专栏首页glm的全栈学习之路Codeforces Round #622 (Div. 2) A.Fast Food Restaurant

Codeforces Round #622 (Div. 2) A.Fast Food Restaurant

Fast Food Restaurant

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Tired of boring office work, Denis decided to open a fast food restaurant.

On the first day he made aa portions of dumplings, bb portions of cranberry juice and cc pancakes with condensed milk.

The peculiarity of Denis's restaurant is the procedure of ordering food. For each visitor Denis himself chooses a set of dishes that this visitor will receive. When doing so, Denis is guided by the following rules:

  • every visitor should receive at least one dish (dumplings, cranberry juice, pancakes with condensed milk are all considered to be dishes);
  • each visitor should receive no more than one portion of dumplings, no more than one portion of cranberry juice and no more than one pancake with condensed milk;
  • all visitors should receive different sets of dishes.

What is the maximum number of visitors Denis can feed?

Input

The first line contains an integer tt (1≤t≤5001≤t≤500) — the number of test cases to solve.

Each of the remaining tt lines contains integers aa, bb and cc (0≤a,b,c≤100≤a,b,c≤10) — the number of portions of dumplings, the number of portions of cranberry juice and the number of condensed milk pancakes Denis made.

Output

For each test case print a single integer — the maximum number of visitors Denis can feed.

Example

input

7
1 2 1
0 0 0
9 1 7
2 2 3
2 3 2
3 2 2
4 4 4

output

3
0
4
5
5
5
7

Note

In the first test case of the example, Denis can feed the first visitor with dumplings, give the second a portion of cranberry juice, and give the third visitor a portion of cranberry juice and a pancake with a condensed milk.

In the second test case of the example, the restaurant Denis is not very promising: he can serve no customers.

In the third test case of the example, Denise can serve four visitors. The first guest will receive a full lunch of dumplings, a portion of cranberry juice and a pancake with condensed milk. The second visitor will get only dumplings. The third guest will receive a pancake with condensed milk, and the fourth guest will receive a pancake and a portion of dumplings. Please note that Denis hasn't used all of the prepared products, but is unable to serve more visitors.

思路:七种情况,需要枚举,a,b,c,ab,ac,ac,bc,abc,但是要注意这其中也有贪心的过程,一开始肯定是先用一个的a或b或c,然后再考虑两个,考虑用两个的情况时应该贪心,先用最多的和次多的,最后如果ab或ac或bc都有剩余就abc一起用

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rg register ll
#define inf 2147483647
#define lb(x) (x&(-x))
ll sz[200005],n;
template <typename T> inline void read(T& x)
{
    x=0;char ch=getchar();ll f=1;
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}x*=f;
}
inline ll query(ll x){ll res=0;while(x){res+=sz[x];x-=lb(x);}return res;}
inline void add(ll x,ll val){while(x<=n){sz[x]+=val;x+=lb(x);}}//第x个加上val
ll t;
int main()
{
	cin>>t;
    for(rg i=1;i<=t;i++)
    {
        ll a,b,c,cnt=0;
        cin>>a>>b>>c;
        if(a>0)a--,cnt++;
        if(b>0)b--,cnt++;
        if(c>0)c--,cnt++;
        if(a>=2&&b>=2&&c>=2)cnt+=3,a-=2,b-=2,c-=2;
        else
        {
            if(a>=2&&b>0&&c>0)
            {
                cnt+=2,a-=2,b-=1,c-=1;
                
            }
            else if(b>=2&&a>0&&c>0)
            {
                cnt+=2,b-=2,a-=1,c-=1;
              
            }
            else if(c>=2&&b>0&&a>0)
            {
                cnt+=2,c-=2,b-=1,a-=1;
             
            }
            else if(a>0&&b>0)
            {
                cnt++,a--,b--;
               
            }
            else if(a>0&&c>0)
            {
                cnt++,a--,c--;
              
            }
            else if(c>0&&b>0)
            {
                cnt++,c--,b--;
              
            }
            else if(a>0&&b>0&&c>0)
            {
                cnt++,a--,b--,c--;
     
            }
        }
        if(a>0&&b>0&&c>0)cnt++;
        cout<<cnt<<endl;
    }
    while(1)getchar();
    return 0;
    
}

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