Codeforces Round #402 (Div. 2) A. Pupils Redistribution_k. pupils redistribution-优快云博客

本文介绍了一个算法问题，涉及两组学生的学术成绩分配调整。通过交换学生来实现每组内各分数段人数相等的目标，并提供了具体的实现代码。

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题目：

A. Pupils Redistribution

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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 group B.

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

Examples
input
4
5 4 4 4
5 5 4 5
output
1
input
6
1 1 1 1 1 1
5 5 5 5 5 5
output
3
input
1
5
3
output
-1
input
9
3 2 5 5 2 3 3 3 2
4 1 4 1 1 2 4 4 1
output
4

题意：

两组学生，每个学生有各自的学术值，对学生进行交换组合，使得每一个分数在每组的学生数都相同
**

思路：

分别统计两组1-5分每个学生的人数，分别存于两个数组。
将两组同分数的人数相减，所得的数字除以2，如果不能整除则无论怎么交换都不能符合条件，直接结束
如果能整除2则可以交换，除以2所得的数字即为该数字需要交换的次数
所有数字需要交换的次数相加最后的得数除以2就得到了需要交换的次数
**

代码：

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
using namespace std;

int a[6];
int b[6];
int aa[105];
int bb[105];

int main(){
    int n;
    scanf("%d",&n);
    int i,j;
    for(i=0;i<n;i++){
        scanf("%d",&aa[i]);
        a[aa[i]]++;
    }
    for(i=0;i<n;i++){
        scanf("%d",&bb[i]);
        b[bb[i]]++;
    }
    int k=0;
    int flag=0;
    for(i=1;i<=5;i++){
        if(a[i]==b[i]){
            continue;
        }else{
            if((a[i]+b[i])%2==0){
                k+=abs(a[i]-b[i])/2;
            }if((a[i]+b[i])%2!=0){
                flag=1;
                break;
            }
        }
    }
    if(flag==0){
        printf("%d\n",(k/2));
    }
    if(flag==1){
        printf("-1\n");
    }

    return 0;
}