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  • LightOJ 1199 Partitioning Game(sg函数)题解

    题意:可以把一堆石子分成不相等的两堆,不能操作为败

    思路:把一个石子拆成两个,变成了两个独立的游戏,mex里加上两者的sg异或。sg打表。

    代码:

    #include<set>
    #include<map>
    #include<stack>
    #include<cmath>
    #include<queue>
    #include<vector>
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    typedef long long ll;
    const int maxn = 1e4 + 10;
    const int seed = 131;
    const ll MOD = 1e9 + 7;
    const int INF = 0x3f3f3f3f;
    using namespace std;
    int sg[maxn], s[maxn];
    int main(){
        sg[0] = 0;
        for(int i = 1; i < maxn; i++){
            memset(s, 0, sizeof(s));
            for(int j = 1; j * 2 < i; j++){
                s[sg[j] ^ sg[i - j]] = 1;
            }
            for(int j = 0; j < maxn; j++){
                if(!s[j]){
                    sg[i] = j;
                    //cout << j << endl;
                    break;
                }
            }
        }
        int T, Case = 1;
        scanf("%d", &T);
        while(T--){
            int n;
            scanf("%d", &n);
            ll ans = 0;
            while(n--){
                ll a;
                scanf("%lld", &a);
                ans ^= sg[a];
            }
            if(ans) printf("Case %d: Alice
    ", Case++);
            else printf("Case %d: Bob
    ", Case++);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/KirinSB/p/9709863.html
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