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  • AC日记——Pupils Redistribution Codeforces 779a

    A. Pupils Redistribution
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard 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

     思路:

      模拟;

    来,上代码:

    #include <cmath>
    #include <cstdio>
    #include <iostream>
    #include <algorithm>
    
    using namespace std;
    
    int if_z,n,ai[6],bi[6],vi[6];
    
    char Cget;
    
    inline void in(int &now)
    {
        now=0,if_z=1,Cget=getchar();
        while(Cget>'9'||Cget<'0')
        {
            if(Cget=='-') if_z=-1;
            Cget=getchar();
        }
        while(Cget>='0'&&Cget<='9')
        {
            now=now*10+Cget-'0';
            Cget=getchar();
        }
        now*=if_z;
    }
    
    int main()
    {
        in(n);int pos,cur=0,kol=0;
        for(int i=1;i<=n;i++) in(pos),ai[pos]++;
        for(int i=1;i<=n;i++) in(pos),bi[pos]++;
        for(int i=1;i<=5;i++)
        {
            vi[i]=ai[i]-bi[i];
            vi[i]/=2,cur+=vi[i];
            if(vi[i]>0) kol+=vi[i];
            pos=abs(ai[i]-bi[i]);
            if(pos%2)
            {
                cout<<"-1";
                return 0;
            }
        }
        if(cur==0) cout<<kol;
        else cout<<"-1";
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/IUUUUUUUskyyy/p/6556113.html
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