ATC DP J Stones

ATC DP J Stones

dp[i] = 1 先手必败 =0 后手必败

当下一个状态中全都不是先手必败状态即当前位置必为先手必败。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <iostream>

using namespace std;
typedef long long ll;
const int N = 1e7;
const ll MOD = 1e9 + 7;
priority_queue<int> que;

int cnt[4];
int dp[N];
int vis[N];
int v[N];
int n, m;

// dp[i] = 1 先手必败 =0 后手必败
int dfs(int k) {
    if (vis[k]) return dp[k];
    int res = 0;
    bool flag = false;
    for (int i = 0; i < n; i++) {
        if (k - v[i] >= 0) {
            res += dfs(k - v[i]);
            flag = true;
        }
    }
    if (!flag) {
        dp[k] = 1;
    } else if (res == 0) {
        dp[k] = 1;
    }
    vis[k] = 1;
    return dp[k];
}

int main() {
    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        int tem;
        cin >> tem;
        v[i] = tem;
    }
    sort(v, v + n);
    vis[0] = 1;
    dp[0] = 1;
    dfs(m);
    if (dp[m]) {
        cout << "Second" << endl;
    } else {
        cout << "First" << endl;
    }
}