$7^7$ 算法理论可过该题，但蒟蒻的我却 TLE 了七个，有任何办法可以优化我的代码使得在不开 O2 的情况下 AC 吗？

代码在下面

#include <bits/stdc++.h>
using namespace std;
int n;
vector <int> p[30];
bool k[30][10];
long long cnt;
long long cou[30][10];
long long ans;
int main(){
//	freopen("bgm.in","r",stdin);
//	freopen("bgm.out","w",stdout);
	cin>>n;
	for(int i=1;i<=n;i++){
		char opt;
		int x;
		cin>>opt>>x;
		x%=7;
		if(k[opt-'@'][x]==0){
			p[opt-'@'].push_back(x);
			k[opt-'@'][x]=true;
			cou[opt-'@'][x]++;
		}
		else{
			cou[opt-'@'][x]++;
		}
	}
	for(int i=1;i<=29;i++){
		for(int j=1;j<=9;j++){
			if(cou[i][j]!=0){
				continue;
			}
			cou[i][j]=1;
		}
	}
	for(int a=0;a<p['B'-'@'].size();a++){
		for(int b=0;b<p['E'-'@'].size();b++){
			for(int c=0;c<p['S'-'@'].size();c++){
				for(int d=0;d<p['I'-'@'].size();d++){
					for(int e=0;e<p['G'-'@'].size();e++){
						for(int f=0;f<p['O'-'@'].size();f++){
							for(int g=0;g<p['M'-'@'].size();g++){
								ans=(p['B'-'@'][a]+p['E'-'@'][b]+p['S'-'@'][c]+p['S'-'@'][c]+p['I'-'@'][d]+p['E'-'@'][b])*(p['G'-'@'][e]+p['O'-'@'][f]+p['E'-'@'][b]+p['S'-'@'][c])*(p['M'-'@'][g]+p['O'-'@'][f]+p['O'-'@'][f]);
								if(ans%7==0){
//									cout<<a<<" "<<b<<" "<<c<<" "<<d<<" "<<e<<" "<<f<<" "<<g<<" "<<ans<<endl;
									cnt+=cou['B'-'@'][p['B'-'@'][a]]*cou['E'-'@'][p['E'-'@'][b]]*cou['S'-'@'][p['S'-'@'][c]]*cou['I'-'@'][p['I'-'@'][d]]*cou['G'-'@'][p['G'-'@'][e]]*cou['O'-'@'][p['O'-'@'][f]]*cou['M'-'@'][p['M'-'@'][g]];
//									cout<<cou[''-'@'][p['I'-'@'][d]]<<endl;
								}
							}
						}
					}
				}
			}
		}
	}
	cout<<cnt<<endl;
	return 0;
}