【2021夏纪中游记】2021.7.19模拟赛

2021.7.19模拟赛

比赛概括：

(mathrm{sum}=8+80+0+0)

唉。

T1 玉米田（加强版）：

见【Luogu P1879】[USACO06NOV]玉米田Corn Fields。

T2 最大公约数：

差不多是这个：【Luogu P3455】 [POI2007]ZAP-Queries。

代码：

const int N = 1e6 + 10;

inline ll Read()
{
	ll x = 0, f = 1;
	char c = getchar();
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') f = -f, c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
	return x * f;
}

bool vis[N];
int pri[N], phi[N], cnt, sum[N];
void Prime()
{
	phi[1] = 1;
	for (int i = 2; i <= N - 10; i++)
	{
		if (!vis[i]) pri[++cnt] = i, phi[i] = i - 1;
		for (int j = 1; j <= cnt && i * pri[j] <= N - 10; j++)
		{
			vis[i * pri[j]] = 1;
			phi[i * pri[j]] = phi[i] * (pri[j] - 1);
			if (!(i % pri[j])) {phi[i * pri[j]] = phi[i] * pri[j]; break;}
		}
	}
	for (int i = 1; i <= N - 10; i++)
		sum[i] = sum[i - 1] + phi[i];
}

int t, n;
ll ans;

int main()
{
	Prime();
	for (t = Read(); t--; )
	{
		n = Read();
		ans = 0;
		for (int l = 1, r; l <= n; l = r + 1)
		{
			 r = n / (n / l);
			 ans += (2 * sum[n / l] - 1) * ((l + r) * (r - l + 1) / 2);
		}
		ans -= n * (n + 1) / 2;
		ans /= 2;
		printf ("%lld
", ans);
	}
	return 0;
}

T3 「JOI 2020 Final」只不过是长的领带：

题目大意：

正文：

贪心，排个序就行了。

代码：

const int N = 2e5 + 10;

inline ll Read()
{
	ll x = 0, f = 1;
	char c = getchar();
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') f = -f, c = getchar();
	while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
	return x * f;
}

int n;
struct node
{
	ll val, id;
	bool operator < (const node &a) const
	{
		return val < a.val;
	}
}a[N];
ll b[N], c[N], ans;
ll mx[N];

int main()
{
	n = Read();
	for (int i = 0; i <= n; i++)
		a[i] = (node) {Read(), i};
	for (int i = 1; i <= n; i++)
		b[i] = Read();
	sort (a, a + 1 + n);
	sort (b + 1, b + 1 + n);
	for (int i = 1; i <= n; i++)
		ans = max(ans, max(0ll, a[i].val - b[i]));
	c[a[0].id] = ans;
	for (int i = n; i; i--)
		mx[i] = max(mx[i + 1], max(0ll, a[i].val - b[i]));
	ans = 0;
	for (int i = 1; i <= n; i++)
	{
		swap(a[0], a[i]);
		ans = max(ans, max(0ll, a[i].val - b[i]));
		c[a[0].id] = max(ans, mx[i + 1]);	
	}
	for (int i = 0; i <= n; i++)
		printf ("%lld ", c[i]);
	return 0;
}

T4 【SHTSC2014】概率充电器(charger)：

详见 【Luogu P4284】[SHOI2014]概率充电器。