# Classes(传送门)

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 94 Accepted Submission(s): 64
Problem Description
The school set up three elective courses, assuming that these courses are A, B, C. N classes of students enrolled in these courses.
Now the school wants to count the number of students who enrolled in at least one course in each class and records the maximum number of students.
Each class uploaded 7 data, the number of students enrolled in course A in the class, the number of students enrolled in course B, the number of students enrolled in course C, the number of students enrolled in course AB, the number of students enrolled in course BC, the number of students enrolled in course AC, the number of students enrolled in course ABC. The school can calculate the number of students in this class based on these 7 data.
However, due to statistical errors, some data are wrong and these data should be ignored.
Smart you must know how to write a program to find the maximum number of students.
Input
The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with one integer N, indicates the number of class.
Then N lines follow, each line contains 7 data: a, b, c, d, e, f, g, indicates the number of students enrolled in A, B, C, AB, BC, AC, ABC in this class.
It's guaranteed that at least one data is right in each test case.
Limits

T≤100
1≤N≤100
0≤a,b,c,d,e,f,g≤100
Output
For each test case output one line contains one integer denotes the max number of students who enrolled in at least one course among N classes.
题意 ：n各班级 给出他们报A B C三门课的情况（报A、B、C、AB、BC、AC、ABC分别有多少人）但有的是错误的 你需要找出正确的班级中 班级人数最多的人数：
思路：
画个

Venn图就很好看出 AC-ABC=只报AC的人 其他…… A-只报AC的和只报AB的=只报A的 这样就可以都求出来了 然后都加起来即可
code：
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
int a,b,c,d,e,f;
int m[8],n[8];
cin>>a;
while(a--)
{
cin>>b;
f=-1;
while(b--)
{
for(c=0;c<7;c++)
cin>>m[c];
m[5]=m[5]-m[6];
m[4]=m[4]-m[6];
m[3]=m[3]-m[6];
m[2]=m[2]-m[5]-m[4]-m[6];
m[1]=m[1]-m[3]-m[4]-m[6];
m[0]=m[0]-m[5]-m[3]-m[6];
if(m[0]<0||m[1]<0||m[2]<0||m[3]<0||m[4]<0||m[5]<0||m[6]<0)
continue;
d=m[0]+m[1]+m[2]+m[3]+m[4]+m[5]+m[6];
if(d>f)
f=d;
}
cout<<f<<endl;
}
return 0;
}