1 条题解
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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; } }
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