20240424codeforces刷题题解

240424刷题题解

Walk on Matrix

CodeForces - 1332D

思路

构造题，每个 $dp_{i,j}$ 都是由其左上方向中的按位与最大值决定的。

我们需要从使得贪心解与正确解的差值为 $k$ 。

为了方便获得 $k$ ，可以考虑构造一个贪心解为 $0$ 的矩阵。

写出例二的二进制表示
在这里插入图片描述

观察可知左边界上的数只能由其上方的数决定（上边界同理）。

由按位与的性质可知， $2^{n}-1$ 即 $1111111\ldots$ 这种形式的数按位与任意数 $x$ 得到的数任然是 $x$ 本身。

利用这两条性质，我们可以利用左边界或者上边界来传递需要的数。

由按位与的性质可知，如果 $2^n>x$ ，那么 $2^n\&x=0$ 。

利用条性质，我们可以构造出如下矩阵，利用上边界和右边界（或左边界和下边界）以及 $2^{n+1}-1$ 传递 $k$ ，利用 $2^n$ 来截断边界上的正确解。

在这里插入图片描述

题解

#define _CRT_SECURE_NO_WARNINGS 1#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <unordered_map>
#include <cstring>
#include <cstdio>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <bitset>
#include <cmath>
#include <numeric>#define endl '\n'#define ft first
#define sd second#define yes cout << "yes\n"
#define no cout << "no\n"#define Yes cout << "Yes\n"
#define No cout << "No\n"#define YES cout << "YES\n"
#define NO cout << "NO\n"#define pb push_back
#define eb emplace_back#define all(x) x.begin(), x.end()
#define unq_all(x) x.erase(unique(all(x)), x.end())
#define sort_all(x) sort(all(x))
#define reverse_all(x) reverse(all(x))#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f3f#define RED cout << "\033[91m"
#define GREEN cout << "\033[92m"
#define YELLOW cout << "\033[93m"
#define BLUE cout << "\033[94m"
#define MAGENTA cout << "\033[95m"
#define CYAN cout << "\033[96m"
#define RESET cout << "\033[0m"// 红色
#define DEBUG1(x)                     \RED;                              \cout << #x << " : " << x << endl; \RESET;// 绿色
#define DEBUG2(x)                     \GREEN;                            \cout << #x << " : " << x << endl; \RESET;// 蓝色
#define DEBUG3(x)                     \BLUE;                             \cout << #x << " : " << x << endl; \RESET;// 品红
#define DEBUG4(x)                     \MAGENTA;                          \cout << #x << " : " << x << endl; \RESET;// 青色
#define DEBUG5(x)                     \CYAN;                             \cout << #x << " : " << x << endl; \RESET;// 黄色
#define DEBUG6(x)                     \YELLOW;                           \cout << #x << " : " << x << endl; \RESET;using namespace std;typedef long long ll;
typedef unsigned long long ull;typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef pair<string, int> psi;
typedef pair<string, ll> psl;typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<string> vs;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<pss> vpss;typedef vector<vi> vvi;
typedef vector<vl> vvl;typedef queue<int> qi;
typedef queue<ll> ql;
typedef queue<pii> qpii;
typedef queue<pll> qpll;
typedef queue<psi> qpsi;
typedef queue<psl> qpsl;typedef priority_queue<int> pqi;
typedef priority_queue<ll> pql;
typedef priority_queue<string> pqs;
typedef priority_queue<pii> pqpii;
typedef priority_queue<psi> pqpsi;
typedef priority_queue<pll> pqpl;
typedef priority_queue<psi> pqpsl;typedef map<int, int> mii;
typedef map<int, bool> mib;
typedef map<ll, ll> mll;
typedef map<ll, bool> mlb;
typedef map<char, int> mci;
typedef map<char, ll> mcl;
typedef map<char, bool> mcb;
typedef map<string, int> msi;
typedef map<string, ll> msl;
typedef map<int, bool> mib;typedef unordered_map<int, int> umii;
typedef unordered_map<ll, ll> uml;
typedef unordered_map<char, int> umci;
typedef unordered_map<char, ll> umcl;
typedef unordered_map<string, int> umsi;
typedef unordered_map<string, ll> umsl;template <typename T>
inline T read()
{T x = 0;int y = 1;char ch = getchar();while (ch > '9' || ch < '0'){if (ch == '-')y = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){x = (x << 3) + (x << 1) + (ch ^ 48);ch = getchar();}return x * y;
}template <typename T>
inline void write(T x)
{if (x < 0){putchar('-');x = -x;}if (x >= 10){write(x / 10);}putchar(x % 10 + '0');
}/*#####################################BEGIN#####################################*/void solve()
{int k;cin >> k;cout << 3 << " " << 2 << endl;int t1 = 131072;int t2 = 262143;cout << t2 << " " << k << endl;cout << t1 << " " << t2 << endl;cout << k << " " << k << endl;
}int main()
{ios::sync_with_stdio(false), std::cin.tie(nullptr), std::cout.tie(nullptr);int _ = 1;// std::cin >> _;while (_--){solve();}return 0;
}//

Peaceful Rooks

CodeForces - 1411C

图，dfs

思路

仔细思考不难发现，如果以相同行号和列号作为一个节点的编号，那么可以轻易的将点在棋盘上的位置关系转化为图上的边，那么要移动棋子使得棋子的行号和列行相同，则表示为要将我们所建的抽象图的每个点转化为自环。因为同一行和列上不能同时出现两个棋子，即一个点的度数不能大于2。所以如果要把一个环上的所有点的边都变成自环，就需要额外一个点辅助存放拆下来的第一条边，所以需要 $n + 1$ 次操作，如果把一条链上的所有点的边变成自环，只需 $n$ 次操作。

所以， $总操作数 = 环上的点数 + 1 + 链上的点数$

只需要用并查集查一下抽象图上的环就行了。

题解

#define _CRT_SECURE_NO_WARNINGS 1#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <unordered_map>
#include <cstring>
#include <cstdio>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <bitset>
#include <cmath>
#include <numeric>#define endl '\n'#define ft first
#define sd second#define yes cout << "yes\n"
#define no cout << "no\n"#define Yes cout << "Yes\n"
#define No cout << "No\n"#define YES cout << "YES\n"
#define NO cout << "NO\n"#define pb push_back
#define eb emplace_back#define all(x) x.begin(), x.end()
#define unq_all(x) x.erase(unique(all(x)), x.end())
#define sort_all(x) sort(all(x))
#define reverse_all(x) reverse(all(x))#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f3f#define RED cout << "\033[91m"
#define GREEN cout << "\033[92m"
#define YELLOW cout << "\033[93m"
#define BLUE cout << "\033[94m"
#define MAGENTA cout << "\033[95m"
#define CYAN cout << "\033[96m"
#define RESET cout << "\033[0m"// 红色
#define DEBUG1(x)                     \RED;                              \cout << #x << " : " << x << endl; \RESET;// 绿色
#define DEBUG2(x)                     \GREEN;                            \cout << #x << " : " << x << endl; \RESET;// 蓝色
#define DEBUG3(x)                     \BLUE;                             \cout << #x << " : " << x << endl; \RESET;// 品红
#define DEBUG4(x)                     \MAGENTA;                          \cout << #x << " : " << x << endl; \RESET;// 青色
#define DEBUG5(x)                     \CYAN;                             \cout << #x << " : " << x << endl; \RESET;// 黄色
#define DEBUG6(x)                     \YELLOW;                           \cout << #x << " : " << x << endl; \RESET;using namespace std;typedef long long ll;
typedef unsigned long long ull;typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef pair<string, int> psi;
typedef pair<string, ll> psl;typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<string> vs;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<pss> vpss;typedef vector<vi> vvi;
typedef vector<vl> vvl;typedef queue<int> qi;
typedef queue<ll> ql;
typedef queue<pii> qpii;
typedef queue<pll> qpll;
typedef queue<psi> qpsi;
typedef queue<psl> qpsl;typedef priority_queue<int> pqi;
typedef priority_queue<ll> pql;
typedef priority_queue<string> pqs;
typedef priority_queue<pii> pqpii;
typedef priority_queue<psi> pqpsi;
typedef priority_queue<pll> pqpl;
typedef priority_queue<psi> pqpsl;typedef map<int, int> mii;
typedef map<int, bool> mib;
typedef map<ll, ll> mll;
typedef map<ll, bool> mlb;
typedef map<char, int> mci;
typedef map<char, ll> mcl;
typedef map<char, bool> mcb;
typedef map<string, int> msi;
typedef map<string, ll> msl;
typedef map<int, bool> mib;typedef unordered_map<int, int> umii;
typedef unordered_map<ll, ll> uml;
typedef unordered_map<char, int> umci;
typedef unordered_map<char, ll> umcl;
typedef unordered_map<string, int> umsi;
typedef unordered_map<string, ll> umsl;template <typename T>
inline T read()
{T x = 0;int y = 1;char ch = getchar();while (ch > '9' || ch < '0'){if (ch == '-')y = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){x = (x << 3) + (x << 1) + (ch ^ 48);ch = getchar();}return x * y;
}template <typename T>
inline void write(T x)
{if (x < 0){putchar('-');x = -x;}if (x >= 10){write(x / 10);}putchar(x % 10 + '0');
}/*#####################################BEGIN#####################################*/const int N = 2e5 + 5;int p[N];int find(int x)
{if (p[x] != x)p[x] = find(p[x]);return p[x];
}void solve()
{int n, m;cin >> n >> m;for (int i = 1; i <= n; i++)p[i] = i;int a, b;int ans = 0;for (int i = 0; i < m; i++){cin >> a >> b;if (a == b)continue;if (find(a) == find(b))ans += 2;else{ans++;p[find(a)] = find(b);}}cout << ans << endl;
}int main()
{ios::sync_with_stdio(false), std::cin.tie(nullptr), std::cout.tie(nullptr);int _ = 1;std::cin >> _;while (_--){solve();}return 0;
}// https://codeforces.com/problemset/problem/1411/C

Mathematical Problem

CodeForces - 1916D

思路

对于 $n\leq3$ 来说，答案由样例直接给出了。

所以考虑 $n\geq5$ 的情况。

若 $a=b^2$ ，则 $b\times100=(c\times10)^2$ ，所以对于任意 $n\geq3$ 来说，我们都可以由 $n - 2$ 的方案乘以 $100$ 获得 $n - 2$ 种方案，所以我们需要只需要再额外寻找两个方案就行了。

由于 $×100 \times100$ 使得数集中多了两个 $0$ ，所以需要寻找一种平方数，在非首位插入两个零之后仍为平方数。

观察样例给出的 $n$ 为 $3$ 的方案： $169, 196, 961$ ，尝试插入两个 $0$ ，发现 $10609=103^2,90601=301^2$ 。

以此类推，发现 $1006009=1003^2,9006001=3001^2$ 。

所以所需的两个平方数可以通过在 $169$ 和 $961$ 的第 $2$ 位和倒数第 $2$ 位插入 $\frac {n-3}{2}$ 个 $0$ 获得。

题解

#define _CRT_SECURE_NO_WARNINGS 1#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <unordered_map>
#include <cstring>
#include <cstdio>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <bitset>
#include <cmath>
#include <numeric>#define endl '\n'#define ft first
#define sd second#define yes cout << "yes\n"
#define no cout << "no\n"#define Yes cout << "Yes\n"
#define No cout << "No\n"#define YES cout << "YES\n"
#define NO cout << "NO\n"#define pb push_back
#define eb emplace_back#define all(x) x.begin(), x.end()
#define unq_all(x) x.erase(unique(all(x)), x.end())
#define sort_all(x) sort(all(x))
#define reverse_all(x) reverse(all(x))#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f3f#define RED cout << "\033[91m"
#define GREEN cout << "\033[92m"
#define YELLOW cout << "\033[93m"
#define BLUE cout << "\033[94m"
#define MAGENTA cout << "\033[95m"
#define CYAN cout << "\033[96m"
#define RESET cout << "\033[0m"// 红色
#define DEBUG1(x)                     \RED;                              \cout << #x << " : " << x << endl; \RESET;// 绿色
#define DEBUG2(x)                     \GREEN;                            \cout << #x << " : " << x << endl; \RESET;// 蓝色
#define DEBUG3(x)                     \BLUE;                             \cout << #x << " : " << x << endl; \RESET;// 品红
#define DEBUG4(x)                     \MAGENTA;                          \cout << #x << " : " << x << endl; \RESET;// 青色
#define DEBUG5(x)                     \CYAN;                             \cout << #x << " : " << x << endl; \RESET;// 黄色
#define DEBUG6(x)                     \YELLOW;                           \cout << #x << " : " << x << endl; \RESET;using namespace std;typedef long long ll;
typedef unsigned long long ull;typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef pair<string, int> psi;
typedef pair<string, ll> psl;typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<string> vs;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<pss> vpss;typedef vector<vi> vvi;
typedef vector<vl> vvl;typedef queue<int> qi;
typedef queue<ll> ql;
typedef queue<pii> qpii;
typedef queue<pll> qpll;
typedef queue<psi> qpsi;
typedef queue<psl> qpsl;typedef priority_queue<int> pqi;
typedef priority_queue<ll> pql;
typedef priority_queue<string> pqs;
typedef priority_queue<pii> pqpii;
typedef priority_queue<psi> pqpsi;
typedef priority_queue<pll> pqpl;
typedef priority_queue<psi> pqpsl;typedef map<int, int> mii;
typedef map<int, bool> mib;
typedef map<ll, ll> mll;
typedef map<ll, bool> mlb;
typedef map<char, int> mci;
typedef map<char, ll> mcl;
typedef map<char, bool> mcb;
typedef map<string, int> msi;
typedef map<string, ll> msl;
typedef map<int, bool> mib;typedef unordered_map<int, int> umii;
typedef unordered_map<ll, ll> uml;
typedef unordered_map<char, int> umci;
typedef unordered_map<char, ll> umcl;
typedef unordered_map<string, int> umsi;
typedef unordered_map<string, ll> umsl;template <typename T>
inline T read()
{T x = 0;int y = 1;char ch = getchar();while (ch > '9' || ch < '0'){if (ch == '-')y = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){x = (x << 3) + (x << 1) + (ch ^ 48);ch = getchar();}return x * y;
}template <typename T>
inline void write(T x)
{if (x < 0){putchar('-');x = -x;}if (x >= 10){write(x / 10);}putchar(x % 10 + '0');
}/*#####################################BEGIN#####################################*/
const int N = 105;vs ans[N];
void solve()
{int n;cin >> n;for (auto an : ans[n]){cout << an << endl;}
}int main()
{ios::sync_with_stdio(false), std::cin.tie(nullptr), std::cout.tie(nullptr);int _ = 1;std::cin >> _;ans[1].eb("1");ans[3].eb("169");ans[3].eb("196");ans[3].eb("961");for (int i = 5; i <= 99; i += 2){for (auto j : ans[i - 2]){ans[i].eb(j + "00");}int len = (i - 3) / 2;string s1 = "1";string s2 = "9";for (int j = 0; j < len; j++){s1 += "0";s2 += "0";}s1 += "6";s2 += "6";for (int j = 0; j < len; j++){s1 += "0";s2 += "0";}s1 += "9";s2 += "1";ans[i].eb(s1);ans[i].eb(s2);}while (_--){solve();}return 0;
}//

Pekora and Trampoline

CodeForces - 1491C

思路

对于 $s_i$ 来说，需要经过的轮数，为 $s_i-f(s_i)$ ， $f(s_i)$ 表示 $j+s_j =i,j\leq i$ ，即 $s_{i}$ 之前会经过 $s_{i}$ 的蹦床。

为了尽可能经过尽量多的蹦床，要尽量从编号小的蹦床开始新的一轮。所以从 $i = 1$ 开始遍历，如果 $s_{i}\neq1$ ，那么对于所有的 $i+2\sim i+s_{i}$ 都会被经过，如果当 $s_{i}=1$ 时这个点被经过，那么 $s_{i+1}$ 也会受影响。

题解

#define _CRT_SECURE_NO_WARNINGS 1#include <iostream>
#include <vector>
#include <algorithm>
#include <set>
#include <unordered_map>
#include <cstring>
#include <cstdio>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <list>
#include <bitset>
#include <cmath>
#include <numeric>#define endl '\n'#define ft first
#define sd second#define yes cout << "yes\n"
#define no cout << "no\n"#define Yes cout << "Yes\n"
#define No cout << "No\n"#define YES cout << "YES\n"
#define NO cout << "NO\n"#define pb push_back
#define eb emplace_back#define all(x) x.begin(), x.end()
#define unq_all(x) x.erase(unique(all(x)), x.end())
#define sort_all(x) sort(all(x))
#define reverse_all(x) reverse(all(x))#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f3f#define RED cout << "\033[91m"
#define GREEN cout << "\033[92m"
#define YELLOW cout << "\033[93m"
#define BLUE cout << "\033[94m"
#define MAGENTA cout << "\033[95m"
#define CYAN cout << "\033[96m"
#define RESET cout << "\033[0m"// 红色
#define DEBUG1(x)                     \RED;                              \cout << #x << " : " << x << endl; \RESET;// 绿色
#define DEBUG2(x)                     \GREEN;                            \cout << #x << " : " << x << endl; \RESET;// 蓝色
#define DEBUG3(x)                     \BLUE;                             \cout << #x << " : " << x << endl; \RESET;// 品红
#define DEBUG4(x)                     \MAGENTA;                          \cout << #x << " : " << x << endl; \RESET;// 青色
#define DEBUG5(x)                     \CYAN;                             \cout << #x << " : " << x << endl; \RESET;// 黄色
#define DEBUG6(x)                     \YELLOW;                           \cout << #x << " : " << x << endl; \RESET;using namespace std;typedef long long ll;
typedef unsigned long long ull;typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<string, string> pss;
typedef pair<string, int> psi;
typedef pair<string, ll> psl;typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<string> vs;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<pss> vpss;typedef vector<vi> vvi;
typedef vector<vl> vvl;typedef queue<int> qi;
typedef queue<ll> ql;
typedef queue<pii> qpii;
typedef queue<pll> qpll;
typedef queue<psi> qpsi;
typedef queue<psl> qpsl;typedef priority_queue<int> pqi;
typedef priority_queue<ll> pql;
typedef priority_queue<string> pqs;
typedef priority_queue<pii> pqpii;
typedef priority_queue<psi> pqpsi;
typedef priority_queue<pll> pqpl;
typedef priority_queue<psi> pqpsl;typedef map<int, int> mii;
typedef map<int, bool> mib;
typedef map<ll, ll> mll;
typedef map<ll, bool> mlb;
typedef map<char, int> mci;
typedef map<char, ll> mcl;
typedef map<char, bool> mcb;
typedef map<string, int> msi;
typedef map<string, ll> msl;
typedef map<int, bool> mib;typedef unordered_map<int, int> umii;
typedef unordered_map<ll, ll> uml;
typedef unordered_map<char, int> umci;
typedef unordered_map<char, ll> umcl;
typedef unordered_map<string, int> umsi;
typedef unordered_map<string, ll> umsl;template <typename T>
inline T read()
{T x = 0;int y = 1;char ch = getchar();while (ch > '9' || ch < '0'){if (ch == '-')y = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){x = (x << 3) + (x << 1) + (ch ^ 48);ch = getchar();}return x * y;
}template <typename T>
inline void write(T x)
{if (x < 0){putchar('-');x = -x;}if (x >= 10){write(x / 10);}putchar(x % 10 + '0');
}/*#####################################BEGIN#####################################*/void solve()
{int n;cin >> n;vl s(n + 5);for (int i = 1; i <= n; i++){cin >> s[i];}vl fs(n + 5); // 储存之前被经过的轮数ll ans = 0;for (int i = 1; i <= n; i++){for (int j = i + 2; j <= n && j <= i + s[i]; j++){fs[j]++;}fs[i + 1] += max(0LL, fs[i] - s[i] + 1);ans += max(0LL, s[i] - fs[i] - 1);}cout << ans << endl;
}int main()
{ios::sync_with_stdio(false), std::cin.tie(nullptr), std::cout.tie(nullptr);int _ = 1;std::cin >> _;while (_--){solve();}return 0;
}// https://codeforces.com/problemset/problem/1491/C