Codeforces Round #242 (Div. 2) A. Squats-优快云博客

本文介绍了一个关于仓鼠锻炼的问题，需要计算使站立和坐着的仓鼠数量相等所需的最少分钟数，并输出调整后的状态。通过遍历仓鼠的状态，算法能够有效地找到最优解。

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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 $\text{[math]}$ 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 test(s)

Input

4
xxXx

Output

1
XxXx

Input

2
XX

Output

1
xX

Input

6
xXXxXx

Output

0
xXXxXx

问最少多少步，使得X与x数量相同并输出

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cmath>
#include<algorithm>
#define LL __int64
#define inf 0x3f3f3f3f
using namespace std;
char now[10];
char s[210];
bool bj;
int main()
{
    int n,m,i,j,k;
    while(scanf("%d",&n)!=EOF)
    {
        bj=false;
        getchar();
        scanf("%s",s);
        k=strlen(s);
        int l=k;
        j=0;
        for(i=0; i<k; i++)
        {
            if(s[i]=='X')
                j++;
        }
        if(n/2==j)
        {
            printf("0\n");
            printf("%s\n",s);
        }
        else if(n/2<j)
        {
            printf("%d\n",j-n/2);
            int sum=0;
            for(i=0;i<l;i++)
            {
                if(sum==(j-n/2))
                break;
                if(s[i]=='X')
                {
                    s[i]='x';
                    sum++;
                }
            }
            printf("%s\n",s);
        }
        else
        {
            printf("%d\n",n/2-j);
            int sum=0;
            for(i=0;i<l;i++)
            {
                if(sum==(n/2-j))
                    break;
                if(s[i]=='x')
                {
                    s[i]='X';
                    sum++;
                }
            }
            printf("%s\n",s);
        }
    }
    return 0;
}