【YBTOJ】【UVA10140】Prime Distance

Prime Distance

链接：

洛谷

题目大意：

给定两个正整数 (l,r)，求 ([l,r]) 间 相邻 的两个差最大的质数和 相邻 的两个差最小的质数。如果区间内质数个数 (le 1)，输出 There are no adjacent primes.。

思路：

水题，通过前 (2^16) 个质数排除 ([l,r]) 之间的合数。

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;
}

int pri[N], tot;
bool vis[N];
void Prework()
{
	for (int i = 2; i <= N - 10; i++)
	{
		if (!vis[i]) pri[++tot] = i;
		for (int j = 1; j <= tot && i * pri[j] <= N - 10; j++)
		{
			vis[i * pri[j]] = 1;
			if (i % pri[j] == 0) break;
		}
	}
}


int main()
{
	Prework();
	for (ll l, r; ~scanf("%lld%lld", &l, &r); )
	{
		ll a, b, dif1 = 1ll << 60, c, d, dif2 = 0;
		memset (vis, 0, sizeof vis);
		if(l == 1) vis[0] = 1;
		for (int i = 1; i <= tot; i++)
			for (ll j = l / pri[i]; j * pri[i] <= r; j++)
				if(j > 1 && j * pri[i] >= l)vis[j * pri[i] - l] = 1;
		
		ll lst = 0;
		for (ll i = l; i <= r; i++)
		{
			if (vis[i - l]) continue;
			if (!lst) {lst = i; continue;} 
			int tmp = i - lst;
			if (tmp < dif1) dif1 = tmp, a = lst, b = i;
			if (tmp > dif2) dif2 = tmp, c = lst, d = i;
			lst = i;
		}
		if (!lst || !dif2) puts("There are no adjacent primes.");
		else printf("%lld,%lld are closest, %lld,%lld are most distant.
", a, b, c, d);
	}
	return 0;
}