agc002E - Candy Piles(博弈论)

题意

sol

orz sovitpower

#include<bits/stdc++.h> 
#define pair pair<int, double>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define ll long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 1e6 + 10, mod = 998244353, inf = 2e9 + 10;
const double eps = 1e-9;
template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}
template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename a> inline void debug(a a){cout << a << '\n';}
template <typename a> inline ll sqr(a x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, a[maxn];
signed main() {
    n = read();
    for(int i = 1; i <= n; i++) a[i] = read();
    sort(a + 1, a + n + 1, greater<int>());
    for(int i = 1; i <= n; i++) {
        if(i + 1 > a[i + 1]) {
            if((a[i] - i) & 1) {puts("first"); return 0;}
            int j;
            for(j = i + 1; a[j] == i; j++); 
            if(!((j - i) & 1)) {puts("first"); return 0;}
            puts("second"); return 0;
        }
    }
    return 0;
}