HDU 6106 Classes【水题】

Classes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 227    Accepted Submission(s): 142

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.

 

Sample Input

2
2
4 5 4 4 3 2 2
5 3 1 2 0 0 0
2
0 4 10 2 3 4 9
6 12 6 3 5 3 2

 

Sample Output

7
15

Hint
In the second test case, the data uploaded by Class 1 is wrong. 
Because we can't find a solution which satisfies the limitation. 
As for Class 2, we can calculate the number of students who only enrolled in course A is 2, 
the number of students who only enrolled in course B is 6, and nobody enrolled in course C,
the number of students who only enrolled in courses A and B is 1, 
the number of students who only enrolled in courses B and C is 3, 
the number of students who only enrolled in courses A and C is 1, 
the number of students who enrolled in all courses  is 2, 
so the total number in Class 2 is 2 + 6 + 0 + 1 + 3 + 1 + 2 = 15.
 

 

Source

2017 Multi-University Training Contest - Team 6

题意：

n个班每个班的学生报名ABC三种课程，求n个班中 max（只报名A的人+只报名B的人+只报名C+只报名AC+只报名BC+只报名AB+只报名ABC的人）

数据有可能是假的，假的数据直接忽略掉，保证一定有一个真的数据

思路：

判断一下单独每个部分是否都大于0即可

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
int a[110];
int calc()
{
    a[3]-=a[6];
    a[5]-=a[6];
    a[4]-=a[6];
    if(a[3]<0||a[5]<0||a[4]<0) return 0;
    a[0]-=a[3]+a[5]+a[6];
    a[1]-=a[3]+a[4]+a[6];
    a[2]-=a[4]+a[5]+a[6];
    if(a[0]<0||a[1]<0||a[2]<0) return 0;
    return a[0]+a[1]+a[2]+a[3]+a[4]+a[5]+a[6];
}
int main()
{
    int t;
    scanf("%d", &t);
    int ans = 0;
    while(t --)
    {
        int n;
        scanf("%d", &n);
        ans = 0;
        for(int i = 0; i < n; i ++)
        {
            for(int i = 0; i < 7; i ++)
                scanf("%d", &a[i]);
            ans = max(ans, calc());
        }
        printf("%d
", ans);
    }
    return 0;
}

第二种写法：

刚开始有个问题，不能直接判ab<=min(a,b) bc<=min(b,c) ac<min(a,c) abc<min(a,b,c)会wa，因为这个判断太松了，

所以应为ab+ac-abc<=a bc+ab-abc<=b ac+bc-abc<=c 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int maxn = 105;

int a[110];
int calc()
{
	if (a[3] + a[5] - a[6] > a[0]) return 0;
	if (a[3] + a[4] - a[6] > a[1]) return 0;
	if (a[4] + a[5] - a[6] > a[2]) return 0;
	if (a[6] > min(a[3], min(a[4], a[5]))) return 0;
	return a[0] + a[1] + a[2] - a[3] - a[4] - a[5] + a[6];
}
int main()
{
	int t;
	scanf("%d", &t);
	int ans = 0;
	while (t--)
	{
		int n;
		scanf("%d", &n);
		ans = 0;
		for (int i = 0; i < n; i++)
		{
			for (int i = 0; i < 7; i++)
				scanf("%d", &a[i]);
			ans = max(ans, calc());
		}
		printf("%d
", ans);
	}
	return 0;
}