Pinely Round 3 (Div. 1 + Div. 2)（A~D）（有意思的题）

A - Distinct Buttons

题意：

思路：模拟从（0,0）到每个位置需要哪些操作，如果总共需要4种操作就输出NO。

// Problem: A. Distinct Buttons
// Contest: Codeforces - Pinely Round 3 (Div. 1 + Div. 2)
// URL: https://codeforces.com/contest/1909/problem/0
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
void solve() 
{cin >> n;int flag[4] = {0 , 0 , 0 , 0};for(int i = 0 ; i < n ; i ++){int x , y;cin >> x >> y;if(x < 0){flag[0] = 1;}else if(x > 0){flag[1] = 1;}if(y < 0){flag[2] = 1;}else if(y > 0){flag[3] = 1;}}	int cnt = 0;for(int i = 0 ; i < 4 ; i ++){cnt += flag[i];}if(cnt <= 3){cout <<"Yes\n";}else{cout <<"NO\n";}
}            
int main() 
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t=1;cin>>t;while(t--){solve();}return 0;
}

B - Make Almost Equal With Mod

思路：比较有意思的题目，可以发现k取2的倍数即可。证明如下：将所有数变为二进制表示，那么某个数模2的结果即二进制最后一位，模4的结果即二进制倒数第二位...如此类推。

由于题目必然存在解，也就是说数组不可能全相等。既然不可能全相等，那一定存在整个数组某一位存在1和0。因此k取2的倍数必然能够满足题意。

// Problem: B. Make Almost Equal With Mod
// Contest: Codeforces - Pinely Round 3 (Div. 1 + Div. 2)
// URL: https://codeforces.com/contest/1909/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
#define int long long
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 1e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
void solve() 
{cin >> n;int cnt1 = 0 , cnt0 = 0;for(int i = 0 ; i < n ; i++){cin >> a[i];}for(int j = 2 ; j <= llinf ; j *= 2){set<int>st;for(int i = 0 ;i < n ; i ++){st.insert(a[i] % j);}if(st.size() == 2){cout << j << endl;return;;}}}            
signed main() 
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t=1;cin>>t;while(t--){solve();}return 0;
}

C - Heavy Intervals

题意：

思路：首先想到对l,r,c数组进行排序。可以发现，无论如何排序，所有区间长度之和是不会改变的。因此要让权值之和最小，需要让小的区间尽可能小。即对于任意 $r_{i}$ 而言， $l_{i}$ 为最靠近它的元素。而从小到大的处理 $r_{i}$ 可以保证不会影响到后面的数。

// Problem: C. Heavy Intervals
// Contest: Codeforces - Pinely Round 3 (Div. 1 + Div. 2)
// URL: https://codeforces.com/contest/1909/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
#define int long long
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
void solve() 
{cin >> n;int l[n] , r[n] , c[n];for(int i = 0 ; i < n ; i ++)cin >> l[i];for(int i = 0 ; i < n ; i ++)cin >> r[i];for(int i = 0 ; i < n ; i ++)cin >> c[i];sort(c , c + n);sort(l , l + n);sort(r , r + n);int pre[n];stack<int>st;int ll = 0;int ans = 0;for(int i = 0 ; i < n ;i ++){while(ll < n && r[i] > l[ll]){st.push(l[ll]);ll++;}int x = st.top();st.pop();pre[i] = r[i] - x;} 	sort(pre , pre + n);for(int i = 0 ; i < n ; i ++){ans += pre[i] * c[n - i - 1];}cout << ans << endl;
}            
signed main() 
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t=1;cin>>t;while(t--){solve();}return 0;
}

D - Split Plus K

题意：

思路：假设最终所有数为 $ans$ ，对于 $a_{i}$ 而言，需要操作 $t$ 次以后能变成ans，需要满足 $a_{i} + t*k = (t + 1) * ans$ 。

转换一下后得到 $t = (a_{i} - ans)/(ans-k)$

即 $ans$ 成立的条件为： $\forall i((a_{i} - k)\%(ans - k)) = 0$

为了方便解释，假设所有数都大于 $k$ 。想要操作数最小，即 $ans-k$ 需要最大。可以发现，最终的 $ans-k$ 的最大值为 $gcd(a_{1} - k , a_{2} - k , a_{3} - k .... a_{n} - k)$ 。求出gcd之后再带回原式子求出操作数 $t$ 即可。相反所有数都小于 $k$ 也是一样的操作。需要注意存在数等于 $k$ 时，需要所有数都等于 $k$ ，否则输出-1。

// Problem: D. Split Plus K
// Contest: Codeforces - Pinely Round 3 (Div. 1 + Div. 2)
// URL: https://codeforces.com/contest/1909/problem/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define endl '\n'
#define int long long
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
LL gcd(LL a, LL b){return b > 0 ? gcd(b , a % b) : a;
}LL lcm(LL a , LL b){return a / gcd(a , b) * b;
}
int n , m;
vector<int>a(N , 0);
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
void solve() 
{// x + tk = (t + 1) * ans
// x - ans = t(ans - k)
// ans - k < 0 ??
// (x - ans / ans - k ) = t // ans 越大越好cin >> n >> m;for(int i = 0 ; i < n ; i++)cin >> a[i];sort(a.begin() , a.begin() + n);for(int i = 0 ; i < n ; i ++){if(a[0] < m && a[i] >= m){cout << -1 << endl;return;}}for(int i = 0 ; i < n ; i ++){a[i] -= m;}int ans = 0;for(int i = 0 ; i < n ; i ++){ans = gcd(ans , abs(a[i]));}int out = 0;if(a[0] == 0 && a[n - 1] != 0 || a[0] != 0 && a[n - 1] == 0){cout << -1 << endl;return;}else if(ans == 0){cout << 0 << endl;return;}for(int i = 0 ; i < n ; i ++){if(a[i] >= 0){out += (a[i] - ans) / ans ;}else{out += (a[i] + ans) / -ans;}}cout << out << endl;
}            
signed main() 
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout.precision(10);int t=1;cin>>t;while(t--){solve();}return 0;
}

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