題目鏈接

題意 : 中文題、點鏈接

分析 :

前置技能是 SG 函數、NIM博弈變形

每次可取石子是約數的情況下、那麽就要打出 SG 函數

才可以去通過異或操作判斷一個局面的勝負

打 SG 函數的時候、由於 N 很大

所以不能使用遞歸的方式打表、會爆棧

還有要預處理每個數的約數

打出 SG 函數之後

暴力判斷初始局面的每堆石子取走約數後是否對答案產生貢獻

#include<bits/stdc++.h>
#define LL long long
#define ULL unsigned long long

#define scl(i) scanf("%lld", &i)
#define
 
 scll(i, j) scanf("%lld %lld", &i, &j)
#define sclll(i, j, k) scanf("%lld %lld %lld", &i, &j, &k)
#define scllll(i, j, k, l) scanf("%lld %lld %lld %lld", &i, &j, &k, &l)

#define scs(i) scanf("%s", i)
#define sci(i) scanf("%d", &i)
#define scd(i) scanf("%lf", &i)
#define
 
 scIl(i) scanf("%I64d", &i)
#define scii(i, j) scanf("%d %d", &i, &j)
#define scdd(i, j) scanf("%lf %lf", &i, &j)
#define scIll(i, j) scanf("%I64d %I64d", &i, &j)
#define sciii(i, j, k) scanf("%d %d %d", &i, &j, &k)
#define scddd(i, j, k) scanf("%lf %lf %lf", &i, &j, &k)
#define
 
 scIlll(i, j, k) scanf("%I64d %I64d %I64d", &i, &j, &k)
#define sciiii(i, j, k, l) scanf("%d %d %d %d", &i, &j, &k, &l)
#define scdddd(i, j, k, l) scanf("%lf %lf %lf %lf", &i, &j, &k, &l)
#define scIllll(i, j, k, l) scanf("%I64d %I64d %I64d %I64d", &i, &j, &k, &l)

#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define lowbit(i) (i & (-i))
#define mem(i, j) memset(i, j, sizeof(i))

#define fir first
#define sec second
#define VI vector<int>
#define ins(i) insert(i)
#define pb(i) push_back(i)
#define pii pair<int, int>
#define VL vector<long long>
#define mk(i, j) make_pair(i, j)
#define all(i) i.begin(), i.end()
#define pll pair<long long, long long>

#define _TIME 0
#define _INPUT 0
#define _OUTPUT 0
clock_t START, END;
void __stTIME();
void __enTIME();
void __IOPUT();
using namespace std;
const int maxn = 1e5 + 10;

int SG[maxn];
VI divisor[maxn];

inline void init()
{
    for(int i=1; i<maxn; i++)
        for(int j=i; j<maxn; j+=i)
            divisor[j].pb(i);

    SG[0] = 0;
    for(int i=1; i<maxn; i++){
        set<int> s;
        for(int j=0; j<(int)divisor[i].size(); j++) s.ins(SG[i-divisor[i][j]]);
        for(int j=0;;j++) if(!s.count(j)){ SG[i] = j; break; }
    }
}

int arr[maxn];
int main(void){__stTIME();__IOPUT();

    init();

    int n;
    sci(n);

    int xorSum = 0;
    for(int i=0; i<n; i++) sci(arr[i]), xorSum ^= SG[arr[i]];

    int ans = 0;
    for(int i=0; i<n; i++){
        for(int j=0; j<(int)divisor[arr[i]].size(); j++){
            if((xorSum ^
                SG[arr[i]] ^
                SG[arr[i] - divisor[arr[i]][j]]) == 0) ans++;
        }
    }

    printf("%d\n", ans);


__enTIME();return 0;}


void __stTIME()
{
    #if _TIME
        START = clock();
    #endif
}

void __enTIME()
{
    #if _TIME
        END = clock();
        cerr<<"execute time = "<<(double)(END-START)/CLOCKS_PER_SEC<<endl;
    #endif
}

void __IOPUT()
{
    #if _INPUT
        freopen("in.txt", "r", stdin);
    #endif
    #if _OUTPUT
        freopen("out.txt", "w", stdout);
    #endif
}

Nowcoder 挑戰賽23 B 遊戲 ( NIM博弈、SG函數打表 )