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  • Nowcoder 挑战赛23 B 游戏 ( NIM博弈、SG函数打表 )

    题目链接

    题意 : 中文题、点链接

    分析 :

    前置技能是 SG 函数、NIM博弈变形

    每次可取石子是约数的情况下、那么就要打出 SG 函数

    才可以去通过异或操作判断一个局面的胜负

    打 SG 函数的时候、由于 N 很大

    所以不能使用递归的方式打表、会爆栈

    还有要预处理每个数的约数

    打出 SG 函数之后

    暴力判断初始局面的每堆石子取走约数后是否对答案产生贡献

    #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
    ", 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
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/qwertiLH/p/9606583.html
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