“科林明伦杯”哈尔滨理工大学暑假训练赛部分题题解

A 寂寞如雪

今天你寂寞吗？

题目内容

题目分析

本题题意即为求最大子段和，由于奇数段和偶数段仅与子段左端点 \(l\) 的奇偶性有关，因此我们扫描两边，一遍奇数段，一遍偶数段即可。此外，本题输入处理也有一些难点，细节请看代码。

代码实现

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 7;
int top, a[N], b[N];

int maxsum(int x, int y, int* a) {
    if (y - x == 1) return a[x];
    int m = (x + y) >> 1;
    int ma = max(maxsum(x, m, a), maxsum(m, y, a));
    int l = a[m - 1], v = 0;
    for (int i = m - 1; i >= x; --i) {
        v += a[i];
        if (v > l) l = v;
    }
    int r = a[m]; v = 0;
    for (int i = m; i < y; ++i) {
        v += a[i];
        if (v > r) r = v;
    }
    return max(ma, l + r);
}

int main() {
    int x = 0;
    for (int i = 0; i < len; ++i) {
        if (ch[i] == '0') {
            // cerr << x << " ";
            if (x) a[++top] = x;
            x = 0;
        } else ++x;
    }
    
    if (x) a[++top] = x;
    
    for (int i = 1; i <= top; ++i) b[i] = a[i] *= a[i] = a[i];
    for (int i = 1; i <= top; ++i) {
        if (i & 1) b[i] = -b[i];
        else a[i] = -a[i];
    }
    
    int m1 = maxsum(1, top, a);
    int m2 = maxsum(1, top, b);
            
    cout << max(m1, m2);
    
    return 0;
}

F 奇怪的魔法

题目内容

传送门

前置知识

组合数
乘法逆元
Modular int模板

题目分析

分析输入与题意，本题关键即在推导有多少个不同数组的计算公式。

设当前数组最大值为 i，则有：

\[ans_i = C_{n + i - 1}^{i} \]

由于最大值的取值范围为 \(0\le x\le k\) 所以从 \(1\) 到 \(k\) 枚举 \(x\) 即可。

因此，

\[ans = \sum_{i = 0}^{k}i\times C_{n + i - 1}^{i} \]

代码实现

// URL: https://ac.nowcoder.com/acm/contest/37895/F
// Memory Limit: 524288 MB
// Time Limit: 2000 ms

#include <bits/stdc++.h>
using namespace std;

// #define int ll
using ll = long long;
using ull = unsigned long long;
#define fastio ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)

const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int N = 2e6;

template <int mod>
struct modint {
    unsigned int _v;
    modint() : _v(0) {}
    template <class T>
    modint(T v) {
        ll x = (ll)(v % (ll)(umod()));
        if (x < 0) x += umod();
        _v = (unsigned int)(x);
    }
    modint(bool v) { _v = ((unsigned int)(v) % umod()); }

    static constexpr unsigned int umod() { return mod; }
    unsigned int val() const { return _v; }

    modint& operator++() {
        _v++;
        if (_v == umod()) _v = 0;
        return *this;
    }
    modint& operator--() {
        if (_v == 0) _v = umod();
        _v--;
        return *this;
    }
    modint operator++(int) {
        modint result = *this;
        ++*this;
        return result;
    }
    modint operator--(int) {
        modint result = *this;
        --*this;
        return result;
    }

    modint& operator+=(const modint& rhs) {
        _v += rhs._v;
        if (_v >= umod()) _v -= umod();
        return *this;
    }
    modint& operator-=(const modint& rhs) {
        _v -= rhs._v;
        if (_v >= umod()) _v += umod();
        return *this;
    }
    modint& operator*=(const modint& rhs) {
        ull z = _v;
        z *= rhs._v;
        _v = (unsigned int)(z % umod());
        return *this;
    }
    modint& operator/=(const modint& rhs) { return *this = *this * rhs.inv(); }

    modint operator+() const { return *this; }
    modint operator-() const { return modint() - *this; }

    modint pow(ll n) const {
        assert(n >= 0);
        modint x = *this, r = 1;
        for (; n; n >>= 1) {
            if (n & 1) r *= x;
            x *= x;
        }
        return r;
    }

    modint inv() const { return pow(umod() - 2); }

    friend modint operator+(const modint& lhs, const modint& rhs) { return modint(lhs) += rhs; }
    friend modint operator-(const modint& lhs, const modint& rhs) { return modint(lhs) -= rhs; }
    friend modint operator*(const modint& lhs, const modint& rhs) { return modint(lhs) *= rhs; }
    friend modint operator/(const modint& lhs, const modint& rhs) { return modint(lhs) /= rhs; }
    friend bool operator==(const modint& lhs, const modint& rhs) { return lhs._v == rhs._v; }
    friend bool operator!=(const modint& lhs, const modint& rhs) { return lhs._v != rhs._v; }
};
typedef modint<mod> mint;

int n, k;
mint ans, fact[N + 7], fact_inv[N + 7];
void init(int n = N + 5) {
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
    fact_inv[n] = fact[n].inv();
    for (int i = n - 1; i >= 0; --i) fact_inv[i] = fact_inv[i + 1] * (i + 1);
}
mint C(int n, int m) { return m > n ? 0 : fact[n] * fact_inv[n - m] * fact_inv[m]; }

signed main() {
    fastio;
    cin >> n >> k;
    init(n + k + 1);
    for (int i = 0; i <= k; ++i) ans += C(n + i - 1, i) * i;
    cout << ans.val();

    return 0;
}