1499A. Domino on Windowsill

题意：给定一个 Educational Codeforces Round 106 (Rated for Div. 2) 简单题解（A~C） - 图1 的空间，Educational Codeforces Round 106 (Rated for Div. 2) 简单题解（A~C） - 图2#card=math&code=k1%E3%80%81k2%20%E8%A1%8C%E8%A6%81%E8%AE%BE%E7%BD%AE%E4%B8%BA%E7%99%BD%E8%89%B2%282%20%5Ctimes%201%29) 然后其他的设置为黑色

思路：为了满足条件需要判断一下白色和黑色的方块是否足够。

  1. int main() {
  2.     ios_base::sync_with_stdio(false), cin.tie(0);
  3.     int _ = 1;
  4.     for (cin >> _; _--;) {
  5.         int n, k1, k2, w, b;
  6.         cin >> n >> k1 >> k2 >> w >> b;
  7.         if (2 * w <= k1 + k2 and 2 * b <= 2 * n - k1 - k2) cout << "YES\n";
  8.         else
  9.             cout << "NO\n";
  10.     }
  11.     return 0;
  12. }

1499B. Binary Removals

找下前缀的 0和后缀的 1的个数进行比较

  1. int main() {
  2.     ios_base::sync_with_stdio(false), cin.tie(0);
  3.     int _ = 1;
  4.     for (cin >> _; _--;) {
  5.         string s;
  6.         cin >> s;
  7.         int L = 0, R = s.size() - 1;
  8.         while (L < s.size() and (s[L] == '0' or (L == 0 or s[L - 1] == '0')))
  9.             L++;
  10.         while (R >= 0 and
  11.                (s[R] == '1' or (R == (int)s.size() - 1 or s[R + 1] == '1')))
  12.             R--;
  13.         if (L >= R) cout << "YES\n";
  14.         else
  15.             cout << "NO\n";
  16.     }
  17.     return 0;
  18. }

学习高 Rank 的大佬写法

  1. int main() {
  2.     ios_base::sync_with_stdio(false), cin.tie(0);
  3.     int _ = 1;
  4.     for (cin >> _; _--;) {
  5.         string s;
  6.         cin >> s;
  7.         int i = s.find("11");
  8.         int j = s.rfind("00");
  9.         cout << (i != -1 && j != -1 && i < j ? "NO" : "YES") << endl;
  10.     }
  11.     return 0;
  12. }

1499C. Minimum Grid Path

  1. using ll = long long;
  2. int main() {
  3.     ios_base::sync_with_stdio(false), cin.tie(0);
  4.     int _ = 1;
  5.     for (cin >> _; _--;) {
  6.         int n;
  7.         cin >> n;
  8.         vector<int> c(n);
  9.         ll ans = LLONG_MAX;
  10.         ll mi[2] = {(ll)1E13, (ll)1E13}, sm[2] = {0, 0}, cnt[2] = {0, 0};
  11.         for (int i = 0; i < n; ++i) {
  12.             ll c;
  13.             cin >> c;
  14.             mi[i & 1] = min(mi[i & 1], c);
  15.             cnt[i & 1] += 1;
  16.             sm[i & 1] += c;
  17.             ans = min(ans, mi[0] * (n - cnt[0]) + sm[0] + mi[1] * (n - cnt[1]) +
  18.                                sm[1]);
  19.         }
  20.         cout << ans << "\n";
  21.     }
  22.     return 0;
  23. }