开O2全re，不开o2Ac#3其他RE（并且同一个代码不同时刻结果不同）？？？

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;
#define PI 3.141593
#define maxm 110
#define IOS ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define __bp __builtin_popcount
#define nxp next_permutation
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
long long lgcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long llcm(long long a, long long b) { return a / gcd(a, b) * b; }
int sgn(int x) {
	if (x > 0) return 1;
	if (x < 0) return -1;
}



int main(){
	int const _count[11]={6,2,5,5,4,5,6,3,7,6};
	int n;
	cin>>n;
	int a[2111]={6,0};
	int ans=0;
	for(int i=1;i<2000;i++){
		int tmp=i;
		int t=0;
		int c[11]={0};
		while(tmp!=0){
			c[t++]=tmp%10;
			cerr<<tmp<<' ';
			tmp/=10;
			
		}
		for(int j=0;j<t;j++){
			a[i]+=_count[c[j]];
			//cerr<<c[j]<<' ';
		}
		//cout<<a[i];
		
	}
	for(int i=0;i<2000;i++){
		for(int j=0;j<2000;j++){
			int k=i+j;
			if(a[i]+a[j]+a[k]+4==n){
				
				ans++;
			}
		}
	}
	cout<<ans;
	return 0;
	
}