高僧斗法

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题意

n 个物品放在阶梯，向上移动物品，不能越过挡在前面的物品，最后物品都挤在高段台阶，不能移动的输。

题解

两两一组，当移动上面的，可以移动下面的相同距离，所以可以把每组之间的间距看作一堆石子，转化为 Nim 博弈。

代码

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#include <bits/stdc++.h>
#define forl(i, l, r) for (int i = l; i <= r; i++)
#define forr(i, r, l) for (int i = r; i >= l; i--)
#define for1(i, n) for (int i = 1; i <= n; i++)
#define for0(i, n) for (int i = 0; i < n; i++)
#define meminf(a) memset(a, inf, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define mem0(a) memset(a, 0, sizeof(a))
#define inlld(lld) scanf("%lld", &lld)
#define inlf(f) scanf("%lf", &f)
#define ind(d) scanf("%d", &d)
#define ins(s) scanf("%s", s)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef std::pair<long long, long long> pll;
typedef std::vector<long long> vll;
typedef std::pair<int, int> pii;
typedef unsigned long long ull;
typedef std::vector<int> vii;
typedef long double db;
typedef long long ll;
const db pi = acos((db)-1);
const ll inf =0x3f3f3f3f;
const ll mod = 1e9+7;
const int N = 1.1e5;
const db eps = 1e-8;
using namespace std;
int sign(db a) { return a < -eps ? -1 : a > eps; }
int db_cmp(db a, db b){ return sign(a-b); }

vii v,vv;
int main() {
#ifdef PerpEternal
    freopen("/Users/perpeternal/Documents/Sketch/data/in.dat", "r", stdin);
    // freopen("/Users/perpeternal/Documents/Sketch/data/out.dat", "w", stdout);
#endif
    //ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    int x;
    while(~ind(x))v.pb(x);
    int Xor=0;
    for(int i=0;i+1<v.size();i+=2){
        vv.pb(v[i+1]-v[i]-1);
        Xor^=vv.back();
    }
    if(Xor){
        for0(i,v.size()-1){
            forl(j,v[i]+1,v[i+1]-1){
                if(i%2){
                    if((Xor^vv[i/2])==vv[i/2]+j-v[i]){
                        printf("%d %d\n",v[i],j);
                        return 0;
                    }
                }else{
                    if((Xor^vv[i/2])==vv[i/2]-j+v[i]){
                        printf("%d %d\n",v[i],j);
                        return 0;
                    }
                }
            }
            
        }
    }else puts("-1");
    return 0;
}
graph TD
1-->3
2-->3
3-->4
3-->5