/////该问题的关键在于颜色的分布是否连续 在不知道颜色分布是否连续的情况下 只能选择将每个颜色均作为目标颜色 而不是人为出现最多次数为目标颜色
`#include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

using namespace std;

const int N = 1e4 + 10;

int main()
{
int t;
cin >> t;
while(t –)
{
int n, k;
cin >> n >> k;
int a[N];

    for(int i = 1; i <= 60; i ++) a[i] = 0;
    for(int i = 0; i < n; i ++) 
    {   
        cin >> a[i];
        //cout << a[i] << " ";
    }
    //cout << endl;

    int min_day = n;
    for(int i = 1; i  <= 60; i ++)
    {
        int day = 0;
        for(int j = 0; j < n; j ++)
        {
            if(a[j] != i)
            {
                day ++;
                j += k - 1;
            }
        }

        min_day = min(min_day, day);
    }

    cout << min_day << endl;
}

}
**//////////////线性dp** ![数字三角形.jpg](https://cdn.acwing.com/media/article/image/2025/04/01/226510_87c2b7fc0e-数字三角形.jpg)#include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

using namespace std;

const int N = 1e4 + 10;

int main()
{
int n;
cin >> n;

int a[110][110];
int dp[110][110];

for(int i = 0; i < n; i ++)
{
    for(int j = 0; j <= i; j ++)
    {
        cin >> a[i][j];

        //cout << a[i][j] << " ";
    }
    //cout << endl;
}

for(int j = 0; j < n; j++) 
    dp[n-1][j] = a[n-1][j];


for(int i = n - 2; i >= 0; i --)
{
    for(int j = 0; j <= i; j ++)
    {
        dp[i][j] = a[i][j] + max(dp[i + 1][j], dp[i + 1][j + 1]);
    }
}


cout << dp[0][0];

}

`

`#include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

using namespace std;

const int N = 1e5 + 10;
const int M = 1e9 + 7;
////使用数组前记得初始化数组
int dp[N];

int main()
{
int n, m;

cin >> n >> m;

bool a[N];

memset(a, false, sizeof(a));
memset(dp, 0, sizeof(dp));


//for(int i = 0; i <= n; i ++) dp[i] = -1;

for(int i = 0; i < m; i ++)
{
//找出破碎楼梯阶数
    int x;
    cin >> x;
    a[x] = true;
}



dp[0] = 1;



for(int i = 1; i <= n; i ++)
{   
    if(a[i]) continue;
    else
    {
        if(i == 1) dp[i] = 1;
        else dp[i] = (dp[i - 1] + dp[i - 2]) % M;

    }

    //cout << dp[i] << " ";
}


cout << dp[n] ;

}
`

`#include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

using namespace std;

typedef long long LL;

const int N = 1e6 + 10;
const int M = 1e9 + 7;

LL dp[N];

int n, k;

int main()
{

memset(dp, 0, sizeof(dp));


cin >> n >> k;
dp[0] = 1;
dp[1] = 2;

//从最后一个空位来分析 最后一个空位存在两个状态 放与不放 如果不放那就是看dp[i-1]个数
放的话 就得看i - k - 1这个位置的个数
for(int i = 2; i <= n; i ++)
{   
    if(i - k - 1 >= 0) dp[i] = (dp[i - 1] + dp[i - k - 1]) % M;
    else dp[i] = (i + 1)  % M ;

}

cout << dp[n];

}
![栈.jpg](https://cdn.acwing.com/media/article/image/2025/04/01/226510_189cb23b0e-栈.jpg) //栈#include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

using namespace std;

typedef long long LL;

const int N = 1e6 + 10;
const int M = 1e9 + 7;

LL dp[N];

int n, k;

int main()
{
int n;
cin >> n;

string s;
cin >> s;

stack<char> t;
bool flag = true;

for(int i = 0; i < n; i ++)
{

    if(s[i] == '(') t.push(s[i]);
    else
    {   
        if(!t.empty()) t.pop();
        //char tmp = t.top();
    }
}

if(!t.empty()) flag = false;

if(flag) cout << "Yes";
else cout << "No";
// while(!t.empty())
// {
//     if(s[i] == '(') t.
// }


//cout << s;

}
![队列.jpg](https://cdn.acwing.com/media/article/image/2025/04/01/226510_bd0d75fc0e-队列.jpg) //队列#include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

include [HTML_REMOVED]

using namespace std;

typedef long long LL;

const int N = 1e6 + 10;
const int M = 1e9 + 7;

int a[1010];
bool st[1010];

int main()
{
int m, n;
cin >> m >> n;
queue[HTML_REMOVED] t;
memset(a, 0, sizeof(a));
memset(st, false, sizeof(st));
for(int i = 0; i < n; i ++) cin >> a[i];

int len = 0;
int cnt = 0;
for(int i = 0; i < n; i ++)
{

    if(!st[a[i]])
    {   
        if(len < m)
        {   
            st[a[i]] = true;
            t.push(a[i]);
            len ++;
            //cnt ++;
        }
        else
        {   
            int k = t.front();
            st[k] = false;

            t.pop();
            t.push(a[i]);
            st[a[i]] = true;
            //cnt ++;
        }
        cnt ++;
    }

}

cout << cnt;

}
`