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Codeforces Round #242 (Div. 2) A. Squats

时间:2014-10-24 11:05:53      阅读:286      评论:0      收藏:0      [点我收藏+]

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 bubuko.com,布布扣 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
题意:将X变成n/2个

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 300;

char str[maxn];

int main() {
	int n;
	scanf("%d", &n);
	scanf("%s", str);
	int ans = 0;
	for (int i = 0; i < n; i++)
		if (str[i] == 'X')
			ans++;
	int flag = ans > (n/2) ? 1 : 0;
	ans = abs(ans - (n/2));
	printf("%d\n", ans);
	int cnt = 0;
	for (int i = 0; i < n; i++) {
		if (cnt >= ans || ans == 0)
			break;
		if (flag && str[i] == 'X') {
			str[i] = 'x';
			ans--;
		}
		else if (!flag && str[i] == 'x') {
			str[i] = 'X';
			cnt++;
		}
	}
	printf("%s\n", str);
	return 0;
}



Codeforces Round #242 (Div. 2) A. Squats

原文:http://blog.csdn.net/u011345136/article/details/40422299

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