Codeforces 984 扫雷check 欧几里得b进制分数有限小数判定 f函数最大连续子段

A

/* Huyyt */
#include <bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define mkp(a,b) make_pair(a,b)
#define pb push_back
using namespace std;
typedef long long ll;
const long long mod = 1e9 + ;
const int N = 2e5 + ;
int num[];
int main()
{
        int n;
        cin >> n;
        for(int i=;i<=n;i++)
        {
                cin >> num[i];
        }
        sort(num+,num++n);
        int now=;
        if(n%)
        {
                cout<<num[n/+]<<endl;
        }
        else
        {
                cout<<num[n/]<<endl;
        }
}

B

/* Huyyt */
#include <bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define mkp(a,b) make_pair(a,b)
#define pb push_back
const int dir[][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }};
using namespace std;
typedef long long ll;
const long long mod = 1e9 + ;
const int N = 2e5 + ;
char f[][];
int n, m;
bool ok(int x, int y)
{
        if (x > n || x < )
        {
                return false;
        }
        if (y > m || y < )
        {
                return false;
        }
        return true;
}
int main()
{
        cin >> n >> m;
        for (int i = ; i <= n; i++)
        {
                scanf("%s", f[i] + );
        }
        for (int i = ; i <= n; i++)
        {
                for (int j = ; j <= m; j++)
                {
                        if (f[i][j] == '*')
                        {
                                continue;
                        }
                        if (f[i][j] == '.')
                        {
                                for (int w = ; w < ; w++)
                                {
                                        int dx = i + dir[w][];
                                        int dy = j + dir[w][];
                                        if (ok(dx, dy))
                                        {
                                                if (f[dx][dy] == '*')
                                                {
                                                        cout << "NO" << endl;
                                                        return ;
                                                }
                                        }
                                }
                        }
                        else
                        {
                                int now = ;
                                int cur = f[i][j] - '';
                                for (int w = ; w < ; w++)
                                {
                                        int dx = i + dir[w][];
                                        int dy = j + dir[w][];
                                        if (ok(dx, dy))
                                        {
                                                if (f[dx][dy] == '*')
                                                {
                                                        now++;
                                                }
                                        }
                                }
                                if (now != cur)
                                {
                                        cout << "NO" << endl;
                                        return ;
                                }
                        }

                }
        }
        cout << "YES" << endl;
        return ;
}

C

好题。。应该是我做过最难的2C了

给你 p q b 三个1e18的数 问你在b进制下p/q能不能被有限地表现出来

首先把p与q约分 考虑1/q能不能在b进制在被表现出来

因为b进制下每退一位所表示的数就小b倍

所以我们可以把被除数乘上b再除q就是这一位小数的值 即\(\frac{3*4}{5}=2\)

同时被除数变为\(\frac{3}{5}\)\(\%\) \(\frac{1}{4}\)=\(\frac{(3*4)\%5}{5}\)=\(\frac{2}{5}\)

这样进行下去看能不能在某一步被除数分子乘上b是q的倍数

即存在n使得 \(b^{n}\ mod\ q=0\) 但是我们不可能枚举n来检验是否存在

由唯一分解定理可知 \(b^{n}\ mod\ q=0\) 等价于b的因子中存在所有q的质因子

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll LLmaxn = 2e18;
const int N = ;
int n;
inline ll readint()
{
        char c = getchar();
        ll ans = ;
        while (c < '' || c > '')
        {
                c = getchar();
        }
        while (c >= '' && c <= '')
        {
                ans = ans * 10LL + c - '', c = getchar();
        }
        return ans;
}
ll gcd(ll a, ll b)
{
        ll t;
        while (b)
        {
                t = b;
                b = a % b;
                a = t;
        }
        return a;
}
int main()
{
        n=readint();
        ll p, q, b;
        for (int i = ; i <= n; i++)
        {
                p=readint(),q=readint(),b=readint();
                ll now = gcd(p, q);
                p /= now, q /= now;
                if (q == )
                {
                        cout << "Finite" << endl;
                        continue;
                }
                p %= q;
                if (p == )
                {
                        cout << "Finite" << endl;
                        continue;
                }
                now = gcd(q, b);
                while (now != )
                {
                        while (q % now == )
                        {
                                q /= now;
                        }
                        now=gcd(q,b);
                }
                if (q == )
                {
                        cout << "Finite" << endl;
                }
                else
                {
                        cout << "Infinite" << endl;
                }

        }
}

D

给你一个f函数(类似于石子合并花费 只是把+变成XOR操作)

直接n²用类似石子合并的思想先把小的区间处理出来dp[i][i + len] = dp[i][i + len - 1] ^ dp[i + 1][i + len]

再汇总dp[i][i + len] = max(max(dp[i][i + len - 1], dp[i + 1][i + len]), dp[i][i + len])

/*Huyyt*/
#include<bits/stdc++.h>
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll LLmaxn = 2e18;
const int N = ;
inline int readint()
{
        char c = getchar();
        int ans = ;
        while (c < '' || c > '')
        {
                c = getchar();
        }
        while (c >= '' && c <= '')
        {
                ans = ans *  + c - '', c = getchar();
        }
        return ans;
}
int number[];
int dp[][];
int main()
{
        int now = ;
        int n;
        n = readint();
        for (int i = ; i <= n; i++)
        {
                number[i] = readint();
        }
        for (int i = ; i <= n; i++)
        {
                dp[i][i] = number[i];
        }
        for (int len = ; len <= n - ; len++)
        {
                for (int i = ; i <= n - len; i++)
                {
                        dp[i][i + len] = dp[i][i + len - ] ^ dp[i + ][i + len];
                }
        }
        for (int len = ; len <= n - ; len++)
        {
                for (int i = ; i <= n - len; i++)
                {
                        dp[i][i + len] = max(max(dp[i][i + len - ], dp[i + ][i + len]), dp[i][i + len]);
                }
        }
        int l, r;
        int q;
        q = readint();
        while (q--)
        {
                l = readint(), r = readint();
                cout << dp[l][r] << endl;
        }
}