Codeforces Round 976 (Div. 2) and Divide By Zero 9.0（A~E）

时间：2024-10-01 07:02:13

A - Find Minimum Operations

思路

对 n n n进行 m m m进制分解，所有位上相加就是答案（参考 m = 2 m=2 m=2时）

代码

// Problem: A. Find Minimum Operations
// Contest: Codeforces - Codeforces Round 976 (Div. 2) and Divide By Zero 9.0
// URL: https://codeforces.com/contest/2020/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
int calc(int n , int m){
	int cnt = 0;
	if(m == 1){
		return n;
	}
	while(n){
		cnt += n % m;
		n /= m;
	}
	return cnt;
}
void solve() 
{
	int n , m;
	cin >> n >> m;
	cout << calc(n , m) << endl;
}            
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 - Brightness Begins

思路

经典关灯，第 x x x位置的灯会被 x x x的所有因数操作，因此当且仅当 x x x为完全平方数时会被关闭，考虑二分答案，找到 m i d mid mid之前有多少个完全平方数就能算出最终有多少个灯是关着的，可以直接用 s q r t sqrt sqrt函数近似的找到最接近 m i d mid mid的完全平方数，然后左右探测一下就能找到有多少个完全平方数小于等于 m i d mid mid.

代码

// Problem: B. Brightness Begins
// Contest: Codeforces - Codeforces Round 976 (Div. 2) and Divide By Zero 9.0
// URL: https://codeforces.com/contest/2020/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 = 5e18;
typedef pair<int,int>pl;
priority_queue<LL , vector<LL>, greater<LL> >mi;//小根堆
priority_queue<LL> ma;//大根堆
void solve() 
{
	int n;
	cin >> n;
	int l = n , r = 2e18;
	while(l < r){
		int mid = (l + r) >> 1;
		//[1 - mid]中有几个完全平方数
		int t = sqrt(mid);
		while(t * t > mid){
			t--;
		}
		while((t + 1) * (t + 1) <= mid){
			t++;
		}
		if(mid - t >= n){
			r = mid;
		}
		else{
			l = mid + 1;
		}
	}	
	cout << l << 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;
}

C - Bitwise Balancing

思路

拆位，对每一位进行分析， [ b , c , d ] [b,c,d] [b,c,d]在某一位上的组合只有 8 8 8种，对每种分类讨论 a a a的取值

代码

// Problem: C. Bitwise Balancing
// Contest: Codeforces - Codeforces Round 976 (Div. 2) and Divide By Zero 9.0
// URL: https://codeforces.com/contest/2020/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 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
void solve() 
{
	//如果d的这一位是0，
	int b , c , d;
	cin >> b >> c >> d;
	int a = 0;
	for(int i = 63 ; i >= 0 ; i --){
		if(d >> i & 1LL){
			if(c >> i & 1LL){
				if(b >> i & 1LL){
					continue;
				}
				else{
					cout << -1 << endl;
					return;
				}
			}
			else{
				a |= (1LL << i);
			}
		}
		else{
			if(b >> i & 1LL){
				if(c >> i & 1LL){
					a |= (1LL << i);
				}
				else{
					cout << -1 << endl;
					return;
				}
			}
			else{
				continue;
			}
		}
	}
	cout << a <<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 - Connect the Dots

思路

注意到 d d d很小，因此如果我们只需要知道某一位置上能否跟前面 d d d个数相连，就能把最终的图连出来。
用数组 c n t [ i ] [ j ] cnt[i][j] cnt[i][j]来表示 a i a_i ai​能否跟 a i − j a_{i - j} ai−j​相连，用差分来维护 c n t cnt cnt数组即可。

代码

// Problem: D. Connect the Dots
// Contest: Codeforces - Codeforces Round 976 (Div. 2) and Divide By Zero 9.0
// URL: https://codeforces.com/contest/2020/problem/D
// Memory Limit: 512 MB
// Time Limit: 2000 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
struct DSU {
    std::vector<int> f, siz;
    DSU() {}
    DSU(int n) {
        init(n);
    }
    void init(int n) {
        f.resize(n);
        std::iota(f.begin(), f.end(), 0);
        siz.assign(n, 1);
    }
    int find(int x) {
        while (x != f[x]) {
            x = f[x] = f[f[x]];
        }
        return x;
    }
    bool same(int x, int y) {
        return find(x) == find(y);
    }
    bool merge(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return false;
        }
        siz[x] += siz[y];
        f[y] = x;
        return true;
    }
    int size(int x) {
        return siz[find(x)];
    }
};

void solve() 
{
	//注意到D很小
	int n , m;
	cin >> n >> m;
	int vis[n + 100][20];
	int cnt[n + 100][20];
	memset(vis , 0 , sizeof vis);
	memset(cnt , 0 , sizeof cnt);
	DSU dsu(n + 5);
	for(int i = 0; i < m ; i ++){
		int a , d , k;
		cin >> a >> d >> k;
		int l = a , r = a + k * d;
		cnt[l + d][d]++;
		cnt[r + d][d]--;
	}
	for(int i = 1 ; i <= n ; i ++){
		int tot = 0;
		for(int j = 1 ; j <= min(10 , i) ; j ++){
			cnt[i][j] += cnt[i - j][j];
		}
	}
	for(int i = 2 ; i <= n ; i ++){
		for(int j = 1 ; j <= min(i - 1 , 10) ; j ++){
			if(cnt[i][j]){
				dsu.merge(i , i - j);
			}
		}
	}
	int ans = 0;
	for(int i = 1 ; i <= n ; i ++){
		ans += (dsu.f[i] == i);
	}
	cout << ans << endl;
}            
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;
}

E - Expected Power

思路

观察到 a a a数组取值很小，考虑 d p dp dp解决问题。用 d p [ i ] dp[i] dp[i]来代表取到当前取到 i i i的概率，那么对于每个 a i a_{i} a

相关文章