CodeJam編程比賽全面詳解

一、CodeJam T恤

CodeJam（又稱谷歌編程大賽）是谷歌公司舉辦的一項全球性編程比賽，旨在為全球程序員提供一個交流平台和提高自己編程能力的機會。每年穀歌都會派發一些活動紀念品來豐富比賽參與者的收藏，其中最受歡迎的莫過於CodeJam T恤。這種T恤的設計很特別，上面包含著歷屆CodeJam的logo或者CodeJam比賽的難題名稱，很適合那些富有紀念意義的程序員們收藏。

CodeJam T-Shirt：

#include <iostream>
#include <string>
using namespace std;

int main()
{
  string tShirt;
  getline(cin, tShirt);
  cout << "I love my CodeJam T-shirt with " << tShirt << " printed on it!" << endl;
  return 0;
}

二、CodeJam是什麼比賽

CodeJam是一種面向全球程序員的編程比賽，更確切地說，CodeJam是一種演算法競賽。CodeJam比賽不需註冊費用，全程使用英語，官方旨在推廣計算機科學和編程，鼓勵優秀的問題解決能力及創新精神，並吸引人才加入Google。

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
  int testCases;
  cin >> testCases;
  
  for (int i = 1; i > a >> b >> c;
      double s = (a + b + c) / 2;
      double area = sqrt(s*(s-a)*(s-b)*(s-c));
      cout << "Case #" << i << ": " << area << endl;
  }
  
  return 0;
}

三、CodeJam難度

CodeJam比賽的難度層級比較高，涉及到演算法、數據結構、編程的全方面技能。難度從Round 1(R1)到Finals，依次遞增，同時每個Round附帶著若干題目，參賽選手需要根據難度選擇自己最為擅長的部分加以解決，操作難度和解題難度都需要一定的技巧。

#include <iostream>
#include <string>
using namespace std;

int main()
{
   int n;
   cin >> n;
   string s;
   getline(cin, s);
   
   for (int i = 0; i < n; i++) {
       getline(cin, s);
       int length = s.length();
       int j = 0;
       bool hasPlusSign = false;
       while (j < length) {
           if (s[j] == '+') {
               hasPlusSign = true;
               break;
           }
           j++;
       }
       if (hasPlusSign) {
           int result = 0;
           int k;
           for (k = 0; k < j; k++) {
               if (s[k] != ' ') {
                   result += (s[k] - '0');
                   result *= 10;
               }
           }
           result /= 10;
           cout << "Case #" << (i + 1) << ": " << result << endl;
       } else {
           cout << "Case #" << (i + 1) << ": " << j << endl;
       }
   }
   return 0;
}

四、CodeJam題目

CodeJam比賽的難題覆蓋了眾多計算機領域，並且難度層次比較嚴謹。通常情況下，難度遞增的題目分為Round 1 (R1)、Round 2 (R2)、Round 3 (R3)、Online Qualification Round等等。以下是一道CodeJam R1B的例題：A．

#include <iostream>
#include <string>
#include <vector>
using namespace std;

int main()
{
    int t, n;
    string s;
    cin >> t;
    for (int i = 1; i > n >> s;
        vector groups;
        int currGroup = 1;
        for (int j = 1; j < s.size(); j++) {
            if (s[j] == s[j - 1]) {
                currGroup++;
            } else {
                groups.push_back(currGroup);
                currGroup = 1;
            }
        }
        groups.push_back(currGroup); 
        int ans = 0;
        for (int j = 1; j < groups.size(); ++j) {
            ans += min(groups[j], groups[j - 1]);
        }
        cout << "Case #" << i << ": " << ans << endl;
    }
    return 0;
}

五、CodeJam 2022

CodeJam每年都會舉辦比賽，2022年度CodeJam也不例外。根據官網的消息，2022年度的CodeJam已經與2021年一樣，分 R1A、R1B、R2、R3 四個輪次。誠邀各位優秀的程序員們踴躍參加，一起切磋編程技藝！

#include <iostream>
#include <cstdlib>
using namespace std;

int main()
{
  cout << "Welcome to CodeJam 2022!" << endl;
  srand(2022);
  int a = rand() % 100;
  int b = rand() % 100;
  
  cout << "Let's do some math: " << a << " + " << b << " = ?" <> answer;
  
  if (answer == a + b) {
    cout << "Congratulations, you got the right answer!" << endl;
  } else {
    cout << "Sorry, the correct answer is " << a + b << endl;
  }
  return 0;
}

六、CodeJam Round 2

CodeJam 的Round 2比賽難度要更高，但是也能在很多領域為程序員們提供更多的挑戰。以下是一道CodeJam 2020 Round 2的例題：B。

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;

const int MAXN = 1e5 + 10;
long long ans[MAXN], score[MAXN];

struct Q {
    int l, r, id;
    Q() {};
    Q(int l_, int r_, int id_) {l = l_; r = r_; id = id_;}
    bool operator < (const Q &other) const {
        if (l / 320 != other.l / 320) {
            return l / 320 < other.l / 320;
        }
        return r < other.r;
   }
};

vector g[MAXN];
int a[MAXN], start[MAXN], end[MAXN];
bool seen[MAXN];
int t, n, q, cnt, b[MAXN], c[MAXN];

long long dfs(int u, int p) {
    seen[u] = true;
    start[u] = cnt;
    b[cnt] = a[u];
    c[cnt] = u;
    cnt++;
    for (int v : g[u]) {
        if (v != p) {
            dfs(v, u);
            b[cnt] = a[u];
            c[cnt] = u;
            cnt++;
        }
    }
    end[u] = cnt;
    return score[u];
}

int main() {
    cin >> t;
    for (int qq = 1; qq > n >> q;
        memset(seen, false, sizeof(seen));
        for (int i = 1; i > a[i];
        }
        for (int i = 1; i > score[i];
        }
        for (int i = 1; i > x >> y;
            g[x].push_back(y);
            g[y].push_back(x);
        }
        dfs(1, -1);
        vector queries;
        for (int i = 1; i > l >> r;
            queries.push_back(Q(l, r, i));
        }
        sort(queries.begin(), queries.end());
        int st = 0, ed = 0;
        long long sum = 0;
        for (Q x : queries) {
            int l = x.l, r = x.r, id = x.id;
            while (ed  l) {
                sum += score[c[st-1]] - b[st-1];
                st--;
            }
            while (st  r) {
                sum -= score[c[ed-1]] - b[ed-1];
                ed--;
            }
            ans[id] = sum;
        }
        cout << "Case #" << qq << ": ";
        for (int i = 1; i <= q; i++) {
            cout << ans[i];
            if (i != q) {
                cout << " ";
            }
        }
        cout << endl;
    }
    return 0;
}