牛客春招刷题训练营-2025.03.24题解

活动地址： 牛客春招刷题训练营 - 编程打卡活动

简单题 统计每个月兔子的总数

该序列为斐波那契数列。
用递推求出。

dp = [1, 1]
for i in range(2, 32):
    dp.append(dp[i - 1] + dp[i - 2])
print(dp[int(input()) - 1])

中等题 高精度整数加法

使用 Python 即可通过本题。

print(int(input())+int(input()))

或使用 C++：

#include <bits/stdc++.h>
using namespace std;
const int base = 1000000000;
const int base_digits = 9; // 分解为九个数位一个数字
struct bigint {
    vector<int> a;
    int sign;

    bigint() : sign(1) {}
    bigint operator-() const {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }
    bigint(long long v) {
        *this = v;
    }
    bigint(const string &s) {
        read(s);
    }
    void operator=(const bigint &v) {
        sign = v.sign;
        a = v.a;
    }
    void operator=(long long v) {
        a.clear();
        sign = 1;
        if (v < 0) sign = -1, v = -v;
        for (; v > 0; v = v / base) {
            a.push_back(v % base);
        }
    }

    // 基础加减乘除
    bigint operator+(const bigint &v) const {
        if (sign == v.sign) {
            bigint res = v;
            for (int i = 0, carry = 0; i < (int)max(a.size(), v.a.size()) || carry; ++i) {
                if (i == (int)res.a.size()) {
                    res.a.push_back(0);
                }
                res.a[i] += carry + (i < (int)a.size() ? a[i] : 0);
                carry = res.a[i] >= base;
                if (carry) {
                    res.a[i] -= base;
                }
            }
            return res;
        }
        return *this - (-v);
    }
    bigint operator-(const bigint &v) const {
        if (sign == v.sign) {
            if (abs() >= v.abs()) {
                bigint res = *this;
                for (int i = 0, carry = 0; i < (int)v.a.size() || carry; ++i) {
                    res.a[i] -= carry + (i < (int)v.a.size() ? v.a[i] : 0);
                    carry = res.a[i] < 0;
                    if (carry) {
                        res.a[i] += base;
                    }
                }
                res.trim();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }
    void operator*=(int v) {
        check(v);
        for (int i = 0, carry = 0; i < (int)a.size() || carry; ++i) {
            if (i == (int)a.size()) {
                a.push_back(0);
            }
            long long cur = a[i] * (long long)v + carry;
            carry = (int)(cur / base);
            a[i] = (int)(cur % base);
        }
        trim();
    }
    void operator/=(int v) {
        check(v);
        for (int i = (int)a.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = a[i] + rem * (long long)base;
            a[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }
    int operator%(int v) const {
        if (v < 0) {
            v = -v;
        }
        int m = 0;
        for (int i = a.size() - 1; i >= 0; --i) {
            m = (a[i] + m * (long long)base) % v;
        }
        return m * sign;
    }

    void operator+=(const bigint &v) {
        *this = *this + v;
    }
    void operator-=(const bigint &v) {
        *this = *this - v;
    }
    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }
    bigint operator/(int v) const {
        bigint res = *this;
        res /= v;
        return res;
    }
    void operator%=(const int &v) {
        *this = *this % v;
    }

    bool operator<(const bigint &v) const {
        if (sign != v.sign) return sign < v.sign;
        if (a.size() != v.a.size()) return a.size() * sign < v.a.size() * v.sign;
        for (int i = a.size() - 1; i >= 0; i--)
            if (a[i] != v.a[i]) return a[i] * sign < v.a[i] * sign;
        return false;
    }
    bool operator>(const bigint &v) const {
        return v < *this;
    }
    bool operator<=(const bigint &v) const {
        return !(v < *this);
    }
    bool operator>=(const bigint &v) const {
        return !(*this < v);
    }
    bool operator==(const bigint &v) const {
        return !(*this < v) && !(v < *this);
    }
    bool operator!=(const bigint &v) const {
        return *this < v || v < *this;
    }

    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }
    void check(int v) { // 检查输入的是否为负数
        if (v < 0) {
            sign = -sign;
            v = -v;
        }
    }
    void trim() { // 去除前导零
        while (!a.empty() && !a.back()) a.pop_back();
        if (a.empty()) sign = 1;
    }
    bool isZero() const { // 判断是否等于零
        return a.empty() || (a.size() == 1 && !a[0]);
    }
    friend bigint gcd(const bigint &a, const bigint &b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint &a, const bigint &b) {
        return a / gcd(a, b) * b;
    }
    void read(const string &s) {
        sign = 1;
        a.clear();
        int pos = 0;
        while (pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if (s[pos] == '-') sign = -sign;
            ++pos;
        }
        for (int i = s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for (int j = max(pos, i - base_digits + 1); j <= i; j++) x = x * 10 + s[j] - '0';
            a.push_back(x);
        }
        trim();
    }
    friend istream &operator>>(istream &stream, bigint &v) {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }
    friend ostream &operator<<(ostream &stream, const bigint &v) {
        if (v.sign == -1) stream << '-';
        stream << (v.a.empty() ? 0 : v.a.back());
        for (int i = (int)v.a.size() - 2; i >= 0; --i)
            stream << setw(base_digits) << setfill('0') << v.a[i];
        return stream;
    }

    /* 大整数乘除大整数部分 */
    typedef vector<long long> vll;
    bigint operator*(const bigint &v) const { // 大整数乘大整数
        vector<int> a6 = convert_base(this->a, base_digits, 6);
        vector<int> b6 = convert_base(v.a, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while (a.size() < b.size()) a.push_back(0);
        while (b.size() < a.size()) b.push_back(0);
        while (a.size() & (a.size() - 1)) a.push_back(0), b.push_back(0);
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        for (int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.a.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.a = convert_base(res.a, 6, base_digits);
        res.trim();
        return res;
    }
    friend pair<bigint, bigint> divmod(const bigint &a1,
                                       const bigint &b1) { // 大整数除大整数，同时返回答案与余数
        int norm = base / (b1.a.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.a.resize(a.a.size());
        for (int i = a.a.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.a[i];
            int s1 = r.a.size() <= b.a.size() ? 0 : r.a[b.a.size()];
            int s2 = r.a.size() <= b.a.size() - 1 ? 0 : r.a[b.a.size() - 1];
            int d = ((long long)base * s1 + s2) / b.a.back();
            r -= b * d;
            while (r < 0) r += b, --d;
            q.a[i] = d;
        }
        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }
    static vector<int> convert_base(const vector<int> &a, int old_digits, int new_digits) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for (int i = 1; i < (int)p.size(); i++) p[i] = p[i - 1] * 10;
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for (int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while (cur_digits >= new_digits) {
                res.push_back((int)(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while (!res.empty() && !res.back()) res.pop_back();
        return res;
    }
    static vll karatsubaMultiply(const vll &a, const vll &b) {
        int n = a.size();
        vll res(n + n);
        if (n <= 32) {
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++) {
                    res[i + j] += a[i] * b[j];
                }
            }
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for (int i = 0; i < k; i++) a2[i] += a1[i];
        for (int i = 0; i < k; i++) b2[i] += b1[i];

        vll r = karatsubaMultiply(a2, b2);
        for (int i = 0; i < (int)a1b1.size(); i++) r[i] -= a1b1[i];
        for (int i = 0; i < (int)a2b2.size(); i++) r[i] -= a2b2[i];

        for (int i = 0; i < (int)r.size(); i++) res[i + k] += r[i];
        for (int i = 0; i < (int)a1b1.size(); i++) res[i] += a1b1[i];
        for (int i = 0; i < (int)a2b2.size(); i++) res[i + n] += a2b2[i];
        return res;
    }

    void operator*=(const bigint &v) {
        *this = *this * v;
    }
    bigint operator/(const bigint &v) const {
        return divmod(*this, v).first;
    }
    void operator/=(const bigint &v) {
        *this = *this / v;
    }
    bigint operator%(const bigint &v) const {
        return divmod(*this, v).second;
    }
    void operator%=(const bigint &v) {
        *this = *this % v;
    }
};
int main() {
    bigint a, b;
    cin >> a >> b;
    cout << a + b << '\n';
    return 0;
}

困难题 喜欢切数组的红

以下数组下标都从 开始。 设 中的元素为正数的下标， 中的元素为可以作为第一刀的下标， 中的元素为可以作为第二刀的下标。
当数组总和不为 的倍数，表示没有位置可切，直接输出 。
然后枚举 中的元素：

• 如果 区间内没有正数，跳过本次循环。
• 否则在 区间内二分查找第一个正数的下标 ：
  • 如果 不存在，跳过本次循环。
  • 否则在 区间内二分查找第一个可以作为第二刀的下标 ：
    • 设 为 的最后一个元素，在 内二分查找第一个不能作为第二刀的下标 (即 内第一个大于 的元素)，答案即加上 。
    • 在 C++ 中，如果 upper_bound 或 lower_bound 二分查找不存在会返回 end() 迭代器，减去 begin() 即为 ，查找不成功也不会影响答案。

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr int p = 1e9 + 7;
int a[200005];
ll qzh[200005];
int n;
void solve() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= n; i++)
        qzh[i] = qzh[i - 1] + a[i];
    ll sum = qzh[n];
    if (sum % 3 != 0) {
        cout << 0 << '\n';
        return;
    }
    vector<int> z;
    vector<int> d1, d2;
    for (int i = 1; i <= n; i++) {
        if (a[i] > 0)
            z.push_back(i);
        if (qzh[i] == sum / 3)
            d1.push_back(i);
        if (qzh[i] == 2 * sum / 3)
            d2.push_back(i);
    }
    ll ans = 0;
    if (z.empty()) {
        cout << 0 << '\n';
        return;
    }
    for (int i = 0; i < d1.size(); i++) {
        int z0 = z[0];
        if (d1[i] < z0)
            continue;
        auto z1 = upper_bound(z.begin(), z.end(), d1[i]);
        if (z1 == z.end())
            continue;
        auto dd2 = lower_bound(d2.begin(), d2.end(), *z1);
        int z2 = z[z.size() - 1];
        auto dd3 = lower_bound(dd2.begin(), d2.end(), z2);
        ans += dd3 - dd2;
    }
    cout << ans << '\n';
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    solve();
    return 0;
}