A - I Scream

根据 奶脂率 和 乳脂率 判断是何种冰淇淋

  1. int main() {
  2.     ios_base::sync_with_stdio(false), cin.tie(0);
  3.     int a, b;
  4.     cin >> a >> b;
  5.     a = a + b;
  6.     if (a >= 15 && b >= 8) cout << 1;
  7.     else if (a >= 10 && b >= 3)
  8.         cout << 2;
  9.     else if (a >= 3)
  10.         cout << 3;
  11.     else
  12.         cout << 4;
  13.     return 0;
  14. }

B - Job Assignment

分开存储，循环判断。如果是同一个人完成任务则总时间累计

  1. int main() {
  2.     ios_base::sync_with_stdio(false), cin.tie(0);
  3.     vector<int> a, b;
  4.     int n, ans = 1e9 + 5;
  5.     cin >> n;
  6.     a.resize(n);
  7.     b.resize(n);
  8.     for (int i = 0; i < n; ++i) {
  9.         cin >> a[i] >> b[i];
  10.     }
  11.     for (int i = 0; i < n; ++i)
  12.         for (int j = 0; j < n; ++j) {
  13.             if (i == j) ans = min(ans, a[i] + b[j]);
  14.             else
  15.                 ans = min(ans, max(a[i], b[j]));
  16.         }
  17.     cout << ans;
  18.     return 0;
  19. }

C - Squared Error

公式推导

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注意用 long long ，溢出了WA了我3次….

  1. using ll = long long;
  2. int main() {
  3.     ios_base::sync_with_stdio(false), cin.tie(0);
  4.     int n;
  5.     cin >> n;
  6.     vector<ll> a(n);
  7.     ll sos = 0, sum = 0;  // sos sum of squares
  8.     for (ll &x : a) cin >> x;
  9.     for (int i = 0; i < n; i++) {
  10.         sos += a[i] * a[i];
  11.         sum += a[i];
  12.     }
  13.     cout << n * sos - sum * sum << endl;
  14.     return 0;
  15. }

D - Journey

按题意来即可，注意输出位数来保持精度

  1. int main() {
  2.     ios_base::sync_with_stdio(false), cin.tie(0);
  3.     int n;
  4.     cin >> n;
  5.     double cnt = 0.0;
  6.     for (int i = 1; i < n; ++i) {
  7.         cnt += 1.0 * n / (n - i);
  8.     }
  9.     cout << setprecision(20) << cnt << "\n";
  10.     return 0;
  11. }

E - Mex Min

熟悉的定义…

  1. using ll = long long;
  2. ll n, m, a[5000005], s[5000005], mina, temp = 1;
  3. int main() {
  4.     ios_base::sync_with_stdio(false), cin.tie(0);
  5.     cin >> n >> m;
  6.     for (int i = 1; i <= n; i++) {
  7.         cin >> a[i];
  8.         if (i <= m) s[a[i]]++;
  9.     }
  10.     for (int i = 0; i <= n; i++) {
  11.         if (s[i] == 0) {
  12.             mina = i;
  13.             break;
  14.         }
  15.     }
  16.     for (int i = m + 1; i <= n; i++) {
  17.         s[a[temp]]--;
  18.         s[a[temp + m]]++;
  19.         if (a[temp] < mina && s[a[temp]] == 0) {
  20.             // cout<<a[temp];
  21.             mina = min(mina, a[temp]);
  22.         }
  23.         temp++;
  24.     }
  25.     cout << mina;
  26.     return 0;
  27. }

F - Digits Paradise in Hexadecimal

不会，先记录下dalao们的解法

__builtin_popcount() 用于计算一个 32 位无符号整数有多少个位为1

  1. #include <bits/stdc++.h>
  2. using namespace std;
  3. #define rep(i, n) for (int i = 0; i < (n); i++)
  4. using ll = long long;
  5. constexpr int MOD = 1e9 + 7;
  6. long long dp[2][17];
  7. string s;
  8. int n, k, m, a, e;
  9. int main() {
  10.     cin >> s >> k;
  11.     for (char c : s) {
  12.         a = '0' <= c && c <= '9' ? c - '0' : c - 'A' + 10;
  13.         rep(i, 16) dp[e][i + 1] =
  14.             (dp[e ^ 1][i] * (16 - i) + dp[e ^ 1][i + 1] * (i + 1)) % MOD;
  15.         dp[e][1] += m ? 15 : a - 1;
  16.         if (m) {
  17.             int t = __builtin_popcount(m), b1 = t - __builtin_popcount(m >> a);
  18.             dp[e][t + 1] += a - b1;
  19.             dp[e][t] += b1;
  20.         }
  21.         m |= 1 << a;
  22.         e ^= 1;
  23.     }
  24.     dp[e ^ 1][__builtin_popcount(m)] += 1;
  25.     cout << dp[e ^ 1][k] % MOD << endl;
  26. }