ACM思维题训练集合
A colored stripe is represented by a horizontal row of n square cells, each cell is pained one of k colors. Your task is to repaint the minimum number of cells so that no two neighbouring cells are of the same color. You can use any color from 1 to k to repaint the cells.
Input
The first input line contains two integers n and k (1 ≤ n ≤ 5·105; 2 ≤ k ≤ 26). The second line contains n uppercase English letters. Letter "A" stands for the first color, letter "B" stands for the second color and so on. The first k English letters may be used. Each letter represents the color of the corresponding cell of the stripe.
Output
Print a single integer — the required minimum number of repaintings. In the second line print any possible variant of the repainted stripe.
Examples
Input
6 3
ABBACC
Output
2
ABCACA
Input
3 2
BBB
Output
1
BAB
题目很简单，类比平面图的四色定理，线性上有3色定理，即如果k>=3那么无论遇到什么情况AABB的时候，我们都可以将相同的变成第三种颜色，所以这里要特判K=2的时候，我一开始想了很久，就是不知道怎么处理，但是k=2的串只有ABABA……和BABABA……这两种情况，判断当前穿变成这两种那种花费少，就是答案了。
#include <bits/stdc++.h>
using namespace std;
int n, m, cnt = 0;
char check(char a, char b)
{
for (char i = 'A'; i <= 'A' + m - 1; i++)
{
if (i == a || i == b)
continue;
else
return i;
}
}
int main()
{
cin >> n >> m;
string s;
cin >> s;
if (m == 2)
{
int cnt1 = 0, cnt2 = 0;
for (int i = 0; i < n; i++)
{
if (s[i] != ('A' + (i % 2)))
cnt1++;
else
cnt2++;
}
if (cnt2 > cnt1)
{
cout << cnt1 << endl;
for (int i = 0; i < n; i++)
if (i % 2 == 0)
putchar('A');
else
putchar('B');
}
else
{
cout << cnt2 << endl;
for (int i = 1; i <= n; i++)
if (i % 2 == 0)
putchar('A');
else
putchar('B');
}
puts("");
return 0;
}
for (int i = 1; i < s.length();)
{
if (s[i] == s[i - 1])
{
s[i] = check(s[i - 1], s[i + 1]);
i = i + 2;
cnt++;
}
else
i++;
}
cout << cnt << endl
<< s << endl;
}