"<<(x)<<endl
  37.  #define tst(a) cout<<a<<" "
  38.  #define tst1(a) cout<<#a<<endl
  39.  #define CINBEQUICKER std::ios::sync_with_stdio(false)
  40.  
  41.  const double eps = 1e-;
  42.  const double PI = atan(1.0)*;
  43.  const int inf =  / ;
  44.  
  45.  typedef vector<int> vi;
  46.  typedef vector<string> vs;
  47.  typedef vector<double> vd;
  48.  typedef pair<int, int> pii;
  49.  typedef long long int64;
  50.  
  51.  inline int Mymod (int a, int b) {int x=a%b; if(x<) x+=b; return x;}
  52.  
  53.  int n, m;
  54.  int p[][], num[];
  55.  char v[][];
  56.  
  57.  int main()
  58.  {
  59.  //    freopen("a.in","r",stdin);
  60.      //freopen("a.out","w",stdout);
  61.  //    std::ios::sync_with_stdio(false);
  62.      while (scanf ("%d%d", &n, &m) != EOF){
  63.          string s;
  64.          for (int i = ; i < n; ++ i)
  65.              scanf ("%s", v[i]);
  66.          for (int i = ; i < n; ++ i){
  67.              int tmp = m;
  68.              for (int j = m-; j >= ; -- j){
  69.                  if (v[i][j] == '')
  70.                      tmp = j, p[i][j] = j;
  71.                  else
  72.                      p[i][j] = tmp;
  73.              }
  74.          }
  75.          int ans = ;
  76.          for (int i = ; i < m; ++ i){
  77.              clr0 (num);
  78.              for (int j = ; j < n; ++ j) num[p[j][i] - i] ++;
  79.              int pos = ;
  80.              while (pos && num[pos] == ) -- pos;
  81.              int cnt = ;
  82.              while (pos){
  83.                  if (num[pos]) cnt += num[pos];
  84.                  ans = max(cnt*pos, ans);
  85.                  -- pos;
  86.              }
  87.          }
  88.          printf ("%d\n", ans);
  89.      }
  90.      return ;
  91.  }

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