D. Tokitsukaze, CSL and Stone Game
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Tokitsukaze and CSL are playing a little game of stones.

In the beginning, there are n piles of stones, the i-th pile of which has ai stones. The two players take turns making moves. Tokitsukaze moves first. On each turn the player chooses a nonempty pile and removes exactly one stone from the pile. A player loses if all of the piles are empty before his turn, or if after removing the stone, two piles (possibly empty) contain the same number of stones. Supposing that both players play optimally, who will win the game?

Consider an example: n=3 and sizes of piles are a1=2, a2=3, a3=0. It is impossible to choose the empty pile, so Tokitsukaze has two choices: the first and the second piles. If she chooses the first pile then the state will be [1,3,0] and it is a good move. But if she chooses the second pile then the state will be [2,2,0] and she immediately loses. So the only good move for her is to choose the first pile.

Supposing that both players always take their best moves and never make mistakes, who will win the game?

Note that even if there are two piles with the same number of stones at the beginning, Tokitsukaze may still be able to make a valid first move. It is only necessary that there are no two piles with the same number of stones after she moves.

Input
The first line contains a single integer n (1≤n≤105) — the number of piles.

The second line contains n integers a1,a2,…,an (0≤a1,a2,…,an≤109), which mean the i-th pile has ai stones.

Output
Print "sjfnb" (without quotes) if Tokitsukaze will win, or "cslnb" (without quotes) if CSL will win. Note the output characters are case-sensitive.

Examples
inputCopy
1
0
outputCopy
cslnb
inputCopy
2
1 0
outputCopy
cslnb
inputCopy
2
2 2
outputCopy
sjfnb
inputCopy
3
2 3 1
outputCopy
sjfnb
Note
In the first example, Tokitsukaze cannot take any stone, so CSL will win.

In the second example, Tokitsukaze can only take a stone from the first pile, and then, even though they have no stone, these two piles will have the same number of stones, which implies CSL will win.

In the third example, Tokitsukaze will win. Here is one of the optimal ways:

Firstly, Tokitsukaze can choose the first pile and take a stone from that pile.
Then, CSL can only choose the first pile, because if he chooses the second pile, he will lose immediately.
Finally, Tokitsukaze can choose the second pile, and then CSL will have no choice but to lose.
In the fourth example, they only have one good choice at any time, so Tokitsukaze can make the game lasting as long as possible and finally win.

题意：
给n堆石子，每一堆有a[i] 个 石头，现在二人玩游戏，每一人每一次选择一个非空的石头堆，从中拿出一个石头。如果当一个人操作之后，这n堆石头中有两堆中的石头数量是相同的，那么这个人就输掉。假设二人决定聪明，请判断是先手赢还是后手赢。
思路：

先根据初始情况判断先手是否必输：

通过思考我们可以知道初始情况是以下情况先手必输

1、 空堆的数量 >=2

2、有三个堆中石头数量相等。

3、 有这种情况， x x y y 即 有相同数量堆的情况>=2

4、 有这种情况， x x x-1

如果先手不必输，那么二人的最优操作之后，石子堆一定会到达这个状态 （无顺序） 0,1,2,3,,,n-1

我们只需要求a[i] 的sum和然后减去 等差数列的前列项和后判断奇偶即可。

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/

int num[maxn], a[maxn];

map<int, int> m1;

int main()
{
    int n;
    scanf("%d", &n);
    int flag3 = 0;
    int flag2 = 0;
    int x;
    int id;
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &num[i]);
        a[i] = num[i];
        m1[a[i]]++;
        if (!flag3 && m1[a[i]] >= 3)
        {
            flag3 = 1;
        } else if (m1[a[i]] >= 2)
        {
            id = i;
            x = a[i];
            flag2++;
        }
    }
    int flag1 = 0;
    if (n == 1)
    {
        if (a[1] & 1)
        {
            cout << "sjfnb" << endl;
        } else
        {
            cout << "cslnb" << endl;
        }
        return 0;
    }
    if (flag3)
    {
        cout << "cslnb" << endl;
        return 0;
    } else if (flag2)
    {
        if (flag2 > 1)
        {
            cout << "cslnb" << endl;
            return 0;
        } else
        {
            if (m1[x - 1] || x == 0)
            {
                cout << "cslnb" << endl;
                return 0;
            }
        }
    }
    ll ans = 0ll;
    repd(i, 1, n)
    {
        ans += a[i];
    }
    if (flag1)
    {
        ans++;
    }
    ans -= 1ll * n * (n - 1ll) / 2;// 注意爆ll 
    if (ans & 1)
        cout << "sjfnb" << endl;
    else
        cout << "cslnb" << endl;

    return 0;
}