【CodeForces 779A】Pupils Redistribution（模拟）

A. Pupils Redistribution

In Berland each high school student is characterized by academic performance — integer value between 1 and 5.

In high school 0xFF there are two groups of pupils: the group A and
the group B. Each group consists of exactly n students.
An academic performance of each student is known — integer value between 1 and 5.

The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1,
the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the
composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.

To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class A and
one student of class B. After that, they both change their groups.

Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance.

Input

The first line of the input contains integer number n (1 ≤ n ≤ 100)
— number of students in both groups.

The second line contains sequence of integer numbers a1, a2, ..., an (1 ≤ ai ≤ 5),
where ai is
academic performance of the i-th student of the group A.

The third line contains sequence of integer numbers b1, b2, ..., bn (1 ≤ bi ≤ 5),
where bi is
academic performance of the i-th student of the groupB.

Output

Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained.





题目大意：给出两组
4000
数，将两组中的数交换，使相同的数在两组内个数相同，求最小交换次数

思路：分别求出个数后取差值的一半，简单模拟

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#include<bits/stdc++.h>  

#define manx 100005  

using namespace std;  

int main()  

{  

    int n,x,a[6]={0},b[6]={0};  

    scanf("%d",&n);  

    for (int i=1; i<=n; i++){  

        scanf("%d",&x);  

        a[x]++;  

    }  

    for (int i=1; i<=n; i++){  

        scanf("%d",&x);  

        b[x]++;  

    }  

    int ans=0;  

    for (int i=1; i<=5; i++){  

        if((a[i]+b[i])==0) continue;  

        if((a[i]+b[i])%2){  //奇数不能均分  

            printf("-1\n");  

            return 0;  

        }  

        int t=abs(a[i]-b[i]);  

        ans+=t/2;  

    }  

    printf("%d\n",ans/2);  

    return 0;  

}