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