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A - Find Minimum Operations

思路

对 $n$ 进行 $m$ 进制分解，所有位上相加就是答案（参考 $m = 2$ 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 $x$ 位置的灯会被 $x$ 的所有因数操作，因此当且仅当 $x$ 为完全平方数时会被关闭，考虑二分答案，找到 $mi d$ 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 $s q r t$ 函数近似的找到最接近 $mi d$ 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 $mi d$ .

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， $[b, c, d]$ 在某一位上的组合只有 $8$ 种，对每种分类讨论 $a$ 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 $d$ 很小，因此如果我们只需要知道某一位置上能否跟前面 $d$ 个数相连，就能把最终的图连出来。
用数组 $c n t [i] [j]$ 来表示 $a_i$ 能否跟 $a_{i - j}$ 相连，用差分来维护 $c n t$ 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 $a$ 数组取值很小，考虑 $d p$ 解决问题。用 $d p [i]$ 来代表取到当前取到 $i$ 的概率，那么对于每个 $a_{i}$ ,我们枚举 $\leq x \leq 1023)$ , $\oplus a_{i}] = dp[x] * p , dp[x] = dp[x] * (1 - p)$ ，总体复杂度为 $O (1024 * N)$ 能通过此题

代码

// Problem: E. Expected Power
// Contest: Codeforces - Codeforces Round 976 (Div. 2) and Divide By Zero 9.0
// URL: https://codeforces.com/contest/2020/problem/E
// Memory Limit: 256 MB
// Time Limit: 4000 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;
#define int long long
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 qpow(LL a , LL b)//快速幂
{LL sum=1;while(b){if(b&1){sum=sum*a%mod;}a=a*a%mod;b>>=1;}return sum;
}
void solve() 
{int t = qpow(10000 , mod - 2);int dp[1030][2];//取到i的概率memset(dp , 0 , sizeof dp);dp[0][0] = 1;int n;cin >> n;int a[n];for(int i = 0 ; i < n ; i ++){cin >> a[i];}for(int i = 0 ; i < n ; i ++){int p;cin >> p;for(int j = 0 ; j < 1024 ; j ++){int k = j ^ a[i];dp[k][1] += dp[j][0] * ((p * t) % mod);dp[k][1] %= mod;		dp[j][1] += dp[j][0] * (((10000 - p) * t) % mod);dp[j][1] %= mod;}for(int j = 0 ; j < 1024 ; j++){dp[j][0] = dp[j][1];dp[j][1] = 0;}}int ans = 0;for(int i = 1 ; i < 1024 ; i ++){ans += dp[i][0] * i * i;ans %= mod;}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;
}