前言

如果有问题欢迎大家评论或私信指出

题解

A.困难数学题

输出 即可，因为相同的数字异或值为。

#include<bits/stdc++.h>

using i64 = long long;
using u64 = unsigned long long;

void solve() {
    int a;
    std::cin >> a;
    std::cout << 0;
}  

signed main() {
    std::ios::sync_with_stdio(0);
    std::cout.tie(0);   
    std::cin.tie(0);

    i64 t = 1; 
    // std::cin >> t;
    while (t--) {
        solve();
    }
}

B.构造序列

正负交替即可

#include<bits/stdc++.h>

using i64 = long long;
using u64 = unsigned long long;

void solve() {
    int a, b;
    std::cin >> a >> b;
    if (a == b) {
        std::cout << a + b;
    }
    else {
        std::cout << std::min(a, b) * 2 + 1;
    }
}  

signed main() {
    std::ios::sync_with_stdio(0);
    std::cout.tie(0);   
    std::cin.tie(0);

    i64 t = 1; 
    // std::cin >> t;
    while (t--) {
        solve();
    }
}

C.连点成线

写成纯模拟了，每行每列的最大值最小值。

#include<bits/stdc++.h>

using i64 = long long;
using u64 = unsigned long long;

void solve() {
    int n, m;
    std::cin >> n >> m;

    std::set<int> st;
    int ans = 0;
    std::vector<int> rmi(n + 1, 1000000), rmx(n + 1), cmx(n + 1), cmi(n + 1, 1000000);
    for (int i = 1; i <= m; i++) {
        int x, y;
        std::cin >> x >> y;
        ans = std::max(ans, x - (rmi[y] ? rmi[y] : x));
        ans = std::max(ans, (rmx[y] ? rmx[y] : x) - x);
        ans = std::max(ans, y - (cmi[x] ? cmi[x] : y));
        ans = std::max(ans, (cmx[x] ? cmx[x] : y) - y);

        rmi[y] = std::min(rmi[y], x);
        rmx[y] = std::max(rmx[y], x);
        cmi[x] = std::min(cmi[x], y);
        cmx[x] = std::max(cmx[x], y);
        // std::cout << rmi[y] << ' ' << rmx[y] << ' ' << cmi[x] << ' ' << cmx[x] << ' ' << '\n';
    }

    std::cout << ans << '\n';
}  

signed main() {
    std::ios::sync_with_stdio(0);
    std::cout.tie(0);   
    std::cin.tie(0);

    i64 t = 1; 
    // std::cin >> t;
    while (t--) {
        solve();
    }
}

D.我们N个真是太厉害了

是dp的一种想法，如果sum表示的是小于等于sum的数都可以表示出来，那枚举到的时候，到+都可以表示出来

#include<bits/stdc++.h>

using i64 = long long;
using u64 = unsigned long long;

void solve() {
    int n;
    std::cin >> n;

    std::vector<int> a(n + 1);
    for(int i=1;i<=n;i++){
        std::cin>>a[i];
    }
    std::sort(a.begin() + 1, a.end());
    i64 sum = 0;
    for (int i = 1; i <= n; i++) {
        if(sum >= a[i] - 1) {
            sum += a[i];
        }
    }
    if (sum >= n) {
        std::cout << "Cool!" << '\n';
    }
    else {
        std::cout << sum + 1 << '\n';
    }
}  

signed main() {
    std::ios::sync_with_stdio(0);
    std::cout.tie(0);   
    std::cin.tie(0);

    i64 t = 1; 
    std::cin >> t;
    while (t--) {
        solve();
    }
}

E.折返跑

转化一下题意就可以变成了球盒问题，两个杆子是和最多可以推次，而且题目要求除了最后一趟每次都推，那不就是要推次，然后就变成什么问题个球要分到个盒里，且每个盒子不能为空。

#include<bits/stdc++.h>

using i64 = long long;
using u64 = unsigned long long;

template<class T>
constexpr T power(T a, i64 b) {
    T res = 1;
    for (; b; b /= 2, a *= a) {
        if (b % 2) {
            res *= a;
        }
    }
    return res;
}

constexpr i64 mul(i64 a, i64 b, i64 p) {
    i64 res = a * b - i64(1.L * a * b / p) * p;
    res %= p;
    if (res < 0) {
        res += p;
    }
    return res;
}

template<int P>
struct MInt {
    int x;
    constexpr MInt() : x{} {}
    constexpr MInt(i64 x) : x{norm(x % getMod())} {}
    
    static int Mod;
    constexpr static int getMod() {
        if (P > 0) {
            return P;
        } else {
            return Mod;
        }
    }
    constexpr static void setMod(int Mod_) {
        Mod = Mod_;
    }
    constexpr int norm(int x) const {
        if (x < 0) {
            x += getMod();
        }
        if (x >= getMod()) {
            x -= getMod();
        }
        return x;
    }
    constexpr int val() const {
        return x;
    }
    explicit constexpr operator int() const {
        return x;
    }
    constexpr MInt operator-() const {
        MInt res;
        res.x = norm(getMod() - x);
        return res;
    }
    constexpr MInt inv() const {
        assert(x != 0);
        return power(*this, getMod() - 2);
    }
    constexpr MInt &operator*=(MInt rhs) & {
        x = 1LL * x * rhs.x % getMod();
        return *this;
    }
    constexpr MInt &operator+=(MInt rhs) & {
        x = norm(x + rhs.x);
        return *this;
    }
    constexpr MInt &operator-=(MInt rhs) & {
        x = norm(x - rhs.x);
        return *this;
    }
    constexpr MInt &operator/=(MInt rhs) & {
        return *this *= rhs.inv();
    }
    friend constexpr MInt operator*(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res *= rhs;
        return res;
    }
    friend constexpr MInt operator+(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res += rhs;
        return res;
    }
    friend constexpr MInt operator-(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res -= rhs;
        return res;
    }
    friend constexpr MInt operator/(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res /= rhs;
        return res;
    }
    friend constexpr std::istream &operator>>(std::istream &is, MInt &a) {
        i64 v;
        is >> v;
        a = MInt(v);
        return is;
    }
    friend constexpr std::ostream &operator<<(std::ostream &os, const MInt &a) {
        return os << a.val();
    }
    friend constexpr bool operator==(MInt lhs, MInt rhs) {
        return lhs.val() == rhs.val();
    }
    friend constexpr bool operator!=(MInt lhs, MInt rhs) {
        return lhs.val() != rhs.val();
    }
};

template<>
int MInt<0>::Mod = 998244353;

template<int V, int P>
constexpr MInt<P> CInv = MInt<P>(V).inv();

constexpr int P = 1000000007;
using Z = MInt<P>;

struct Comb {
    int n;
    std::vector<Z> _fac;
    std::vector<Z> _invfac;
    std::vector<Z> _inv;
    
    Comb() : n{0}, _fac{1}, _invfac{1}, _inv{0} {}
    Comb(int n) : Comb() {
        init(n);
    }
    
    void init(int m) {
        m = std::min(m, Z::getMod() - 1);
        if (m <= n) return;
        _fac.resize(m + 1);
        _invfac.resize(m + 1);
        _inv.resize(m + 1);
        
        for (int i = n + 1; i <= m; i++) {
            _fac[i] = _fac[i - 1] * i;
        }
        _invfac[m] = _fac[m].inv();
        for (int i = m; i > n; i--) {
            _invfac[i - 1] = _invfac[i] * i;
            _inv[i] = _invfac[i] * _fac[i - 1];
        }
        n = m;
    }
    
    Z fac(int m) {
        if (m > n) init(2 * m);
        return _fac[m];
    }
    Z invfac(int m) {
        if (m > n) init(2 * m);
        return _invfac[m];
    }
    Z inv(int m) {
        if (m > n) init(2 * m);
        return _inv[m];
    }
    Z binom(int n, int m) {
        if (n < m || m < 0) return 0;
        return fac(n) * invfac(m) * invfac(n - m);
    }
} comb;

void solve() {
    int n, m;
    std::cin >> n >> m;

    Z ans = comb.binom(n - 2, m - 1);
    // for (int i = m - 1; i <= n - 2; i++) {
    //     ans += comb.binom(i - 1, m - 2);
    // }

    std::cout << ans << '\n';
}  

signed main() {
    std::ios::sync_with_stdio(0);
    std::cout.tie(0);   
    std::cin.tie(0);

    i64 t = 1; 
    std::cin >> t;
    while (t--) {
        solve();
    }
}

F.口吃

期望dp是我不会的东西了，跟群友学的，建议大家去看视频讲题吧，我仅提供一份代码。

#include<bits/stdc++.h>

using i64 = long long;

const int mod = 1000000007;

template<class T>
constexpr T power(T a, i64 b) {
    T res = 1;
    for (; b; b /= 2, a *= a) {
        if (b % 2) {
            res *= a;
        }
    }
    return res;
}

constexpr i64 mul(i64 a, i64 b, i64 p) {
    i64 res = a * b - i64(1.L * a * b / p) * p;
    res %= p;
    if (res < 0) {
        res += p;
    }
    return res;
}

template<int P>
struct MInt {
    int x;
    constexpr MInt() : x{} {}
    constexpr MInt(i64 x) : x{norm(x % getMod())} {}
    
    static int Mod;
    constexpr static int getMod() {
        if (P > 0) {
            return P;
        } else {
            return Mod;
        }
    }
    constexpr static void setMod(int Mod_) {
        Mod = Mod_;
    }
    constexpr int norm(int x) const {
        if (x < 0) {
            x += getMod();
        }
        if (x >= getMod()) {
            x -= getMod();
        }
        return x;
    }
    constexpr int val() const {
        return x;
    }
    explicit constexpr operator int() const {
        return x;
    }
    constexpr MInt operator-() const {
        MInt res;
        res.x = norm(getMod() - x);
        return res;
    }
    constexpr MInt inv() const {
        assert(x != 0);
        return power(*this, getMod() - 2);
    }
    constexpr MInt &operator*=(MInt rhs) & {
        x = 1LL * x * rhs.x % getMod();
        return *this;
    }
    constexpr MInt &operator+=(MInt rhs) & {
        x = norm(x + rhs.x);
        return *this;
    }
    constexpr MInt &operator-=(MInt rhs) & {
        x = norm(x - rhs.x);
        return *this;
    }
    constexpr MInt &operator/=(MInt rhs) & {
        return *this *= rhs.inv();
    }
    friend constexpr MInt operator*(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res *= rhs;
        return res;
    }
    friend constexpr MInt operator+(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res += rhs;
        return res;
    }
    friend constexpr MInt operator-(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res -= rhs;
        return res;
    }
    friend constexpr MInt operator/(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res /= rhs;
        return res;
    }
    friend constexpr std::istream &operator>>(std::istream &is, MInt &a) {
        i64 v;
        is >> v;
        a = MInt(v);
        return is;
    }
    friend constexpr std::ostream &operator<<(std::ostream &os, const MInt &a) {
        return os << a.val();
    }
    friend constexpr bool operator==(MInt lhs, MInt rhs) {
        return lhs.val() == rhs.val();
    }
    friend constexpr bool operator!=(MInt lhs, MInt rhs) {
        return lhs.val() != rhs.val();
    }
};

template<>
int MInt<0>::Mod = 998244353;

template<int V, int P>
constexpr MInt<P> CInv = MInt<P>(V).inv();

constexpr int P = 1000000007;
using Z = MInt<P>;

void solve() {
    int n;
    std::cin >> n;

    std::vector<Z> e(n + 1), p(n + 1), q(n + 1);
    for (int i = 1; i < n; i++) {
        std::cin >> p[i];
    }
    for (int i = 1; i < n; i++) {
        std::cin >> q[i];
    }

    e[0] = 1;
    e[1] = (p[1] + q[1]) / p[1];
    for (int i = 2; i < n; i++) {
        e[i] = (q[i] * q[i] * e[i - 1] + (p[i] + q[i]) * (p[i] + q[i])) / (p[i] * p[i]);
    }

    Z ans = 0;
    for (int i = 0; i < n; i++) {
        ans += e[i];
    }

    std::cout << ans;
}   

signed main() {
    std::ios::sync_with_stdio(0);
    std::cout.tie(0);   
    std::cin.tie(0);

    i64 t = 1; 
    // std::cin >> t;
    while (t--) {
        solve();
    }
}