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:
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
using namespace std;
#define ll long long
#define rg register ll
#define inf 2147483647
#define lb(x) (x&(-x))
ll sz[200005],n;
template 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<