篮球杯小白+强者

2. 宣读数字【算法赛】

思维题，注意到完全平方数的约数是奇数个，其余都是偶数个。

	#include <bits/stdc++.h>
	using namespace std;
	#define LL long long
	#define pb push_back
	#define x first
	#define y second 
	#define int long long 
	#define endl '\n'
	const LL maxn = 4e05+7;
	const LL N = 5e05+10;
	const LL mod = 988244353;
	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;
		}
	}
	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;
	}
	std::vector<int> minp, primes;
	void sieve(int n) {
	    minp.assign(n + 1, 0);
	    primes.clear();
	    
	    for (int i = 2; i <= n; i++) {
	        if (minp[i] == 0) {
	            minp[i] = i;
	            primes.push_back(i);
	        }
	        
	        for (auto p : primes) {
	            if (i * p > n) {
	                break;
	            }
	            minp[i * p] = p;
	            if (p == minp[i]) {
	                break;
	            }
	        }
	    }
	}
	
	void solve() 
	{
		cin >> n;
    	if((int)sqrt(n) * (int)sqrt(n) == n){
    		cout <<"L\n";
    	}
    	else{
    		cout <<"Q\n";
    	}
	}            
	signed main() 
	{
	    ios::sync_with_stdio(false);
	    cin.tie(0);
	    cout.tie(0);
	    cout.precision(10);
	    int t=1;
	    sieve(N); 
		cin>>t;
	    while(t--)
	    {
	    	solve();
	    }
	    return 0;
	}

3. 最大质因子个数【算法赛】

贪心：用尽可能多的质数来构造这个数。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define int long long 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 988244353;
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;
	}
}
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;
}
std::vector<int> minp, primes;
void sieve(int n) {
    minp.assign(n + 1, 0);
    primes.clear();
    
    for (int i = 2; i <= n; i++) {
        if (minp[i] == 0) {
            minp[i] = i;
            primes.push_back(i);
        }
        
        for (auto p : primes) {
            if (i * p > n) {
                break;
            }
            minp[i * p] = p;
            if (p == minp[i]) {
                break;
            }
        }
    }
}

void solve() 
{
	int n;
	cin >> n;
	int cnt = 0;
	int tmp = 1;
	for(auto it : primes){
		if(n / it >= tmp){
			cnt++;
			tmp *= it;		
		}
		else{
			break;
		}
	}
	cout << cnt << endl;
}            
signed main() 
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cout.precision(10);
    int t=1;
    sieve(N); 
	cin>>t;
    while(t--)
    {
    	solve();
    }
    return 0;
}

4. 物流选址【算法赛】

注意到无论怎么改变，这两个数的差值不会变，因此考虑到差值的每个约数能否满足题意，记录最小值即可。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define int long long 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 988244353;
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;
	}
}
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 n , m;
	cin >> n >> m;
	int k = m - n;
	if(k == 0 || m % n == 0){
		cout << 0 << endl;
	}
	else if(k == 1 || n >= k){
		cout << -1 << endl;
	}
	else{
		int ans = llinf;
		for(int i = 2 ; i * i <= k ; i ++){//可能的倍数
			if(k % i == 0){
				if(n % i == m % i){
					int tmpn = n + i - (n % i);
					int tmpm = m + i - (m % i);
					if(tmpm % tmpn == 0){
						ans = min(ans , i - n % i);
					}
				}
				if(n % (k / i) == m % (k / i)){
					int tmpn = n + (k / i) - (n % (k / i));
					int tmpm = m + (k / i) - (m % (k / i));
					if(tmpm % tmpn == 0){
						ans = min(ans , k / i - m % (k / i));
					}
				}
			}
		}
		int i = k;
		int tmpn = n + i - (n % i);
		int tmpm = m + i - (m % i);
		if(tmpm % tmpn == 0){
			ans = min(ans , i - n % i);
		}		
		if(ans == llinf){
			cout << -1 << endl;
		}
		else{
			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;
}

5. 小蓝的MEX问题【算法赛】

计数问题,对于每次询问，大于k的数全部可以选或者不选,而小于k的数至少选一个，然后可以预处理出所有的MEX取值情况，最后输出即可。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define int long long 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 998244353;
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;
	}
}
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 n , m;
	cin >> n >> m;
	vector<int>cnt(n + 5 ,0);
	for(int i = 0 ; i < n ; i ++){
		cin >> a[i];
		cnt[a[i]] ++;
	}	
	int MEX = 0;
	while(cnt[MEX] > 0){
		MEX++;
	}
	int pre = 1;
	vector<int>ans(n + 5 , 0);
	int tot = n;
	for(int i = 0 ; i <= MEX ; i ++){
		tot -= cnt[i];//这些随便选
		if(i == 0){
			ans[i] = qpow(2 , tot);
			ans[i]--;
			ans[i] += mod;
			ans[i] %= mod;
		}
		else{
			ans[i] = pre * qpow(2 , tot);
			ans[i] %= mod;
		}
		pre *= ((qpow(2 , cnt[i]) - 1 + mod) % mod);
		pre %= mod;
	}
	for(int i = 0 ; i < m ; i ++){
		int x;
		cin >> x;
		cout << ans[x] << 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;
}

6. 平摊购买费用【算法赛】

首先发现排序后没影响，因此先排个序，然后发现若要使得 l - f 最小，必然选取的是前y个数和后m-y个数。 $pre[c - y] + y * x - (pre[n] - pre[n - y] - y * x)$

构建关于y的函数，发现这是一个有波谷的函数，因此考虑三分求波谷即可。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define int long long 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 988244353;
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;
	}
}
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 n , m;
	cin >> n >> m;
	int a[n + 5];
	set<int>st;
	vector<int>pre(n + 5 , 0);
	for(int i = 1 ; i <= n ; i ++)
		cin >> a[i] , st.insert(a[i]);
	sort(a + 1, a + n + 1);
	for(int i = 1 ; i <= n ; i ++){
		pre[i] = pre[i - 1] + a[i];
	}
	map<int,int>mp;
	int idx = 1;
	for(auto it :st){
		mp[it] = idx++;
	}
	map<int,int>pm;
	for(int i = 1 ; i <= n ; i ++){
		int id = mp[a[i]];
		if(!pm.count(id)){
			pm[id] = i;
		}
	}
	//先找比x大的位置
	for(int i = 0 ; i < m ; i ++){
		int x , c;
		cin >> x >> c;
		auto it = st.lower_bound(x);
		if(it == st.end()){
			cout << pre[c] << endl;
		}
		else{
			int tmp = *it;
			//取前几个跟最后几个
			int ans = pre[c];
			auto check =[&] (int t){
				return pre[c - t] + t * x - (pre[n] - pre[n - t] - t * x); 
 			};
 			int l = 0 , r = c;
 			while(l < r){
 				int mid = (r - l) / 3;
 				if(r - l < 3){
 					for(int j = l ; j <= r ; j ++){
 						ans = min(ans , check(j));
 					}
 					break;
 				}
 				int m1 = l + mid;
 				int m2 = m1 + mid;
 				if(check(m1) > check(m2)){
 					l = m1;
 				}
 				else{
 					r = m2;
 				}
 			}
			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;
}

4. 电力之城【算法赛】

观察到一次只会使得电能增加1/2，而最终总的电能是可以确定的，因此变成了一个NIM问题，每次能拿一个或两个石头，求最终谁拿走了最后的石头。

#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() 
{
	int n;
	cin >> n;
	string s;
	cin >> s;
	int cnt = 0;
	for(int i = 1 ; i < n ;i ++){
		cnt += (s[i] == s[i - 1]);
	}	
	if(cnt % 3 == 0){
		cout << "qiao\n";
	}
	else{
		cout <<"lan\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;
}

5. 价值共性度【算法赛】

此题类似于昆明邀请赛的E题，需要知道这么一个事实：一个长度为n的数列，前缀GCD的数量不会超过logn个，因此我们只需要维护以某个数结尾，向前能够组成多少个GCD即可，并且记录这些GCD的最左侧位置，然后暴力求答案即可。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second 
#define int long long 
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 1e6+10;
const LL mod = 988244353;
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;
	}
}
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;
}
std::vector<int> minp, primes;
void sieve(int n) {
    minp.assign(n + 1, 0);
    primes.clear();
    
    for (int i = 2; i <= n; i++) {
        if (minp[i] == 0) {
            minp[i] = i;
            primes.push_back(i);
        }
        
        for (auto p : primes) {
            if (i * p > n) {
                break;
            }
            minp[i * p] = p;
            if (p == minp[i]) {
                break;
            }
        }
    }
}

void solve() 
{
	set< pair<int,int> >st;//前缀gcd
	set< pair<int,int> >pre;
	int n , k;
	cin >> n >> k;
	for(int i = 1 ; i <= n ; i ++){
		cin >> a[i];
	}
	vector<int>S(n + 5 , 0);
	for(int i = 1 ; i <= n ; i ++){
		S[i] = S[i - 1] + a[i];
	}
	int ans = 0;
	for(int i = 1 ; i <= n ; i ++){
		st.empty();
		set< pair<int,int> >tmp;
		tmp.insert({a[i] , i});
		for(auto it : pre){
			tmp.insert({gcd(a[i] , it.first) , it.second});
		}
    pre.clear();
		map<int,int>mp;
		for(auto it : tmp){
			if(mp.count(it.first)){
				continue;
			}
			else{
				mp[it.first] = 1;
				if(i - it.second + 1 >= k){
					ans = max(ans , it.first * (S[i] - S[it.second - 1]));
				}
				st.insert(it);
			}
		}
		swap(st , pre);
	}
	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;
}

6. 小蓝的逆序对问题【算法赛】

（不是正解..但卡过去了）

如此复杂度，想到用根号分治来解决问题，考虑交换两数后的逆序对该如何变化，然后想办法维护每个区间的信息

#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 = 2e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
const int B = 800;
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>num(N , 0);
long long sum = 0;
void merge(int s1, int e1, int s2, int e2){
    vector<int> temp;
    int p1 = s1;
    int p2 = s2;
    while(p1 <= e1&&p2 <= e2){
        if(num[p1] <= num[p2]){
            temp.push_back(num[p1++]);
        }
        else{
            sum += (e1-p1+1);
            temp.push_back(num[p2++]);
        }
    }
    while(p1 <= e1){
        temp.push_back(num[p1++]);
    }
    while(p2 <= e2){
        temp.push_back(num[p2++]);
    }
    for(int i = 0;i < (int)temp.size();i++){
        num[s1+i] = temp[i];
    }
}
void mergesort(int str, int end){
    if(str < end){
        int mid = (str + end)/2;
        mergesort(str,mid);
        mergesort(mid+1,end);
        merge(str,mid,mid+1,end);
    }
} 
struct BIT{//Binary indexed Tree(树状数组)
    int n;
    vector<int> t;
    BIT(int n) : n(n) , t(n + 1 , 0){
    }
    int lowbit(int x){
        return x & -x;
    }
    void modify(int k, int v) {
        while (k <= n) {
            t[k] += v;
            k += lowbit(k);
        }
    }
    void modify(int l, int r, int v) {
        modify(l, v), modify(r + 1, -v);  // 将区间加差分为两个前缀加
    }
    int query(int k) {
        int ret = 0;
        while(k) {
            ret += t[k];
            k -= lowbit(k);
        }
        return ret;
    }
    int query(int l , int r){
        return query(r) - query(l - 1);
    }
};
int ans[500][N];
void solve() 
{
    int n , k;
    cin >> n >> k;
    vector<int>idx(n + 5 , 0);
    memset(ans , -1 , sizeof ans);
    int tot = 505;
    vector<BIT>bit;
    for(int i = 0 ; i < tot ; i ++){
        BIT tmp(N);
        bit.pb(tmp);
    }
    for(int i = 1 ; i <= n ; i ++){
        cin >> num[i];
        idx[i] = (i - 1) / B;
    }
    int a[n + 5];
    for(int i = 1 ; i <= n ; i ++){
        a[i] = num[i];
    }
    unordered_map<int,int>mp;
    set<int>st;
    for(int i = 1 ; i <= n ; i ++){
        st.insert(num[i]);
    }
    int id = 1;
    for(auto it : st){
        mp[it] = id++;
    }
    for(int i = 1 ; i <= n ; i ++){
        int id = idx[i];
        int tp = mp[a[i]];
        bit[id].modify(tp + 1, 1);
    }
    mergesort(1,n);
//    cout << bit[0].query(4) << endl;
    for(int i = 0 ; i < k; i ++){
        int l , r;
        cin >> l >> r;
        long long tmp = sum;
        int ll = idx[l] , rr = idx[r];
    //    cout << ll << " " << rr << endl;
        if(ll == rr){
            for(int i = l ; i <= r ; i ++){
                if(a[r] < a[i]){
                    tmp--;
                }
                if(a[l] > a[i]){
                    tmp--;
                }
                if(a[l] < a[i]){
                    tmp++;
                }
                if(a[r] > a[i]){
                    tmp++;
                }
            }
        }
        else{
            for(int i = ll ; i <= rr ; i ++){
                if(i == ll){
                    for(int j = l ; j <= (ll + 1) * B ; j ++){
                        if(a[r] < a[j]){
                            tmp--;
                        }
                        if(a[l] > a[j]){
                            tmp--;
                        }
                        if(a[l] < a[j]){
                            tmp++;
                        }
                        if(a[r] > a[j]){
                            tmp++;
                        }
                    }
                }
                else if(i == rr){
                    for(int j = rr * B + 1 ; j <= r ; j ++){
                        if(a[r] < a[j]){
                            tmp--;
                        }
                        if(a[l] > a[j]){
                            tmp--;
                        }
                        if(a[l] < a[j]){
                            tmp++;
                        }
                        if(a[r] > a[j]){
                            tmp++;
                        }
                    }                    
                }
                else{
                    int id1 = mp[a[r]];
                    if(ans[i][id1] != -1){
                        tmp += ans[i][id1];
                    }
                    else{
                        ans[i][id1] = bit[i].query(id1);
                           tmp += ans[i][id1];                        
                    }
                    if(ans[i][id1 + 1] != -1){
                            tmp -= B - ans[i][id1 + 1];//比a[r]大的                        
                    }
                    else{
                        ans[i][id1 + 1] = bit[i].query(id1 + 1);
                                   tmp -= B - ans[i][id1 + 1];//比a[r]大的                               
                    }
                    int id2 = mp[a[l]];
                    if(ans[i][id2] != -1){
                        tmp -= ans[i][id2];
                    }
                    else{
                        ans[i][id2] = bit[i].query(id2);
                           tmp -= ans[i][id2];                        
                    }
                    if(ans[i][id2 + 1] != -1){
                            tmp += B - ans[i][id2 + 1];//比a[r]大的                        
                    }
                    else{
                        ans[i][id2 + 1] = bit[i].query(id2 + 1);
                                   tmp += B - ans[i][id2 + 1];//比a[r]大的                               
                    }
                }
                //cout << tmp << endl;
            }
        }
        if(a[l] > a[r]) tmp++;
        if(a[l] < a[r]) tmp--;
        cout << tmp << 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;
}