Codeforces Round 973 (Div. 2)

合集 - 题解(20)

1.[博客迁移20190713]题解 P4169 【[Violet]天使玩偶/SJY摆棋子】2024-05-272.模拟题2024-06-263.SP666 VOCV - Con-Junctions 题解2024-08-254.Codeforces Round #900 (Div. 3)2024-08-255.Codeforces Round 962 (Div. 3)2024-08-276.题解 CF993E 【Nikita and Order Statistics】2024-09-167.题解 P6891 【[JOISC2020]ビルの飾り付け 4】2024-09-188.P2163 [SHOI2007] 园丁的烦恼2024-09-189.[POI2011] LIZ-Lollipop2024-09-19

10.Codeforces Round 973 (Div. 2)2024-09-21

11.Codeforces Round 974 (Div. 3)2024-09-2212.NEERC2013题解2024-09-2313.CF1446C Xor Tree2024-09-2514.Codeforces Round 709 (Div. 1)2024-09-2715.Codeforces Round 975 (Div. 2)2024-09-2816.NEERC2014题解2024-09-2917.题解 P2726 【[SHOI2005]树的双中心】2024-10-0118.CF2019 F. Max Plus Min Plus Size2024-10-0319.Codeforces Round 757 (Div. 2)03-0520.大工之星第一周题解03-16

收起

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;
}