Xiaosha's lunch

// Problem:  Xiaosha's lunch 
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/problem/231178
// Memory Limit: 2 MB
// Time Limit: 231178000 ms
// 2022-02-15 23:23:06
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fir first
#define pb push_back
#define sec second
#define sortall(x) sort((x).begin(),(x).end())
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
const int N = 1e3 + 10;
void solve() {
	int n;
	int a[N];
	ll f[N] = {0};
	ll suf[N] = {0};
	int cnt = 0;
	cin >> n;
	ll x = 0, y ;
	for (int i = 1; i<= n ;i++){
		 int t;
		 cin >> t;
		 if (t == 1) x ++;
		else {
			++cnt;
			a[cnt] = t;	
		}
	}
	sort(a +1, a + 1 + cnt);
	for (int i = cnt; i >= 1;i --) {
		suf[i] = suf[i + 1] + a[i];
		f[i] = suf[i] - f[i + 1];
		if (a[i] >= 4) f[i] = max (f[i], f[i + 1] + a[i] - 4);
	}
	 x = suf[1] - f[1] + x, y = f[1];
	if (x > y) puts("xiaohonwin");
	else if (x == y) puts("loss");
	else puts("xiaoshawin"); 
}
int main () {
    int t;
    cin >> t;
    while (t --) solve();
    return 0;
}

First, we need to understand a key property , By looking at a ring , We can find that for less than 4 We can't get it , For greater than or equal to 4 Can get a[i] -4 At most Then, for multiple rings, we need to consider the following for the current number of rings , So we go from big to small dp, In this way, you can be greedy for the current when you know the later . And then take max That's it