1 条题解
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思路与算法
令a≥b。若a=b,可一步取空而胜;若a≥2b,也一定有胜策。否则仅有减一次b的合法选择,变成(a-b,b),轮到对手。反复执行这个强制过程并记录当前轮次,遇到必胜终态时判断它属于谁。使用大整数避免给未限定石子位数附加机器整数范围。
正确性说明
a≥2b时,设q=floor(a/b)、r=a mod b。如果直接减qb到r是让对方处于必败态,就这样取;否则减(q-1)b,留(b+r,b),此时对方若r>0只有减一次b可把状态交回(b,r),使原先对方的必胜态由自己执手;r=0则直接取空。故该情形必胜。q=1且a≠b时只有唯一一步,胜负必翻转。每次强制转移减少石子总量,最终必到必胜情形,轮次翻转给出初态结论。
复杂度
设D为石子十进制位数,强制步骤不超过欧几里得算法数量级O(D),每步比较减法O(D),总O(D²)时间、O(D)空间;含输入和数位归一化。
易错点
这不是任意取数的Nim或威佐夫游戏。输入以0 0结束,该行不输出;两个初始堆都是正整数。不能每次机械地取最大倍数而不分析胜负。
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 a, b; while (cin >> a >> b) { if (a == 0 && b == 0) break; bool turn = true; while (true) { if (a < b) swap(a, b); if (a == b || a >= b * 2) { cout << (turn ? "win" : "lose") << "\n"; break; } a -= b; turn = !turn; } } }
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