Permutation (greedy strategy)

link ：https://ac.nowcoder.com/acm/contest/10746/J
source ： Cattle from

Title Description
give n Numbers , Is there a calculation order for these numbers , So that the number will not exceed 3 Not less than 0？

Input description
The first line gives a positive integer $t,(1\le t\le 1000)$ Number of representative test data groups
The first row of each set of test data is a positive integer $n,(1\le n\le 500)$
The second line contains $n$ A number separated by a space
Input to ensure that each number is $-3,-2,-1,+0,+1,+2,+3$ One of them .

Output description
Each group of test data output one line ,“Yes” or “No”

The sample input
2
4
+3 +2 -1 -2
5
+3 +2 +1 +0 +2

Sample output
Yes
No

Their thinking
greedy , Think about it first $-3$, Judge whether the $+3$ or $+2-1+2$ or $+2+1$ or $+1+1+1$ Neutralize it , Until there is no $-3$ until ; If nothing can neutralize in the middle $-3$, be “NO”;
Then consider $+3$, When $+3$ The number of is greater than 1 when , Judge whether the $-3$ or $-2+1-2$ or $-2-1$ or $-1-1-1$ Neutralize it , until $+3$ The number is less than or equal to 1 until ; If nothing can neutralize in the middle $+3$, be “NO”;
If none of them “NO”, be “YES”.

AC Code

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int T, n, a[7], *cnt;
bool check() {
    
	int t = 0;
	for (int i = -3; i <= 3; i++) {
    
		t += i * cnt[i];
	}
	if (t > 3 || t < 0) return false;

	//for -3, +3 || +2 +2 -1 || +2 +1 || +1 +1 +1
	t = min(cnt[-3], cnt[3]);
	cnt[-3] -= t;
	cnt[3] -= t;
	
	t = min(cnt[2], cnt[-1]);
	cnt[2] -= t;
	cnt[-1] -= t;
	cnt[1] += t;
	
	t = min(cnt[-3], cnt[1]);
	cnt[-3] -= t;
	cnt[1] -= t;
	cnt[-2] += t;
	
	if (cnt[-3] > 0) return false;
	
	// for 3, -3 || -2 -2 +1 || -2 -1 || -1 -1 -1
	t = min(cnt[-2], cnt[1]);
	cnt[-2] -= t;
	cnt[1] -= t;
	cnt[-1] += t;
	
	t = min(cnt[3], cnt[-1]);
	cnt[3] -= t;
	cnt[-1] -= t;
	cnt[2] += t;
	
	if (cnt[3] > 1) return false;
	return true;
}
int main() {
    
	//freopen(" Permutation formula .in", "r", stdin);
	scanf("%d", &T);
	cnt = &a[3];
	while (T--) {
    
		memset(a, 0, sizeof(a));
		scanf("%d", &n);
		int t;
		for (int i = 1; i <= n; i++) {
    
			scanf("%d", &t);
			cnt[t]++;
		}
		if (check()) {
    
			printf("Yes\n");
		} else {
    
			printf("No\n");
		}
	}
	return 0;
}