Codeforces Round #334 E. Lieges of Legendre （组合游戏）

题意:

给定n,k≤109,然后组合游戏的规则,不同的是只有2种操作
1.任选1堆拿走一个,2.选一个2x个数堆变成k堆x个的
问谁赢

分析:

只看第二种操作,发现k分奇偶
奇数的话sg(2x)=sg(x)⊕sg(x)⋯⊕sg(x)k times=sg(x)
偶数的话sg(2x)=sg(x)⊕sg(x)⋯⊕sg(x)k times=0
n,k这么大显然打表找规律了
根据这个打个表发现很有规律,除了前几项

k->oddi :  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20sg:  0  1  0  1  2  0  2  0  1  0  1  0  1  0  1  0  2  0  1  0  2k->eveni :   0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20sg:  0  1  2  0  1  0  1  0  1  0  1  0  1  0  1  0  1  0  1  0  1  

k奇数是前5项没规律,然后后面的奇数项都是0,偶数项不是1就是2
k偶数前3项没规律,然后后面的奇数项都是1,偶数项都是0
证明也很简单,induction就好了

代码:

////  Created by TaoSama on 2015-12-02//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, k;/*  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20  0  1  0  1  2  0  2  0  1  0  1  0  1  0  1  0  2  0  1  0  2  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20  0  1  2  0  1  0  1  0  1  0  1  0  1  0  1  0  1  0  1  0  1*/int odd[5] = {0, 1, 0, 1, 2}, even[3] = {0, 1, 2};int getSg(int x) {    if(k & 1) {        if(x < 5) return odd[x];        if(x & 1) return 0;        return getSg(x >> 1) == 1 ? 2 : 1;    } else {        if(x < 3) return even[x];        return (x & 1) ^ 1;    }}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%d%d", &n, &k) == 2) {        int ans = 0;        for(int i = 1; i <= n; ++i) {            int x; scanf("%d", &x);            ans ^= getSg(x);        }        puts(ans ? "Kevin" : "Nicky");    }    return 0;}/*int getSg(int x) {    if(k & 1) {        if(!x) return 0;        if(~sg[x]) return sg[x];        set<int> s;        s.insert(getSg(x - 1));        if(!(x & 1)) s.insert(getSg(x >> 1));        for(int i = 0; ; ++i) if(!s.count(i)) return sg[x] = i;    } else {        if(!x) return 0;        if(~sg[x]) return sg[x];        set<int> s;        s.insert(getSg(x - 1));        if(!(x & 1)) s.insert(0);        for(int i = 0; ; ++i) if(!s.count(i)) return sg[x] = i;    }}*/