1 条题解

  • 0
    @ 2026-10-27 6:55:30

    思路与算法

    每支队伍按入场顺序编号为1到本队人数。只需维护两个当前队首编号male和female,开始均为1。每首舞曲输出一对编号,然后各自加1;某个编号超过本队人数时独立回到1。每名舞伴完成当前曲后相当于重新排到本队队尾,所以队列周期各自独立。参考程序按流输出,不按人数创建队列;使用高精度整数保留未给位数上限的人数和曲数。

    正确性说明

    第一曲队首就是两队的1号,初始化正确。若第i曲前两个编号等于真实队首,配对后该人进入队尾;队首应为下一编号,最后一个人之后回到1。程序的两个独立更新恰好模拟该FIFO循环操作。因此由归纳可知每曲输出正确;循环恰运行K次,既不遗漏也不多输出舞曲。

    复杂度

    设舞曲数值为K,输入及过程中最大整数位数为D。读入初始化O(D),每曲进行常数次高精度加减、比较和输出,O(D)时间,总时间O((K+1)D),与必要输出量相称;辅助空间O(D),不创建与人数等长的数组,也不把全部输出留在内存。

    易错点

    不是每一轮把男女编号同时清零;两支队伍独立循环。编号从1开始,超过人数才回到1。每曲只有一对,不能一次输出min(男,女)对。人数可能远大于实际输出曲数。

    参考代码(C++11)

    #include <bits/stdc++.h>
    using namespace std;
    struct cpp_int {
        string s;
        bool neg;
        cpp_int(long long v = 0) : s(to_string(v)), neg(v < 0) {
            if (neg)
                s.erase(s.begin());
        }
        void norm() {
            auto p = s.find_first_not_of('0');
            s = p == string::npos ? "0" : s.substr(p);
            if (s == "0")
                neg = false;
        }
        static int cmpabs(const cpp_int &a, const cpp_int &b) {
            if (a.s.size() != b.s.size())
                return a.s.size() < b.s.size() ? -1 : 1;
            return a.s == b.s ? 0 : a.s < b.s ? -1 : 1;
        }
        friend bool operator<(const cpp_int &a, const cpp_int &b) {
            if (a.neg != b.neg)
                return a.neg;
            int c = cmpabs(a, b);
            return a.neg ? c > 0 : c < 0;
        }
        friend bool operator==(const cpp_int &a, const cpp_int &b) {
            return a.neg == b.neg && a.s == b.s;
        }
        friend bool operator!=(const cpp_int &a, const cpp_int &b) {
            return !(a == b);
        }
        friend bool operator>(const cpp_int &a, const cpp_int &b) {
            return b < a;
        }
        friend bool operator<=(const cpp_int &a, const cpp_int &b) {
            return !(b < a);
        }
        friend bool operator>=(const cpp_int &a, const cpp_int &b) {
            return !(a < b);
        }
        cpp_int operator-() const {
            cpp_int r = *this;
            if (r.s != "0")
                r.neg = !r.neg;
            return r;
        }
        friend cpp_int operator+(const cpp_int &a, const cpp_int &b) {
            cpp_int r;
            r.s.clear();
            int carry = 0;
            if (a.neg == b.neg) {
                int i = (int)a.s.size() - 1, j = (int)b.s.size() - 1;
                while (i >= 0 || j >= 0 || carry) {
                    int v = carry + (i >= 0 ? a.s[i--] - '0' : 0) + (j >= 0 ? b.s[j--] - '0' : 0);
                    r.s += char('0' + v % 10);
                    carry = v / 10;
                }
                r.neg = a.neg;
            } else {
                const cpp_int *x = &a, *y = &b;
                if (cmpabs(a, b) < 0)
                    swap(x, y);
                int i = (int)x->s.size() - 1, j = (int)y->s.size() - 1;
                while (i >= 0) {
                    int v = x->s[i--] - '0' - carry - (j >= 0 ? y->s[j--] - '0' : 0);
                    carry = v < 0;
                    if (carry)
                        v += 10;
                    r.s += char('0' + v);
                }
                r.neg = x->neg;
            }
            reverse(r.s.begin(), r.s.end());
            r.norm();
            return r;
        }
        friend cpp_int operator-(const cpp_int &a, const cpp_int &b) {
            return a + (-b);
        }
        friend cpp_int operator*(const cpp_int &a, int b) {
            cpp_int r;
            r.s.clear();
            long long carry = 0;
            long long z = b;
            if (z < 0)
                z = -z;
            for (int i = (int)a.s.size() - 1; i >= 0; i--) {
                carry += (a.s[i] - '0') * z;
                r.s += char('0' + carry % 10);
                carry /= 10;
            }
            while (carry) {
                r.s += char('0' + carry % 10);
                carry /= 10;
            }
            reverse(r.s.begin(), r.s.end());
            r.neg = a.neg ^ (b < 0);
            r.norm();
            return r;
        }
        friend cpp_int operator/(const cpp_int &a, int b) {
            cpp_int r;
            r.s.clear();
            long long rem = 0, z = b;
            if (z < 0)
                z = -z;
            for (char c : a.s) {
                rem = rem * 10 + c - '0';
                r.s += char('0' + rem / z);
                rem %= z;
            }
            r.neg = a.neg ^ (b < 0);
            r.norm();
            return r;
        }
        cpp_int &operator+=(const cpp_int &b) {
            *this = *this + b;
            return *this;
        }
        cpp_int &operator-=(const cpp_int &b) {
            *this = *this - b;
            return *this;
        }
        cpp_int &operator*=(int b) {
            *this = *this * b;
            return *this;
        }
        friend istream &operator>>(istream &in, cpp_int &a) {
            string v;
            if (in >> v) {
                a.neg = v[0] == '-';
                if (v[0] == '-' || v[0] == '+')
                    v.erase(v.begin());
                a.s = v;
                a.norm();
            }
            return in;
        }
        friend ostream &operator<<(ostream &out, const cpp_int &a) {
            if (a.neg)
                out << '-';
            return out << a.s;
        }
    };
    
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cpp_int men, women, music;
        cin >> men >> women >> music;
        cpp_int male = 1, female = 1;
        while (music > 0) {
            cout << male << ' ' << female << '\n';
            male += 1;
            female += 1;
            if (male > men)
                male = 1;
            if (female > women)
                female = 1;
            music -= 1;
        }
    }
    
    • 1

    信息

    ID
    2032
    时间
    1000ms
    内存
    64MiB
    难度
    进阶
    标签
    递交数
    0
    已通过
    0
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