Lyft Level 5 Challenge 2018 - Elimination Round
A. King Escape
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 int n, x[3], y[3]; 5 6 int f1(int X, int Y) 7 { 8 return X - Y - x[2] + y[2]; 9 } 10 11 int f2(int X, int Y) 12 { 13 return x[2] + y[2] - X - Y; 14 } 15 16 bool ok() 17 { 18 //if (f1(x[0], y[0]) * f1(x[1], y[1]) < 0) return false; 19 //if (f2(x[0], y[0]) * f2(x[1], y[1]) < 0) return false; 20 if (x[0] > x[1]) swap(x[0], x[1]); 21 if (y[0] > y[1]) swap(y[0], y[1]); 22 if (y[2] >= y[0] && y[2] <= y[1]) return false; 23 if (x[2] >= x[0] && x[2] <= x[1]) return false; 24 return true; 25 } 26 27 int main() 28 { 29 while (scanf("%d", &n) != EOF) 30 { 31 scanf("%d%d", x + 2, y + 2); 32 for (int i = 0; i < 2; ++i) 33 scanf("%d%d", x + i, y + i); 34 puts(ok() ? "YES" : "NO"); 35 } 36 return 0; 37 }
B. Square Difference
签.
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 int t; ll a, b; 6 7 bool ok(ll x) 8 { 9 ll limit = sqrt(x); 10 for (ll i = 2; i <= limit && i < x; ++i) 11 if (x % i == 0) 12 return false; 13 return true; 14 } 15 16 int main() 17 { 18 scanf("%d", &t); 19 while (t--) 20 { 21 scanf("%lld%lld", &a, &b); 22 if (a - b != 1) puts("NO"); 23 else 24 puts(ok(a + b) ? "YES" : "NO"); 25 } 26 return 0; 27 }
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
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 100010 5 int n, a[N], ans[N], pos[N]; 6 7 int main() 8 { 9 while (scanf("%d", &n) != EOF) 10 { 11 for (int i = 1; i <= n; ++i) scanf("%d", a + i), pos[a[i]] = i; 12 for (int i = n; i >= 1; --i) 13 { 14 int id = pos[i]; 15 bool flag = 0; 16 for (int j = id - i; j >= 1; j -= i) 17 if (a[j] > i && ans[a[j]] == 0) 18 { 19 flag = 1; 20 break; 21 } 22 if (flag == 0) for (int j = id + i; j <= n; j += i) 23 if (a[j] > i && ans[a[j]] == 0) 24 { 25 flag = 1; 26 break; 27 } 28 ans[i] = flag; 29 } 30 for (int i = 1; i <= n; ++i) 31 putchar(ans[a[i]] ? 'A' : 'B'); 32 puts(""); 33 } 34 return 0; 35 }
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$3-5个因数$
$平方数、立方数、四次方数、n = p \cdot q (p, q 是不同的质数)$
$对于前三类数，可以暴力破出，考虑第四类$
$如果它的p, q在序列中是唯一的，那么我们不需要管它具体是多少$
$直接得到p, q的数量就是这个数的数量$
$否则，拿这个数和别的数作gcd就可以破出p, q$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 1010 6 const ll MOD = (ll)998244353; 7 int n; ll a[N]; 8 map <ll, int> mp, num; 9 10 void work(ll a) 11 { 12 ll limit = pow(a, 1.0 / 4); 13 for (ll i = limit + 10; i >= limit - 10 && i >= 1; --i) 14 if (i * i * i * i == a) 15 { 16 mp[i] += 4; 17 return; 18 } 19 limit = pow(a, 1.0 / 3); 20 for (ll i = limit + 10; i >= limit - 10 && i >= 1; --i) 21 if (i * i * i == a) 22 { 23 mp[i] += 3; 24 return; 25 } 26 limit = pow(a, 1.0 / 2); 27 for (ll i = limit + 10; i >= limit - 10 && i >= 1; --i) 28 if (i * i == a) 29 { 30 mp[i] += 2; 31 return; 32 } 33 ++num[a]; 34 } 35 36 ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } 37 38 int main() 39 { 40 while (scanf("%d", &n) != EOF) 41 { 42 mp.clear(); num.clear(); 43 for (int i = 1; i <= n; ++i) 44 { 45 scanf("%lld", a + i); 46 work(a[i]); 47 } 48 ll res = 1; 49 for (auto it : num) 50 { 51 ll tmp; 52 bool flag = true; 53 for (int i = 1; i <= n; ++i) 54 if (a[i] != it.first && (tmp = gcd(it.first, a[i])) != 1) 55 { 56 mp[tmp] += it.second; 57 mp[it.first / tmp] += it.second; 58 flag = false; 59 break; 60 } 61 if (flag) res = (res * (it.second + 1) % MOD * (it.second + 1)) % MOD; 62 } 63 for (auto it : mp) 64 res = (res * (it.second + 1)) % MOD; 65 printf("%lld\n", res); 66 fflush(stdout); 67 } 68 return 0; 69 }
