第32次CSP认证(持续更新)

第32次CSP认证(持续更新)

仓库规划

思路

数据范围较小，直接暴力判断

对每一个仓库，遍历其他每一个仓库，看看哪个是它的上级，如果都没有就输出$0$

时间复杂度: $O(n^{2}m)$

代码

神奇的代码

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
using i64 = long long;
using u64 = unsigned long long;
struct custom_hash 
{
	static uint64_t splitmix64(uint64_t x) 
    {
		x ^= x << 13;
		x ^= x >> 7;
		x ^= x << 17;
		return x; 
	}
	size_t operator () (uint64_t x) const 
    {
		static const uint64_t FIXED_RANDOM = std::chrono::steady_clock::now().time_since_epoch().count(); // 时间戳
		return splitmix64(x + FIXED_RANDOM);
	}
};

int nums[1010][12];

void solve() 
{
    int n = 0, m = 0;
    std::cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            std::cin >> nums[i][j];
        }
    }
    for (int i = 1; i <= n; i++)
    {
        bool flag = false;
        for (int j = 1; j <= n; j++)
        {
            bool f = true;
            if(i == j) continue;
            for (int k = 1; k <= m; k++)
            {
                if (nums[j][k] <= nums[i][k])
                {
                    f = false;
                    break;
                }
            }
            if (f)
            {
                std::cout << j << endl;
                flag = true;
                break;
            }
        }
        if (!flag)
        {
            std::cout << 0 << endl;
        }
    }
}

signed main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr); std::cout.tie(nullptr);
    int t = 1;
    //std::cin >> t;
    while(t--)
    {
        solve();
    }
    return 0;
}

因子化简

思路

注意到$2\le k\le 10$,而$({10}^5{)}^2={10}^{10}$,那么${10}^5$之外的素数都不用考虑了,一定不能保留。

那么我们可以先把$1o^5$以内的素数预处理出来，一个一个判断就好

${10}^5$以内的素数不超过${10}^4$个，询问不超过$10$个，判断一个数不超过$40$次，最大时间复杂度$O(4\ast {10}^6)$

代码

神奇的代码

#include<bits/stdc++.h>
#define int long long
#define endl '\n'
using i64 = long long;
using u64 = unsigned long long;
const int maxn = 1e5 + 2;
struct custom_hash 
{
	static uint64_t splitmix64(uint64_t x) 
    {
		x ^= x << 13;
		x ^= x >> 7;
		x ^= x << 17;
		return x; 
	}
	size_t operator () (uint64_t x) const 
    {
		static const uint64_t FIXED_RANDOM = std::chrono::steady_clock::now().time_since_epoch().count(); // 时间戳
		return splitmix64(x + FIXED_RANDOM);
	}
};

inline int read()
{
    int x = 0, flag = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')
        {
            flag = -1;
        }
        ch = getchar();
    }
    while(ch <= '9' && ch >= '0')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * flag;
}

int prime[maxn];
int cnt = 0;
int v[maxn]; // 存的是最小质因子

int power(int y, int x)
{
    int res = 1;
    while(x)
    {
        if (x & 1)
        {
            res *= y;
        }
        y *= y;
        x >>= 1;
    }
    return res;
}

void prime_filter() // 欧拉筛
{
    for (int i = 2; i <= maxn; i++)
    {
        if (v[i] == 0)
        {
            prime[++cnt] = i;
            v[i] = i;
        }
        for (int j = 1; j <= cnt; j++)
        {
            if (prime[j] > v[i] || i * prime[j] > maxn) break;
            v[i * prime[j]] = prime[j];
        }
    }
}

bool check(int pos, int k, int n)
{
    return power(prime[pos], k) <= n;
}

void solve() 
{
    int n = 0, k = 0;
    n = read(), k = read();
    int ans = 1;
    for (int i = 1; i <= cnt; i++)
    {
        if (n < prime[i]) break;
        int cnt = 0;
        while(n % prime[i] == 0)
        {
            cnt++;
            n /= prime[i];
        }
        if (cnt >= k)
        {
            ans *= power(prime[i], cnt);
        }
    }
    std::cout << ans << endl;
}

signed main()
{
    // std::ios::sync_with_stdio(false);
    // std::cin.tie(nullptr); std::cout.tie(nullptr);
    prime_filter();
    int t = 1;
    t = read();
    while(t--)
    {
        solve();
    }
    return 0;
}