Codeforces Round #690 (Div. 3)

1462A. Favorite Sequence

简单看懂题即可，左边输出一个然后右边输出一个。

  1. void solve() {
  2.     int n;
  3.     cin >> n;
  4.     vector<ll> a(n + 1);
  5.     // ll a[n + 1]; //两种写法均可
  6.     for (int i = 1; i <= n; ++i)
  7.         cin >> a[i];
  8.     int l = 1, r = n;
  9.     bool f = true;
  10.     for (int i = 1; i <= n; ++i) {
  11.         if (f)
  12.             cout << a[l++] << " ", f = false;
  13.         else
  14.             cout << a[r--] << " ", f = true;
  15.     }
  16.     cout << endl;
  17. }

1462B. Last Year’s Substring

一开始想错了，正确的思路是拆分字符串看是否能组成 2020

  1. void solve() {
  2.     int n;
  3.     string s;
  4.     cin >> n >> s;
  5.     bool f = false;
  7.     for (int fir = 0; fir <= 4 && !f; fir++) {
  8.         int sec = 4 - fir;  //定位
  10.         if (s.substr(0, fir) + s.substr(n - sec) == "2020")
  11.             f = true;
  12.     }
  14.     cout << (f ? "YES\n" : "NO\n");
  15. }

1462C. Unique Number

😂打表。注意一下n > 45的话直接输出-1（因为0~9都被使用了最多到45）

  1. ll a[51] = {
  2.     0,       1,       2,       3,        4,        5,         6,      7,
  3.     8,       9,       19,      29,       39,       49,        59,     69,
  4.     79,      89,      189,     289,      389,      489,       589,    689,
  5.     789,     1789,    2789,    3789,     4789,     5789,      6789,   16789,
  6.     26789,   36789,   46789,   56789,    156789,   256789,    356789, 456789,
  7.     1456789, 2456789, 3456789, 13456789, 23456789, 123456789, -1,      -1,
  8.     -1,       -1,       -1,
  9. };
  10. void solve() {
  11.     int n, cnt = 0;
  12.     string s;
  13.     cin >> n;
  14.     cout << a[n] << endl;
  15. }

当然也可以正常分析：

  1. void solve() {
  2.     int n;
  3.     cin >> n;
  4.     if (n > 45) {
  5.         cout << -1 << endl;
  6.         return;
  7.     }
  8.     string s;
  9.     int nxt = 9;
  10.     while (n > 0) {
  11.         if (n >= nxt) {
  12.             s += '0' + nxt;
  13.             n -= nxt--;
  14.         } else {
  15.             s += '0' + n;
  16.             break;
  17.         }
  18.     }
  19.     reverse(s.begin(), s.end());
  20.     cout << s << endl;
  21. }

D. Add to Neighbour and Remove

  1. void solve() {
  2.     int n;
  3.     cin >> n;
  4.     vector<ll> a(n);
  5.     for (int i = 0; i < n; ++i)
  6.         cin >> a[i];
  7.     ll ans = INT_MAX, sum = 0;
  8.     for (int i = 0; i < n; ++i) {
  9.         sum += a[i];
  10.         ll cur = 0, toadd = 0;
  11.         bool f = true;
  12.         for (int j = 0; j < n; ++j) {
  13.             if (cur)
  14.                 toadd++;
  15.             cur += a[j];
  16.             if (cur > sum) {
  17.                 f = false;
  18.                 break;
  19.             } else if (cur == sum)
  20.                 cur = 0;
  21.         }
  22.         if (f && cur == 0) {
  23.             // cout << toadd << " " << i << endl;
  24.             ans = min(ans, toadd);
  25.         }
  26.     }
  27.     cout << ans << endl;
  28. }

E1. Close Tuples (easy version)

  1. //unsolved