https://codeforces.com/contest/1708/problem/D
题意

思路
这样的操作下,数列减的速度是非常快的,也就是说,易出现很多的0,0的操作没啥意义,所以我们要找到第一个 >0 的数对其后的序列进行排序,就能大大减少复杂度
这道题告诉我们实在没思路可以猜个暴力剪枝的做法
Code:
- #include
-
- #define int long long
-
- constexpr int N = 2e5 + 10;
- constexpr int mod = 998244353;
-
- int n;
- int a[N];
-
- void solve() {
- std::cin >> n;
- for (int i = 1; i <= n; i ++) {
- std::cin >> a[i];
- }
- int pos = 0;
- for (int i = 1; i <= n; i ++) {
- if (a[i]) {
- pos = i - 1;
- break;
- }
- }
- while(n > 1) {
- for (int i = std::max(pos, 1ll); i <= n; i ++) {
- a[i] = a[i + 1] - a[i];
- }
- n --;
- std::sort(a + pos, a + 1 + n);
- pos = std::upper_bound(a + 1, a + 1 + n, 0) - a - 1;
- if (pos == n - 1) break;
- }
- std::cout << a[n] << "\n";
- }
- signed main() {
- std::ios::sync_with_stdio(false);
- std::cin.tie(nullptr);
-
- int t = 1;
- std::cin >> t;
- while (t--) {
- solve();
- }
- return 0;
- }