Lyft Level 5 Challenge 2018

A. King Escape

签.

#include <bits/stdc++.h>
using namespace std;

int n, x[3], y[3];

int f1(int X, int Y)
{
    return X - Y - x[2] + y[2];
}

int f2(int X, int Y)
{
    return x[2] + y[2] - X - Y;
}

bool ok()
{
    //if (f1(x[0], y[0]) * f1(x[1], y[1]) < 0) return false;
    //if (f2(x[0], y[0]) * f2(x[1], y[1]) < 0) return false;
    if (x[0] > x[1]) swap(x[0], x[1]);
    if (y[0] > y[1]) swap(y[0], y[1]);
    if (y[2] >= y[0] && y[2] <= y[1]) return false;
    if (x[2] >= x[0] && x[2] <= x[1]) return false;
    return true;
}

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        scanf("%d%d", x + 2, y + 2);
        for (int i = 0; i < 2; ++i) 
            scanf("%d%d", x + i, y + i);
        puts(ok() ? "YES" : "NO");
    }
    return 0;
}

B. Square Difference

签.

#include <bits/stdc++.h>
using namespace std;

#define ll long long
int t; ll a, b;

bool ok(ll x)
{
    ll limit = sqrt(x);
    for (ll i = 2; i <= limit && i < x; ++i)
        if (x % i == 0)
            return false;
    return true;
}

int main()
{
    scanf("%d", &t);
    while (t--)
    {
        scanf("%lld%lld", &a, &b);
        if (a - b != 1) puts("NO");
        else
            puts(ok(a + b) ? "YES" : "NO");
    }
    return 0;
}

C. Permutation Game

Solved.

题意：

$A和B玩游戏，一个人能从i移动到j$

$当且仅当a[i] < a[j] 并且|i - j| equiv 0 pmod a[i]$

$判断以每个数为下标作起点,A先手能否必胜$

思路：

我们考虑一个位置什么时候必败

• $它下一步没有可移动的位置$
• $它的下一步状态没有一处是必败态$

倒着处理出每个位置的状态即可

#include <bits/stdc++.h>
using namespace std;

#define N 100010
int n, a[N], ans[N], pos[N];

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; ++i) scanf("%d", a + i), pos[a[i]] = i;
        for (int i = n; i >= 1; --i)
        {
            int id = pos[i];
            bool flag = 0;
            for (int j = id - i; j >= 1; j -= i)
                if (a[j] > i && ans[a[j]] == 0)
                {
                    flag = 1;
                    break;
                }
            if (flag == 0) for (int j = id + i; j <= n; j += i)
                if (a[j] > i && ans[a[j]] == 0)
                {
                    flag = 1;
                    break;
                }
            ans[i] = flag; 
        }
        for (int i = 1; i <= n; ++i) 
            putchar(ans[a[i]] ? 'A' : 'B');
        puts("");
    }
    return 0;
}

D. Divisors

Upsolved.

题意：

给出一些$a_i, 求 prod a_i 的因子个数$

$保证a_i 有3-5个因数$

思路：

对一个数求因子个数 假设它质因数分解之后是$n = p_1^{t_1} cdot p_2^{t_2} cdots p_n^{t_n}$

那么因子个数就是$(t_1 + 1) cdot (t_2  + 1) cdots (t_n + 1)$

我们考虑什么样的数有$3-5个因数$

$平方数、立方数、四次方数、n = p cdot q (p, q 是不同的质数)$

$对于前三类数，可以暴力破出，考虑第四类$

$如果它的p, q在序列中是唯一的，那么我们不需要管它具体是多少$

$直接得到p, q的数量就是这个数的数量$

$否则，拿这个数和别的数作gcd就可以破出p, q$

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define N 1010
const ll MOD = (ll)998244353;
int n; ll a[N];
map <ll, int> mp, num;  

void work(ll a)
{
    ll limit = pow(a, 1.0 / 4);
    for (ll i = limit + 10; i >= limit - 10 && i >= 1; --i)
        if (i * i * i * i == a)
        {
            mp[i] += 4;
            return;
        }
    limit = pow(a, 1.0 / 3);
    for (ll i = limit + 10; i >= limit - 10 && i >= 1; --i)
        if (i * i * i == a)
        {
            mp[i] += 3;
            return;
        }
    limit = pow(a, 1.0 / 2);
    for (ll i = limit + 10; i >= limit - 10 && i >= 1; --i)
        if (i * i == a)
        {
            mp[i] += 2;
            return;
        }
    ++num[a]; 
}

ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

int main()
{
    while (scanf("%d", &n) != EOF) 
    {
        mp.clear(); num.clear(); 
        for    (int i = 1; i <= n; ++i)
        {
            scanf("%lld", a + i);
            work(a[i]);
        }
        ll res = 1;  
        for (auto it : num)
        {
            ll tmp;
            bool flag = true;
            for (int i = 1; i <= n; ++i) 
                if (a[i] != it.first && (tmp = gcd(it.first, a[i])) != 1)
                {
                    mp[tmp] += it.second;
                    mp[it.first / tmp] += it.second; 
                    flag = false;
                    break;
                }
            if (flag) res = (res * (it.second + 1) % MOD * (it.second + 1)) % MOD;
        }
        for (auto it : mp)
            res = (res * (it.second + 1)) % MOD;
        printf("%lld
", res);
        fflush(stdout);
    }
    return 0;
}