#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

int main()
{
    string str;
    int n;
    cin >> n >> str;
    int cnt = 0, ans = 0;
    for(int i = 0; i < n; i++)
    {
        if(str[i] == 'x')
        {
            cnt++;
            if(cnt == 3)
            {
                cnt--;
                ans++;
            }
        }
        else cnt = 0;
    }
    cout << ans;
    return 0;
}

#include<iostream>

using namespace std;
const int N = 1e5+10;
int a[N], b[N];

int main()
{
    int n, m, x;
    cin >> n >> m >> x;
    for(int i = 0; i < n; i++) scanf("%d", &a[i]);
    for(int i = 0; i < m; i++) scanf("%d", &b[i]);

    int i = 0, j = m-1;
    while(a[i] + b[j] != x)
    {
        if(a[i] + b[j] > x) j--;
        else i++;
    }
    cout << i << " " << j;
    return 0;
}

#include<iostream>

using namespace std;
const int N = 1e5+10;
int a[N],idx[N];

int main()
{
    int n;
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        scanf("%d", &a[i]);
    }

    int res = 0;
    for(int i = 0, j = 0; i < n; i++)
    {
        idx[a[i]]++;
        while(idx[a[i]] > 1)
        {
            idx[a[j]]--;
            j++;
        }
        res = max(res, i-j+1);
    }
    cout << res;
    return 0;
}

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;
const int N = 1010;
int n, k;

struct Stu
{
    int id, ok;
}s[N];

bool check(int x)
{
    if(x % k == 0 || x % 10 == k) return true;
    return false;
}

int main()
{
    cin >> n >> k;
    int res;
    int cnt = n;
    for(int i = 1; i <= n; i++)
    {
        s[i] = {i, 1};
    }

    int num = 1;
    while(cnt != 1)
    {
        for(int i = 1; i <= n; i++)
        {
            if(s[i].ok)
            {
                if(check(num))
                {
                    s[i].ok = 0; // 删除该元素
                    //printf("id:%d num:%d\n", s[i].id, num);
                    cnt--;
                    if(cnt == 1) break;
                }
            }
            else continue; // 当前元素已经被删除了直接跳到下一重循环
            num++;
            //printf("cnt:%d num:%d\n", cnt, num)

        }
    }
    for(int i = 1; i <= n; i++)
    {
        if(s[i].ok) cout << s[i].id;
    }
    return 0;
}

#include<iostream>

using namespace std;
const int N = 1e5+10;
int nums[N];

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

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

    while(q--)
    {
        int k;
        cin >> k;
        int l = 0, r = n-1;
        while(l < r)
        {
            int mid = (l+r)>>1;
            if(nums[mid] >= k) r = mid;
            else l = mid + 1;
        }

        if(nums[l] != k) cout << "-1 -1" << endl;
        else
        {
            cout << l << " ";
            int l = 0, r = n - 1;
            while(l < r)
            {
                int mid = (l+r+1) >> 1;
                if(nums[mid] <= k) l = mid;
                else r = mid-1;
            }
            cout << l << endl;
        }
    }

    return 0;
}

#include<iostream>

using namespace std;

const int N = 1010;
int a[N][N], b[N][N];

// 二维最核心的插入操作
void insert (int x1, int y1, int x2, int y2, int c)
{
    b[x1][y1] += c;
    b[x2+1][y1] -= c;
    b[x1][y2+1] -= c;
    b[x2+1][y2+1] += c;
}

int main()
{
    int n, m, q;
    scanf("%d%d%d", &n, &m, &q);

    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++) scanf("%d", &a[i][j]);
    }

    // 初始化差分数组b
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++) insert(i, j, i, j, a[i][j]);
    }

    while(q--)
    {
        int x1, y1, x2, y2, c;
        scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &c);
        // 每一次进行一次操作
        insert(x1,y1,x2,y2,c);
    }

    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
            // 求二维前缀和
            b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];

    for (int i = 1; i <= n; i ++ )
    {
        for (int j = 1; j <= m; j ++ ) printf("%d ", b[i][j]);
        printf("\n");
    }

    return 0;
}

#include<iostream>

using namespace std;

const int N = 1e5+10;
int a[N], b[N];

// 差分中唯一操作 在区间【l, r】中对每一个元素进行 + c
void insert(int l, int r, int c)
{
    b[l] += c;
    b[r+1] -= c;
}

int main()
{
    int m, n;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
    }

    // 初始化差分数组
    for(int i = 1; i <= n; i++)
    {
        insert(i, i, a[i]); // b[N]是全局变量，在进行insert操作时就已经将b数组弄成了差分数组
    }

    while(m--)
    {
        int l, r, c;
        scanf("%d%d%d", &l, &r, &c);
        insert(l, r, c);
    }

    for(int i = 1; i <= n; i++)
    {
        b[i] += b[i-1]; // 求前缀和
        printf("%d ", b[i]);
    }
    return 0;
}

根据@ S7former佬的题解理解了一下，打了点注释，存一下思路

/*
    用dijkstra算出原来每个点到 1的最短距离
    然后遍历所有点，找到中转点，
    使得中转点到当前点距离 + 中转点到 1距离 == 原来的最短距离 
    其中 d[i]表示 i号点到 1号点的距离 
*/

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>

using namespace std;

#define f first
#define s second

typedef pair<int, int> PII;
const int N = 1e4+10, M = 2e5+10;
int h[N], ne[M], e[M], idx, w[M];
int n, m;
int d[N];
bool st[N];

void add(int a, int b, int c)
{
    e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

void dijkstra()
{
    memset(d, 0x3f, sizeof d);
    d[1] = 0;

    priority_queue<PII, vector<PII>, greater<PII> > q;
    q.push({0, 1});

    while(q.size())
    {
        auto t = q.top();
        q.pop();
        if(st[t.s]) continue;
        st[t.s] = true;
        for(int i = h[t.s]; i != -1; i = ne[i])
        {
            int j = e[i];
            if(d[j] > w[i] + t.f)
            {
                d[j] = w[i] + t.f;
                q.push({d[j], j});
            }
        }
    }
}

int main()
{
    cin >> n >> m;
    memset(h, -1, sizeof h);
    while(m--)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c), add(b, a, c);
    }

    dijkstra();

    int res = 0;
    for(int k = 2; k <= n; k++)
    {
        int road = 0x3f3f3f3f;
        for(int i = h[k]; i != -1; i = ne[i])
        {
            int j = e[i];
            // j作为中转点， 1-k的距离 == 1-j的距离 + j-k的距离 
            if(d[k] == d[j] + w[i]) road = min(road, w[i]);
        }
        res += road;
    }
    cout << res;
    return 0;
}

#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;
const int N = 1010;
int a[N];

struct Op
{
    int id, type, time;
}op[N*2]; // 操作有借有还，所以 * 2 


int cmp(Op a, Op b)
{
    /*
        此处比较函数应该按照以下规则
        1、时间从小到大
        2、时间相同，还操作在前
        3、时间、操作类型都一样，按 id从小到大   
    */
    if(a.time < b.time) return 1;
    else if(a.time == b.time)
    {
        if(a.type > b.type) return 1;
        else if(a.type == b.type)
        {
            if(a.id < b.id) return 1;
        }
    }
    return 0;
}

int main()
{
    int n, m;
    cin >> n >> m;
    for(int i = 1; i <= n; i++) a[i] = i;
    int cnt = 0;
    while(m--)
    {
        int id, sta, len;
        cin >> id >> sta >> len;
        op[cnt++] = {id, 0, sta}; // 取钥匙 
        op[cnt++] = {id, 1, sta + len}; // 还钥匙 
    }

    sort(op, op+cnt, cmp);

    for(int i = 0; i < cnt; i++)
    {
        if(!op[i].type)
        {
            for(int j = 1; j <= n; j++)
            {
                if(a[j] == op[i].id)
                {
                    a[j] = 0;
                    break;
                }
            }
        }
        else
        {
            for(int j = 1; j <= n; j++)
            {
                if(!a[j])
                {
                    // 从小到大有空位直接插入即可 
                    a[j] = op[i].id;
                    break;
                }
            }
        }
    }
    for(int i = 1; i <= n; i++) cout << a[i] << " ";
    return 0;
}