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蓝桥杯第四场双周赛（1~6）

news 来源：原创 2024/5/11 1:08:05

1、水题

2、模拟题，写个函数即可

#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 = 1e09+7;
const int inf = 0x3f3f;
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;
int a[N];
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
int qc(int a, int b , int c){return (a + b + c)/2;
}
int alg(int a , int b , int c){int cc = qc(a , b , c);return (cc * (cc - a) * (cc - b) * (cc - c));
}
void solve() 
{int a , b , c;cin >> a >> b >> c;if(a + b <= c || a + c <= b || b + c <= a){cout << -1;}elsecout << alg(a , b , c);
}            
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;
}

3、模拟题，找规律，第一行和最后一行只有两个数，其余行都是三个数。

第一行特殊处理，其余行： $(x + 1) / 3 + 1$ 就是当前所在行 $r$ ， $x - ((r - 1) * 3 - 1)$ 就是所在行第s个数 , 每行第一个数是 $r - 1$ ，因此所在列就是r - 1 + s。

#include <iostream>
using namespace std;
int main()
{long long n , m;cin >> n >> m;for(int i = 0 ; i < m ; i ++){long long x;cin >> x;if(x <= 1){cout << 1 << " " << x + 1 << endl;}else{long long r = (x + 1) / 3 + 1;long long st = x - ((r - 1) * 3 - 1);long long dc = r - 1 + st;cout << r << " " << dc << endl;}}return 0;
}

4、考虑找到 $x^a , y^b , z^c$ 的所有可能取值，取值上界应该为 $10^{12} * 10^5 = 10^{17}$ 。由于 $2^{64}>10^{18}$ ,因此每个肯定不超过64种取值。用三重循环找到所有 $x^a + y^b + z^c$ 的所有取值，复杂度为 $O(64^3)$ 。注意 $x^a*x<inf$ 判断可能会爆long long ，所以在判断是否到达上界需要用 $x^a < inf/x$ 。用数组或者set去存每种取值，然后从小到大排序。按照题目条件对每个询问搜索即可（二分/暴力）。整体复杂度 $O(q * 64^3)/O(q*log(64^3))$

#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 = 0x3f3f;
const LL llinf = 2e18;
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;
LL a , b , c;
set<LL>st;
void solve() 
{cin >> a >> b >> c;vector<LL>aa , bb , cc;aa.pb(1);bb.pb(1);cc.pb(1);LL x = 1;while(a != 1 && x < llinf / a){x *= a;aa.pb(x);}LL y = 1;while(b != 1 && y < llinf / b){y *= b;bb.pb(y);}LL z = 1;while(c != 1 && z < llinf / c){z *= c;cc.pb(z);}for(int i = 0 ; i < aa.size() ; i ++){for(int j = 0 ; j < bb.size() ; j ++){for(int z = 0 ; z < cc.size() ; z ++){st.insert(aa[i] + bb[j] + cc[z]);}}}int m;cin >> m;for(int i = 0 ; i < m ; i ++){LL que;cin >> que;auto it = st.upper_bound(que);while(*it - que == 1){que = *it;it = st.upper_bound(que);}cout << que + 1 << " " << (*it - que - 1) << 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;
}

5、方法很多，大体思路为将类型一样的宝石放到一起，将他们的作用区间进行合并，然后对整个数组进行区间修改。

区间合并：将所有区间按照左端点排序，遍历区间，若当前左端点与前一个区间右端点有重合部分，则将他们合并成一个区间，否则将前一个区间存下来，当前区间为一个新的区间。

区间修改：差分/树状数组/线段树。

#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 = 0x3f3f;
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;
int a[N];
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
struct BIT{//Binary indexed Tree(树状数组)int n;vector<int> tr;BIT(int n) : n(n) , tr(n + 1 , 0){}int lowbit(int x){return x & -x;}void modify(int x , int modify_number){for(int i = x ; i <= n ; i += lowbit(i)){tr[i] += modify_number;}}void modify(int l , int r , int modify_number){modify(l , modify_number);modify(r + 1 , -modify_number);}int query(int x){int res = 0;for(int i = x ; i ;  i -= lowbit(i))res += tr[i];return res;}int query(int x , int y){return query(y) - query(x);}
};
void solve() 
{int n , m , q;cin >> n >> m >> q;vector<int>len(m + 5);for(int i = 1 ; i <= m ; i++){cin >> len[i];}BIT bit(n);vector<pair<int,int>>que;for(int i = 0 ; i < q ; i ++){int x , y;cin >> x >> y;que.pb({x , y});}sort(que.begin() , que.end());int r = 0 , pos = 0;for(int i = 0 ;i < q ; i ++){if(que[i].x != pos){pos = que[i].x;r = 0;}bit.modify( max(r + 1, que[i].y) , min(que[i].y + len[pos] - 1 , n) , 1);r = min(que[i].y + len[pos] - 1 , n);}for(int i = 1 ; i <= n ; i ++){cout << bit.query(i)<<" ";}
}            
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;
}

6、删除区间求中位数比较困难。相反，增加数求区间中位数就是一道对顶堆的板子题了。因此考虑逆着做题，先将所有会飘走的气球放弃，将其余气球加入对顶堆。然后再从后往前依次添加气球，维护对顶堆找答案即可（对顶堆网上一大堆模板）。

#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 = 0x3f3f;
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;
int a[N];
void init(int n){for(int i = 0 ; i <= n ; i ++){a[i] = 0;}
}
void solve() 
{cin >> n;for(int i = 1 ; i <= n ; i ++){cin >> a[i];}cin >> m;double ans[m + 5];int que[m + 5];int vis[n + 5];memset(vis,0,sizeof vis);for(int i = 1 ; i <= m ; i ++){cin >> que[i];vis[que[i]] = 1;}for(int i = 1 ;i <= n ; i ++){if(!vis[i]){ma.push(a[i]);}}while(ma.size() > mi.size()){mi.push(ma.top());ma.pop();}for(int i = m ; i > 0 ; i --){if((mi.size() + ma.size()) % 2 == 0){//偶数int x = mi.top();int y = ma.top();ans[i] = (double)(1.0 * x + y) / 2;}else{double x = mi.top();ans[i] = (double)(1.0 * x);}int yy = mi.top();if(a[que[i]] > yy){mi.push(a[que[i]]);}else{ma.push(a[que[i]]);}while(mi.size() > ma.size() + 1){ma.push(mi.top());mi.pop();}while(ma.size() > mi.size()){mi.push(ma.top());ma.pop();}}for(int i = 1 ; i <= m ; i++){printf("%.1f " , ans[i]);}
}            
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;
}

7、边数据较小，网络流问题。