codeforces 424A Squats

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本文介绍了一种算法，用于解决仓鼠队列中站立与坐着的仓鼠数量调整问题，目的是达到指定数量的站立与坐着的仓鼠，通过最少的操作步骤实现目标配置。

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Description
Pasha has many hamsters and he makes them work out. Today, n hamsters (n is even) came to work out. The hamsters lined up and each hamster either sat down or stood up.

For another exercise, Pasha needs exactly hamsters to stand up and the other hamsters to sit down. In one minute, Pasha can make some hamster ether sit down or stand up. How many minutes will he need to get what he wants if he acts optimally well?

Input
The first line contains integer n (2 ≤ n ≤ 200; n is even). The next line contains n characters without spaces. These characters describe the hamsters’ position: the i-th character equals ‘X’, if the i-th hamster in the row is standing, and ‘x’, if he is sitting.

Output
In the first line, print a single integer — the minimum required number of minutes. In the second line, print a string that describes the hamsters’ position after Pasha makes the required changes. If there are multiple optimal positions, print any of them.

Sample Input
Input
4
xxXx
Output
1
XxXx
Input
2
XX
Output
1
xX
Input
6
xXXxXx
Output
0
xXXxXx

#include <iostream>

using namespace std;

int main()
{
    int n;
    while(cin>>n)
    {
        char a[n];
        cin>>a;
        int c=0,d=0;
        for(int i=0; i<n; i++)
        {
            if(a[i]=='X')
                c++;
            else if(a[i]=='x')
                d++;
        }
        int num=n/2-min(c,d);
        int t=num;
        if(c>d)
        {
            for(int j=0; j<n&&t!=0; j++)
            {
                if(a[j]=='X')
                {
                    a[j]='x';
                    t--;
                }
            }
        }
        if(c<d)
        {
            for(int j=0; j<n&&t!=0; j++)
            {
                if(a[j]=='x')
                {
                    a[j]='X';
                    t--;
                }
            }
        }
        cout<<num<<endl;
        for(int i=0; i<n; i++)
            cout<<a[i];
        cout<<endl;
    }
    return 0;
}