补一下下午组队赛的题

ICPC 2018青岛D Magic Multiplication

题目链接

先枚举 a 的第一位是哪个数字，然后通过 a 的第一位可以推出 b 的整个串，再靠 b 去推 a 的整个串

代码有点难写

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<PII, int> PIII;const int N = 1e8 + 3;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;void solve()
{int n, m;cin >> n >> m;string s;cin >> s;vector<int> c(s.size());for (int i = 0; i < s.size(); i ++ ) c[i] = s[i] - '0';vector<int> a, b;int p = 0;auto set = [&](int x){a.push_back(x);p = 0;for (int i = 0; i < m; i ++ ){if (p < s.size() && c[p] % x == 0 && c[p] / x < 10) b.push_back(c[p] / x), p ++ ;else if (p + 1 < s.size() && (c[p] * 10 + c[p + 1]) % x == 0 && (c[p] * 10 + c[p + 1]) / x < 10) b.push_back((c[p] * 10 + c[p + 1]) / x), p += 2;else return false;}return true;};auto check = [&](){for (int i = 1; i < n; i ++ ){if (p < s.size() && c[p] % b[0] == 0 && c[p] / b[0] < 10) a.push_back(c[p] / b[0]), p ++ ;else if (p + 1 < s.size() && (c[p] * 10 + c[p + 1]) % b[0] == 0 && (c[p] * 10 + c[p + 1]) / b[0] < 10) a.push_back((c[p] * 10 + c[p + 1]) / b[0]), p += 2;else return false;for (int j = 1; j < m; j ++ ){if (p < s.size() && a[i] * b[j] == c[p]) p ++ ;else if (p + 1 < s.size() && c[p] * 10 + c[p + 1] == a[i] * b[j]) p += 2;else return false;}}return p == s.size();};for (int i = 1; i <= 9; i ++ ){a.clear();b.clear();if (set(i) && check()){for (auto t : a) cout << t;cout << ' ';for (auto t : b) cout << t;cout << '\n';return;}}cout << "Impossible\n";
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

ICPC 2018青岛E Plants vs. Zombies

题目链接

二分最终答案，要注意的点是，如果在某个位置已经没有移动步数了，但是在他之后的位置已经满足条件了，那也是符合条件的

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<PII, int> PIII;const int N = 1e8 + 3;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;void solve()
{int n, m;cin >> n >> m;vector<int> a(n + 1);for (int i = 1; i <= n; i ++ ) cin >> a[i];int l = 0, r = 1e18;auto check = [&](int x){int cnt = m;vector<int> b(n + 2);for (int i = 1; i <= n; i ++ ){if (cnt > 0){b[i] += a[i];cnt -- ;}if (b[i] >= x) continue;cnt -= ((x - b[i] + a[i] - 1) / a[i]) * 2;b[i + 1] += ((x - b[i] + a[i] - 1) / a[i]) * a[i + 1];if (cnt < 0) return false;}return true;};while (l < r){int mid = l + r + 1 >> 1;if (check(mid)) l = mid;else r = mid - 1;}cout << r << '\n';
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}

ICPC 2018青岛F Tournament

题目链接

非常歹毒的数独，没有小脑的找规律

首先要发现 n 个人最多进行 lowbit(n) 场比赛，所以如果 k 不满足条件就直接退出

然后还是找规律，举个例子：

i
0	[1 2] [3 4] [5 6] [7 8]
1	[2 1 4 3] [6 5 8 7]
2	[3 4] [1 2] [7 8] [5 6]
3	[4 3 2 1 8 7 6 5]
4	[5 6] [7 8] [1 2] [3 4]
5	[6 5 8 7] [2 1 4 3]
6	[7 8] [5 6] [3 4] [1 2]
7	8 7 6 5 4 3 2 1

每次翻转的序列长度是 lotbit(i)

所以翻转一下就好了

#include <bits/stdc++.h>using namespace std;#define int long long
using i64 = long long;typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef pair<PII, int> PIII;const int N = 1e8 + 3;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
const int mod1 = 954169327;
const int mod2 = 906097321;
const int INF = 0x3f3f3f3f3f3f3f3f;int lowbit(int x)
{return x & -x;
}void solve()
{int n, k;cin >> n >> k;if (n % 2 != 0 || k >= lowbit(n)){cout << "Impossible\n";return;}vector<int> a(n + 1);for (int i = 1; i <= n; i ++ ) a[i] = i;for (int i = 1; i <= k; i ++ ){int lb = lowbit(i);for (int j = 1; j <= n; j += 2 * lb){for (int len = 0; len < lb; len ++ ){swap(a[j + len], a[j + lb * 2 - 1 - len]);}}for (int j = 1; j <= n; j ++ ){cout << a[j];if (j != n) cout << ' ';}cout << '\n';}
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t = 1;cin >> t;while (t -- ){solve();}
}