Squats CodeForces - 424A

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.
题意：一个字符串有n个字符，每个字符不是x就是X，现要求让x和X的个数相等，输出最小的转换次数和转换后的字符串。
思路：统计x和X的个数设为a,b，c=min(a,b)，转换最小次数为n/2-a，随便转换几个x或者X然后输出就行了。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
    int n;
    char c[205];
    int s=0;
    while(cin>>n&&n!=0)
    {
        int a=0,b=0,t=0;
        for(int i=0;i<n;i++)
        {
            cin>>c[i];
            if(c[i]=='x')
                a++;
            else
                b++;
        }
        if(a<n/2)
        {
            t=n/2-a;
            s=t;
            for(int i=0;i<n;i++)
            {
                if(t==0)
                    break;
                if(c[i]=='X')
                {
                    c[i]='x';
                    t--;
                }
            }
        }
        else
        {
            t=n/2-b;
            s=t;
            for(int i=0;i<n;i++)
            {
                if(t==0)
                    break;
                if(c[i]=='x')
                {
                    c[i]='X';
                    t--;
                }
            }
        }
        cout<<s<<endl;
        for(int i=0;i<n;i++)
        {
            cout<<c[i];
        }
        cout<<endl;
    }
    return 0;
}