【Atcoder K - Stones】dp

Atc K
意思是有 n 个石块 每个石块有 a[i]个东西 然后你初始体积是 K
从First 先选 谁不能从当前的块里继续选 谁就输了
输出赢的人
我们知道 设dp[k][t]代表当前有k个石头 t代表人 true则赢 false则输
那么dp[k][t] |= !dp[k-arr[i]][!t] 因为如果另一个人 k-arr[i] 必输 那么你只要取 arr[i] 让他达到这个状态 你就必赢了

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<#define dbg3(x1,x2,x3) cout<<#x1<<" = "<#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define lc (rt<<1)
#define rc (rt<<11)
#define mid ((l+r)>>1)

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
const int MAX_N = 305;
bool dp[10000025][2];
int arr[MAX_N],n;
int dfs(int k,int t)
{
    if(dp[k][t]>0) return dp[k][t];
    int ans = 0;
    for(int i = 1;i<=n;++i)
        if(k>=arr[i])   ans |= !dfs(k-arr[i],!t);
    return dp[k][t] = ans;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int k;
    scanf("%d%d",&n,&k);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]);
    if(dfs(k,0)==false) printf("Second\n");
    else printf("First\n");
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}