PAT乙级1041-1060

1041 考试座位号

#include <iostream>
#include <map>
#include <string>
#include <utility>

using namespace std;

int main() {

  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  map<int, pair<string, int>> maps;

  for (int i = 0; i != N; ++i) {
    string a;
    int b, c;
    cin >> a >> b >> c;
    maps[b] = {a, c};
  }

  cin >> N;

  for (int i = 0; i != N; ++i) {
    int b;
    cin >> b;
    cout << maps[b].first << " " << maps[b].second << endl;
  }
  return 0;
}

1042 字符统计

#include <iostream>
#include <map>
#include <string>

using namespace std;

int main(){

  ios::sync_with_stdio(false);
  map<char,int> counts;
  int max = 0;
  char maxchar = 'a' - 1;
  char c;
  while( (c = cin.get()) != '\n'){
    if (!isalpha(c))
      continue;
    c = tolower(c);
    ++counts;
    if (counts > max){
      max = counts;
      maxchar = c;
    }else if (counts == max && c < maxchar){
      maxchar = c;
    }
  }

  cout << maxchar << " " << max << endl;
  return 0;

}

1043 输出PATest

#include <iostream>
#include <map>
#include <vector>

using namespace std;

int main() {

  ios::sync_with_stdio(false);
  map<char, int> counts;
  char c;
  while ((c = cin.get()) != '\n') {
    ++counts;
  }

  vector<char> PATest = {'P', 'A', 'T', 'e', 's', 't'};

  while (true) {
    bool allempty = true;
    for (int i = 0; i != 6; ++i)
      if (counts[PATest[i]] > 0) {
        cout << PATest[i];
        --counts[PATest[i]];
        allempty = false;
      }
    if (allempty)
      break;
  }

  return 0;
}

1044 火星数字

// 13进制转换的题目啊
#include <iostream>
#include <map>
#include <string>
#include <vector>

using namespace std;

std::map<string, int> store = {
    {"tret", 0},     {"jan", 1},      {"feb", 2},       {"mar", 3},
    {"apr", 4},      {"may", 5},      {"jun", 6},       {"jly", 7},
    {"aug", 8},      {"sep", 9},      {"oct", 10},      {"nov", 11},
    {"dec", 12},     {"tam", 13 * 1}, {"hel", 13 * 2},  {"maa", 13 * 3},
    {"huh", 13 * 4}, {"tou", 13 * 5}, {"kes", 13 * 6},  {"hei", 13 * 7},
    {"elo", 13 * 8}, {"syy", 13 * 9}, {"lok", 13 * 10}, {"mer", 13 * 11},
    {"jou", 13 * 12}};

vector<string> gewei = {"tret", "jan", "feb", "mar", "apr", "may", "jun",
                        "jly",  "aug", "sep", "oct", "nov", "dec"};

vector<string> shiwei = {"",    "tam", "hel", "maa", "huh", "tou", "kes",
                         "hei", "elo", "syy", "lok", "mer", "jou"};

int main() {

  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  cin.get();
  for (int i = 0; i != N; ++i) {
    string input;
    getline(cin, input);
    if (isdigit(input[0])) { //说明这是个数字
      int num = stoi(input);
      if (num >= 13) {
        cout << shiwei[num / 13] ;
        num = num % 13;
        if (num != 0)
          cout << " " << gewei[num];
        cout << endl;
      }else{
        cout << gewei[num] << endl;
      }
    } else {
      int num = 0;
      if (input.find(" ") != string::npos) { //有高位
        string shi = input.substr(0, input.find(" "));
        num += store[shi];
        string ge = input.substr(input.find(" ") + 1, input.size());
        num += store[ge];
      } else {
        num += store[input];
      }
      cout << num << endl;
    }
  }

  return 0;
}

1045 快速排序

动态规划法

//想到动态规划。因为实际上你是在找这么一个值，什么值呢？左边的数字都比他小，右边的数字都比他大。
//这样想到的暴力方法是做循环，比如说你拿出第N个元素来，你把他左边的和他全部比一遍，右边的再比一遍
//这样重复的比较其实特别多，这些重复的子比较都可以纳入一个数组来管理，减少数组对比的时间
//因此考虑用一个max[i]表示从头到i元素的最大值，用一个min[i]表示从i元素到尾部的最小值。
//不包括i元素本身。

#include <algorithm>
#include <iostream>
#include <limits>
#include <vector>

#define INT_MAX numeric_limits<int>::max()

using namespace std;

int main() {

  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<int> list(N, 0);
  vector<int> min(N, INT_MAX);
  vector<int> max(N, 0);
  cin >> list[0];
  if (N == 1) {
    cout << 1 << endl;
    cout << list[0] << endl;
    return 0;
  }
  int lastnum = list[0];
  for (int i = 1; i != N; ++i) {
    int num;
    cin >> num;
    if (lastnum > max[i - 1])
      max[i] = lastnum;
    else
      max[i] = max[i - 1];
    lastnum = num;
    list[i] = num;
  }
  max[0] = 0;

  for (int i = N - 1; i > 0; --i) {
    if (list[i] < min[i]) {
      min[i - 1] = list[i];
    } else
      min[i - 1] = min[i];
  }
  min[N - 1] = INT_MAX;
  vector<int> result;
  for (int i = 0; i != N; ++i) {
    if (max[i] < list[i] and min[i] > list[i]){
      result.push_back(list[i]);
    }
  }
  sort(result.begin(),result.end());

  cout << result.size() << endl;
  if(result.empty()){   //即使是空的，也必须输出一个空行，这坑是节点2，总之猝不及防。
    cout << endl;
    return 0;
  }
  for( int i = 0 ; i != result.size() - 1 ; ++i){
    cout << result[i] << " ";
  }
  cout << result.back() << endl;
  return 0;

}

暴力解法

#include <iostream>
#include <vector>

using namespace std;

int main() {

  vector<int> input;
  int k, num;
  cin >> k;
  while (k--) {
    cin >> num;
    input.push_back(num);
  }
  vector<int> result;
  int max = input[0] - 1;
  for (auto i : input) {
    //因为保证输入中各数字互不相同，因此不存在i == max的情况
    if (i > max) {
      max = i;
      result.push_back(max);
    } else if (i < max) {
      for (auto it = result.begin(); it != result.end(); ++it) {
        if (*it > i)
          it = result.erase(it);
        if (it == result.end())
          break;
      }
    }
  }

  if (result.size() == 0) {
    cout << "0" << endl;
    cout << endl;
    return 0;
  }

  cout << result.size() << endl;
  for (auto it = result.begin(); it != result.end() - 1; ++it)
    cout << *it << " ";
  cout << result.back();

  return 0;
}

再提供一种解法

#include <iostream>
#include <vector>
//#include <fstream>

using namespace std;

int main() {
  //    ifstream in("in.txt");
  //    cin.rdbuf( in.rdbuf() );
  size_t N;
  cin >> N;
  vector<int> Arr(N);
  vector<int> Tmp(N, 0);
  for (size_t i = 0; i != N; ++i) {
    cin >> Arr[i];
  }
  int Max = Arr[0];
  for (size_t i = 0; i != N; ++i) {
    if (Max <= Arr[i]) {
      Max = Arr[i];
      ++Tmp[i];
    }
  }
  int Min = Arr[N - 1];
  for (size_t i = N - 1; i != -1; --i) {
    if (Min >= Arr[i]) {
      Min = Arr[i];
      ++Tmp[i];
    }
  }
  vector<int> Res;
  for (size_t i = 0; i != N; ++i) {
    if (2 == Tmp[i]) {
      Res.push_back(Arr[i]);
    }
  }
  cout << Res.size() << endl;
  if (0 == Res.size()) {
    cout << endl;
    return 0;
  }
  for (size_t i = 0; i != Res.size(); ++i) {
    if (i != Res.size() - 1)
      cout << Res[i] << " ";
    else
      cout << Res[i] << endl;
  }
  return 0;
}

1046 划拳

#include<iostream>
#include<vector>

using namespace std;

int main(){

  vector<int> drink(2,0);
  int N;
  cin >> N;
  for(int i = 0 ; i != N ; ++i){
    int jiahan,jiahua,yihan,yihua;
    cin >> jiahan >> jiahua>>yihan>>yihua;
    if (jiahua == yihua)
      continue;
    else if (jiahua == jiahan + yihan){
      ++drink[1];
    }else if (yihua == jiahan + yihan){
      ++drink[0];
    }
  }
  cout << drink[0] << " " <<drink[1] << endl;
  return 0;

}

1047 编程团体赛

python

#就这都能用85ms
maxscore = 0
maxteam  = -1

N = int(input())

total_score = dict()

for i in range(0,N):
  team_player,score = input().split()
  score = int(score)
  team,player = team_player.split('-')
  team = int(team)
  if team in total_score:
    total_score[team] += score
  else:
    total_score[team] = score;
  if total_score[team] > maxscore:
    maxscore = total_score[team]
    maxteam = team

print(maxteam,maxscore)

cpp

//9ms
#include<string>
#include<iostream>
#include<map>

using namespace std;

int main() {

    string input;
    int k;
    cin >> k;
    std::getline(cin, input);
    std::map<string, int> result;
    string winner;
    int max = 0, this_team;

    while (k--) {
        std::getline(cin, input);
        this_team = (result[input.substr(0, input.find('-'))] += std::stoi(input.substr(input.find(' ') + 1)));
        if (this_team > max) {
            winner = input.substr(0, input.find('-'));
            max = this_team;
        }
    }

    cout << winner << " " << result[winner] << endl;

    return 0;

}

1048 数字加密

#include<iostream>
#include<string>

//查了网上的分享才知道当A长度大于B时，B前面要补齐为0，所以怎么排BUG都排不出来……
//我认为这样的坑毫无意义，纯粹是题意不清。

using namespace std;

inline char process_jishu(char a, char b) {

    int i = (a - '0' + b - '0') % 13;
    if (i == 12)
        return 'K';
    else if (i == 11)
        return 'Q';
    else if (i == 10)
        return 'J';
    else return i + '0';

}

inline char process_oushu(char a, char b) {

    int i = b - a;
    return i < 0 ? i + 10 + '0' : i + '0';

}

int main() {

    string A, B;
    cin >> A >> B;
    int i = 1;
    if (A.size() > B.size())
        B = string( A.size() - B.size(),'0') + B;
    for (auto itB = B.rbegin(), itA = A.rbegin(); itB != B.rend() && itA != A.rend(); ++itB, ++itA) {
        if (i % 2) {//奇数
            *itB = process_jishu(*itA, *itB);
        }
        else {
            *itB = process_oushu(*itA, *itB);
        }
        ++i;
    }

    cout << B << endl;

    return 0;
}

1049 数列的片段和

//这道题肯定是用乘法，考点在于每个数乘多少次
//稍微观察一下就能看出规律。
#include <iomanip>
#include <iostream>

using namespace std;

int main() {

  ios::sync_with_stdio(false);
  int k;
  double num;
  cin >> k;
  double result = 0.0;
  for (int i = 0; i != k; ++i) {
    cin >> num;
    result += num * (i + 1) * (k - i);
  }

  cout << std::setiosflags(std::ios::fixed) << std::setprecision(2) << result;

  return 0;
}

1050 螺旋矩阵

//无他，但心细尔。
#include <algorithm>
#include <iostream>
#include <math.h>
#include <vector>

using namespace std;

int getm(int N) {

  for (int i = sqrt(N); i <= N; ++i) {
    if (N % i == 0)
      return i;
  }
}

int main() {

  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  int m = getm(N);
  int n = N / m;
  if (m < n) {
    int tmp = m;
    m = n;
    n = tmp;
  }
  vector<vector<int>> output(m, vector<int>(n, 0));
  vector<int> input(N);
  for (int i = 0; i != N; ++i) {
    cin >> input[i];
  }
  sort(input.rbegin(), input.rend());
  int row = 0;
  int column = n - 1;
  int row_up = m - 1;
  int column_left = 0;
  int idx = 0;
  while (true) {
    for (int j = column_left; j <= column; ++j) {
      output[row][j] = input[idx++];
    }
    ++row;
    if (idx == N)
      break;
    for (int j = row; j <= row_up; ++j) {
      output[j][column] = input[idx++];
    }
    --column;
    if (idx == N)
      break;
    for (int j = column; j >= column_left; --j) {
      output[row_up][j] = input[idx++];
    }
    --row_up;
    if (idx == N)
      break;
    for (int j = row_up; j >= row; --j) {
      output[j][column_left] = input[idx++];
    }
    ++column_left;
    if (idx == N)
      break;
  }

  for (int i = 0; i != m; ++i) {
    for (int j = 0; j != n - 1; ++j) {
      cout << output[i][j] << " ";
    }
    cout << output[i].back() << endl;
  }
  return 0;
}

1051 复数乘法

//哎呦妈呀，要是这道题不能AC，不用怀疑，肯定是输出有问题。而不是计算有问题。
//这么复杂精细恶心的输出，简直别无分店呦。

#include<iostream>
#include<iomanip>
#include<math.h>

using namespace std;

class fushu {

    friend std::ostream &operator<< (std::ostream &os, const fushu &a);
    double shi = 0;
    double xu = 0;
public:
    fushu() {};
    fushu(double a, double b) :shi(a), xu(b) {};
    fushu(double a, double b, int c) {
        shi = a * cos(b);
        xu = a *  sin(b);
    };
    fushu operator *(const fushu &a) {
            fushu b; b.shi = shi * a.shi - xu * a.xu; b.xu = shi *a.xu + xu * a.shi; return b;
        };
};

std::ostream &operator<< (std::ostream &os, const fushu &a) {

    if (-0.005 < a.shi && a.shi< 0) {
        cout << "0.00";
    }
    else {
        cout << std::setiosflags(std::ios::fixed) << std::setprecision(2) << a.shi;
    }

    if (a.xu >= 0) {
        cout << "+" << std::setiosflags(std::ios::fixed) << std::setprecision(2) << a.xu << "i";
    }
    else if (-0.005 < a.xu && a.xu < 0) {
        cout << "+0.00i";
    }else {
        cout << std::setiosflags(std::ios::fixed) << std::setprecision(2) << a.xu << "i";
    }

    return os;
}

int main() {

    double R1, P1, R2, P2;
    cin >> R1 >> P1 >> R2 >> P2;
    fushu f1(R1, P1, 0), f2(R2, P2, 0);

    cout << f1 * f2 << endl;

    return 0;

}

1052 卖个萌

#include <iostream>
#include <string>
#include <vector>

using namespace std;

int main() {

  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  string line;
  vector<string> fuhao[3]; //分别是手足口
  for (int i = 0; i != 3; ++i) {
    getline(cin, line);
    for (auto it = line.begin(); it != line.end();) {
      string tmp;
      if (*it == '[') {
        ++it;
        while (*it != ']') {
          tmp += *it;
          ++it;
        }
        fuhao[i].push_back(tmp);
      } else {
        ++it;
      }
    }
  }
  int N;
  cin >> N;
  string output;
  for (int i = 0; i != N; ++i) {
    for (int j = 0; j != 5; ++j) {
      int index = j > 2 ? 4 - j : j;
      int tmp;
      cin >> tmp;
      if (tmp > fuhao[index].size() || tmp < 1) {
        output.clear();
        output = "Are you kidding me? @\\/@";
        break;
      } else {
        output +=
            (j == 1 ? "(" : "") + fuhao[index][tmp - 1] + (j == 3 ? ")" : "");
      }
    }
    cout << output << endl;
    output.clear();
  }
  return 0;
}

1053 住房空置率

#include <iostream>

using namespace std;

int main() {

  ios::sync_with_stdio(false);
  int house_num, date_num;
  float e;
  cin >> house_num >> e >> date_num;
  int maybe_empty = 0, empty = 0;
  for (int i = 0; i != house_num; ++i) {
    int K, low_day_num = 0;
    cin >> K;
    for (int j = 0; j != K; ++j) {
      float tmp;
      cin >> tmp;
      if (tmp < e) {
        ++low_day_num;
      }
    }
    if (low_day_num > K / 2) {
      if (K > date_num)
        ++empty;
      else
        ++maybe_empty;
    }
  }
  cout.flags(ios::fixed);
  cout.precision(1);
  cout << maybe_empty * 100 / 1.0 / house_num << "% " << empty * 100 / 1.0 / house_num << "%" << endl;
}

1054 求平均值

#include <iostream>
#include <string>
#include <vector>

using namespace std;

bool islegal(const string &a) {
  int dotnum = 0;
  int digit_after_dot_num = 0;
  bool digit = false;
  for (int i = 0; i != a.size(); ++i) {
    if (!isdigit(a[i]) && a[i] != '+' && a[i] != '-' && a[i] != '.') {
      return false;
    } else {
      if (isdigit(a[i])) {
        digit = true;
        if (dotnum != 0) {
          ++digit_after_dot_num;
          if (digit_after_dot_num > 2)
            return false;
        }
      } else if (digit && (a[i] == '+' || a[i] == '-')) {
        return false;
      } else if (a[i] == '.') {
        ++dotnum;
        if (dotnum > 1)
          return false;
      }
    }
  }
  //到这里至少看起来是个合法的数字了。
  if (stof(a) > 1000 || stof(a) < -1000)
    return false;
  else return true;
}

int main(){

  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  string inp;
  float total = 0.0;
  int legalnum = 0;
  vector<string> illegal;
  for(int i = 0 ; i != N ; ++i){
    cin >> inp;
    if (islegal(inp)){
      total += stof(inp);
      ++legalnum;
    }else{
      illegal.push_back(inp);
    }
  }

  for(auto it = illegal.begin() ; it != illegal.end() ; ++it){
    cout << "ERROR: " << *it <<" is not a legal number" << endl;
  }
  if (legalnum != 0){
    cout.flags(ios::fixed);
    cout.precision(2);
    cout << "The average of " << legalnum
         << (legalnum > 1 ? " numbers" : " number") << " is "
         << total / legalnum << endl;
  }else{
    cout << "The average of 0 numbers is Undefined" << endl;
  }
  return 0;
}

1055 集体照

//这道理理解起来最大的歧义在于，什么叫人身高相同，按名字的字典升序排列。
//这牵扯两个方面，一是身高相同的恰好在一列，那简单，按照升序排列就行了。
//关键是身高相同的如果不在同一列怎么办。到底是名字小的在前一排还是名字大的在前一排？
//答案会告诉你，实际上是名字大的在前一排。

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>

using namespace std;

struct Person {
  int height;
  string name;
};

int main() {
  ios::sync_with_stdio(false);
  int studentnum, linenum;
  cin >> studentnum >> linenum;
  vector<Person> students(studentnum);
  for (int i = 0; i != studentnum; ++i) {
    cin >> students[i].name >> students[i].height;
  }
  sort(students.begin(), students.end(), [](const Person &a, const Person &b) {
    return a.height < b.height || (a.height == b.height && a.name > b.name);
  }); //名字大的在前一排
  int persons_per_line = studentnum / linenum;
  vector<vector<string>> ordered;
  int lastindex = 0;
  for (int i = 0; i != linenum; ++i) {
    if (i == linenum - 1)
      persons_per_line = studentnum - (linenum - 1) * persons_per_line;
    vector<string> perline(persons_per_line);
    int pos = persons_per_line / 2; //不加1是因为从零开始
    int pianyi = -1;
    sort(students.begin() + lastindex,
         students.begin() + lastindex + persons_per_line,
         [](const Person &a, const Person &b) {
           return a.height > b.height ||
                  (a.height == b.height && a.name < b.name);
         });
    //这里是名字小的靠前
    for (int j = lastindex; j != lastindex + persons_per_line; ++j) {
      perline[pos] = students[j].name;
      pos += pianyi;
      pianyi = pianyi > 0 ? (pianyi + 1) * -1 : (pianyi - 1) * -1;
    }
    ordered.push_back(perline);
    lastindex += persons_per_line;
  }
  for (auto it = ordered.rbegin(); it != ordered.rend(); ++it) {
    for (auto it2 = it->begin(); it2 != it->end() - 1; ++it2) {
      cout << *it2 << " ";
    }
    cout << it->back() << endl;
  }

  return 0;
}

1056 组合数的和

#include<iostream>

using namespace std;

int main(){

  int N;
  cin >> N;
  int sum = 0;
  for(int i = 0 ; i != N ; ++i){
    int tmp;
    cin >> tmp;
    sum += tmp;
  }

  cout << sum * (N -1) * 11 << endl;
  return 0;

}

1057 数零壹

#include <iostream>
#include <string>

using namespace std;

int main() {

  string line;
  getline(cin, line);
  int sum = 0;
  for (int i = 0; i != line.size(); ++i) {
    if (isalpha(line[i])) {
      sum += tolower(line[i]) - 'a' + 1;
    }
  }
  int num[] = {0, 0};
  while (sum != 0) {
    int r = sum % 2;
    sum /= 2;
    ++num[r];
  }
  cout << num[0] << " " << num[1] << endl;
  return 0;
}

1058 选择题

//这道题算是1073的简化版吗？
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <sstream>
#include <string>
#include <utility>
#include <vector>

#include <fstream>

using namespace std;

#define RIGHT 0
#define WRONG 1

struct Timu {
  int score;    //分值
  int xx_num;   //选项数量
  int r_xx_num; //正确选项数量
  vector<char> xxs;
};

void parasline(const string &line, const int timunum,
               vector<vector<char>> &res) {

  istringstream is(line);
  char kuohao;
  for (int i = 0; i != timunum; ++i) {
    int num = 0;
    is >> kuohao;
    is >> num;
    for (int j = 0; j != num; ++j) {
      char c;
      is >> c;
      res[i].push_back(c);
    }
    sort(res[i].begin(),res[i].end());
    is >> kuohao;
  }
}

int pigai(const vector<vector<char>> &choose, const vector<Timu> &timus,
          const int timunum, map<int, int> &wrongs, int &mostwrong) {
  int res = 0;
  for (int i = 0; i != timunum; ++i) {
    int status = RIGHT;
    if (choose[i] == timus[i].xxs) {
      res += timus[i].score;
    } else {
      ++wrongs[i];
      if (wrongs[i] > mostwrong)
        mostwrong = wrongs[i];
    }
  }
  return res;
}

int main() {

  int student_num, timunum;
  cin >> student_num >> timunum;
  vector<Timu> timus(timunum);
  for (int i = 0; i != timunum; ++i) {
    cin >> timus[i].score >> timus[i].xx_num >> timus[i].r_xx_num;
    for (int j = 0; j != timus[i].r_xx_num; ++j) {
      char c;
      cin >> c;
      timus[i].xxs.push_back(c);
    }
    sort(timus[i].xxs.begin(),timus[i].xxs.end());
  }
  map<int, int> wrongs; // key是题目序号和错误选项，int是错误次数
  cin.get();
  int mostwrong = 0;
  vector<int> students(student_num, 0.0);
  for (int i = 0; i != student_num; ++i) {
    string line;
    getline(cin, line);
    vector<vector<char>> choose(timunum);
    parasline(line, timunum, choose);
    students[i] += pigai(choose, timus, timunum, wrongs, mostwrong);
  }
  for (auto it = students.begin(); it != students.end(); ++it) {
    cout << *it << endl;
  }
  if (wrongs.empty()) {
    cout << "Too simple" << endl;
    return 0;
  }
  cout << mostwrong;
  for (auto it = wrongs.begin(); it != wrongs.end(); ++it) {
    if (it->second == mostwrong) {
      cout << " " << it->first + 1;
    }
  }

  return 0;
}

1059 C语言竞赛

#include <cstdio>
#include <map>
#include <math.h>
#include <vector>

using namespace std;

bool issu(int a) {

  if (a == 2 || a == 3)
    return true;
  else if (a % 2 == 0)
    return false;

  for (int i = 3; i <= sqrt(a) + 1; i += 2) {
    if (a % i == 0)
      return false;
  }
  return true;
}

int main() {

  int compnents, query;
  scanf("%d", &compnents);
  map<int, int> reward;
  for (int i = 0; i != compnents; ++i) {
    int index;
    scanf("%d", &index);
    if (i == 0) {
      reward[index] = 1;
    } else if (issu(i + 1)) {
      reward[index] = 2;
    } else {
      reward[index] = 3;
    }
  }
  scanf("%d", &query);
  for (int i = 0; i != query; ++i) {
    int n;
    scanf("%d", &n);
    switch (reward[n]) {
    case 0:
      printf("%04d: Are you kidding?\n", n);
      break;
    case 1:
      printf("%04d: Mystery Award\n", n);
      reward[n] = 4;
      break;
    case 2:
      printf("%04d: Minion\n", n);
      reward[n] = 4;
      break;
    case 3:
      printf("%04d: Chocolate\n", n);
      reward[n] = 4;
      break;
    case 4:
      printf("%04d: Checked\n", n);
      break;
    }
  }

  return 0;
}

1060 爱丁顿数

//先考虑暴力解法，我先把任何一公里的天数求出来，比如A[i]表示超过i公里的天数
//之后再去遍历这个数组，找i和A[i]这个值对中最小值的最大值。
//但是问题在于，每次有一个大数出现后，就要更新之前全部的数组，这更新下来压力很大。
//该怎么办呢。所以肯定不行。考虑用一个数组保存每天跑了几公里。然后从大到小排个序
//之后从头开始数，如果公里数大于天数（天数从1开始，所以是index+1），那就继续往下数
//这就相当于数超过某公里的天数。结果当公里数不能大于天数了，就不能继续下去了。
//比如提供这么一个序列
//10
//6 6 6 9 3 10 8 2 7 8
//排序后得到[10, 9, 8, 8, 7, 6, 6, 6, 3, 2]从头往后数
//问题关键在于能不能数第一个6，也就是说当gongli[i] == i + 1的时候该怎么处理
//这么显然是不能继续往下数，因为题目要求是超过E公里，而不是等于E公里了。

#include <algorithm>
#include <cstdio>
#include <map>
#include <vector>

using namespace std;

int main() {

  // freopen("input","r",stdin);
  int days;
  scanf("%d", &days);
  vector<int> gongli(days, 0);
  int E = 0;
  for (int i = 0; i != days; ++i) {
    scanf("%d", &gongli[i]);
  }
  sort(gongli.rbegin(), gongli.rend());
  for (int i = 0; i != gongli.size(); ++i) {
    if (gongli[i] > i + 1)
      ++E;
    else break;
  }
  printf("%d\n", E);
  return 0;
}