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UVA10140Prime Distance参与者 2已保存回复 2

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@mlhw8h7u
此快照首次捕获于
2026/02/11 18:35
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此快照最后确认于
2026/02/11 19:25
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CPP
#include<bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
const int maxn = 1e6 + 5;
int l, r, top, q[maxn];
bool vis[maxn], isp[maxn];
void Euler(int n)
{
    isp[0] = false;
    isp[1] = false;
    for (int i = 2; i <= n; ++ i)
    {
        if (isp[i] == false)
        {
            continue;
        }
        for (int j = i; i * j <= n; ++ j)
        {
            isp[i * j] = false;
        }
    }
    return;
}
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	fill(isp, isp + maxn, true);
	Euler(maxn - 2);
	while (cin >> l >> r)
	{
		top = 0;
		memset(q, 0, sizeof q);
		memset(vis, 0, sizeof vis);
		for (int i = 2; i * i <= r; ++ i)
		{
			for (int j = l; j <= r; ++ j)
			{
				if (j % i == 0 && i != j)
				{
					vis[j - l + 1] = true;
				}
			}
		}
		for (int i = 1; i <= r - l + 1; ++ i)
		{
			if (vis[i] == false)
			{
				q[++ top] = i + l - 1;
			}
		}
		if (top < 2)
		{
			cout << "There are no adjacent primes." << endl;
			continue;
		}
		int minn = INT_MAX, maxx = INT_MIN;
		bool appi = false, appa = false;
		for (int i = 2; i <= top; ++ i)
		{
			minn = min(minn, q[i] - q[i - 1]);
			maxx = max(maxx, q[i] - q[i - 1]);
		}
		for (int i = 2; i <= top; ++ i)
		{
			if (appi == false && (q[i] - q[i - 1]) == minn)
			{
				appi = true;
				cout << q[i - 1] << "," << q[i] << " are closest, ";
				break;
			}
		}
		for (int i = 2; i <= top; ++ i)
		{
			if (appa == false && (q[i] - q[i - 1]) == maxx)
			{
				appa = true;
				cout << q[i - 1] << "," << q[i] << " are most distant." << endl;
				break;
			}
		}
	}
	return 0;
}

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