今天这场没准备打的，但是lbw打，我就倒开看看题目了。非常意外的是这场只要做到E就可以100名，而我感觉除了F都是垃圾题。
D. Minimize the Difference
这种求极差的题目很常规，由他那个操作可以看出单调性。故可以两遍二分分别求最小值的最大值和最大值的最小值然后做差，也可以用单调栈去做。
const int N = 2e5 + 5;
int T, n; ll a[N];inline bool check(ll x) {ll sum = 0;for (int i = 1; i <= n; i++) {if (a[i] > x) sum += a[i] - x;if (a[i] < x) {if (x - a[i] > sum) return false;sum -= x - a[i];}} return true;
}inline bool checkII(ll x) {ll sum = 0;for (int i = 1; i <= n; i++) {if (a[i] > x) sum += a[i] - x;if (a[i] < x) sum -= x - a[i], sum = max(sum, 0ll);}if (sum > 0) return false;return true;
}signed main(void) {for (read(T); T; T--) {read(n); ll mx = 0, mn = 1ll << 62;for (int i = 1; i <= n; i++)read(a[i]), chkmax(mx, a[i]), chkmin(mn, a[i]);ll l = mn, r = mx, ans1 = l;while (l <= r) {ll mid = l + r >> 1;if (check(mid)) ans1 = mid, l = mid + 1;else r = mid - 1;}l = mn, r = mx; ll ans2 = r;while (l <= r) {ll mid = l + r >> 1;if (checkII(mid)) ans2 = mid, r = mid - 1;else l = mid + 1;}
// writeln(ans1, ' '); writeln(ans2);writeln(ans2 - ans1);}//fwrite(pf, 1, o1 - pf, stdout);return 0;
}
E. Prefix GCD
我们知道在gcd变小的过程中至少要除以二，则gcd收敛的次数是log级别的。故可以贪心找每次填入后前缀gcd最小的数，时间复杂度 \(O(n\log(A))\)
inline int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);
}const int N = 1e5 + 5;
int T, n, a[N];signed main(void) {for (read(T); T; T--) {read(n);for (int i = 1; i <= n; i++) read(a[i]);if (n == 1) { writeln(a[1]); continue; }sort(a + 1, a + n + 1);int d = gcd(a[1], a[2]);for (int i = 3; i <= n; i++) d = gcd(d, a[i]);int D = a[1] / d, j = 1; ll ans = D;while (D > 1 && j < n) {int M = INT_MAX, k = j + 1;for (int i = j + 1; i <= n; i++)if (M > gcd(a[i] / d, D)) {M = gcd(a[i] / d, D);k = i;}ans += M; swap(a[j + 1], a[k]); D = M; j++;
// writeln(D, ' '); writeln(j); }ans += n - j;writeln(1ll * ans * d);}//fwrite(pf, 1, o1 - pf, stdout);return 0;
}
