2024.12.11 周三

2024.12.11 周三

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Q1. 1100

给定一长度为 $n$ 的数组，你需要执行 $k$ 次操作：每次选择一连续子数组(可为空)，将和作为一元素放到到数组的任意位置。问最后数组和的最大值。

Q2. 1100

给你一长度为$2n$的数组$a$，$1$~$n$各出现2次。让你找出两个大小为$2k$集合$l$,$r$,其中$l$属于$a$₁至$a$_n，$r$属于$a$_n+1至$a$_2n。满足$l$异或和等于$r$异或和。

Q3. 1000

Rudolf has an array $a$ of $n$ integers.
In one operation, he can choose an index $i$ ($2 \le i \le n - 1$) and assign:

• $a_{i - 1} = a_{i - 1} - 1$
• $a_i = a_i - 2$
• $a_{i + 1} = a_{i + 1} - 1$

Rudolf can apply this operation any number of times. Any index $i$ can be used zero or more times.

Can he make all the elements of the array equal to zero using this operation?

Q4. 1100

给你一长度为$n$的数组，$x$为0，$i$:1~n,找到大于$x$且为$a$_i倍数的数,更新$x$。

------------------------独自思考分割线------------------------

A1. 2点

1.第一次操作必然是选择最大连续子数组，然后将其放到最大连续子数组内。

2.以后的操作就可以看作最大和一变二，二变四的过程。

A2. 2点

1.发现$a$₁至$a$_n这一侧单独出现的数字与右侧相同。

2.同时出现$2$次的数异或和为0。可以先使用出现$2$次的数构造，不够再使用单次出现的数构造。

A3. 1点

1.遍历贪心去减，不够减/最后没减为0则NO

A4. 2点

1.快速找到大于$x$的$a$_i的倍数，直接枚举显然不可行。二分当然可以。

2.数学当然ok，$k$$a$_i>x,=>$k$>=$x$/$a$_i,==>k=$x$/$a$_i+1，取整的时候注意一下，答案就是$k$$a$_i。

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A1.

#include <bits/stdc++.h>
#define int long long //
#define endl '\n'     // 交互/调试 关
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
void _();
signed main()
{ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);cout << fixed << setprecision(6);int T = 1;cin >> T;while (T--)_();return 0;
}
const int mod = 1e9 + 7;
int qm(int a, int b)
{int res = 1;while (b){if (b & 1)res = res * a % mod; // 不要取模时记得取消a = a * a % mod;b >>= 1;}return res;
}
int inv(int n)
{return qm(n, mod - 2);
}void _()
{int n, k;cin >> n >> k;int pre = 0, min_pre = 0, max_rangesum = 0;for (int i = 0; i < n; i++){int x;cin >> x;pre += x;max_rangesum = max(max_rangesum, pre - min_pre);min_pre = min(min_pre, pre);}int res = (max_rangesum % mod * qm(2, k)) % mod;res = (res + pre % mod - max_rangesum % mod + (int)(1e7) * mod) % mod;cout << res << endl;
}

A2.

#include <bits/stdc++.h>
#define int long long //
#define endl '\n'     // 交互/调试 关
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
void _();
signed main()
{ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);cout << fixed << setprecision(6);int T = 1;cin >> T;while (T--)_();return 0;
}
void _()
{int n, k;cin >> n >> k;vector<int> l(n + 1);for (int i = 0; i < n; i++){int x;cin >> x;l[x]++;}for (int i = 0; i < n; i++){int x;cin >> x;}set<int> db_l, db_r, sgal;for (int i = 1; i <= n; i++)if (!l[i])db_r.insert(i);else if (l[i] == 1)sgal.insert(i);elsedb_l.insert(i);vector<int> res_l, res_r;auto get = [&](vector<int> &res, set<int> &l){for (auto x : l)if (res.size() < k << 1)res.push_back(x), res.push_back(x);for (auto x : sgal)if (res.size() < k << 1)res.push_back(x);};get(res_l, db_l);get(res_r, db_r);for (auto v : res_l)cout << v << ' ';cout << endl;for (auto v : res_r)cout << v << ' ';cout << endl;
}

A3.

#include <bits/stdc++.h>
#define int long long //
// #define endl '\n'     // 交互/调试 关
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
void _();
signed main()
{ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);cout << fixed << setprecision(6);int T = 1;cin >> T;while (T--)_();return 0;
}void _()
{int n;cin >> n;vector<int> a(n + 1);for (int i = 1; i <= n; i++)cin >> a[i];int f = 1;for (int i = 1; i + 2 <= n; i++){if (a[i + 1] < a[i] << 1 || a[i + 2] < a[i])f = 0;a[i + 1] -= a[i] << 1;a[i + 2] -= a[i];}if (a[n - 1] || a[n])f = 0;cout << (f ? "YES" : "NO") << endl;
}

A4.

#include <bits/stdc++.h>
#define int long long //
#define endl '\n'     // 交互/调试 关
using namespace std;
#define bug(BUG) cout << "bug:# " << (BUG) << endl
#define bug2(BUG1, BUG2) cout << "bug:# " << (BUG1) << " " << (BUG2) << endl
#define bug3(BUG1, BUG2, BUG3) cout << "bug:# " << (BUG1) << ' ' << (BUG2) << ' ' << (BUG3) << endl
void _();
signed main()
{ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);cout << fixed << setprecision(6);int T = 1;cin >> T;while (T--)_();return 0;
}void _()
{int n;cin >> n;int x;cin >> x;int res = x;for (int i = 2; i <= n; i++){cin >> x;int k = res / x + 1;res = x * k;}cout << res << endl;
}