你的任务是求出有多少头牛被除自己之外的所有牛认为是受欢迎的。
如果只一个点自环，那对于“被除自己之外的”这句话就错了吧？
洛谷上相同的题

某大佬题解

请教一个问题，这篇题解里“否则所有不等式组表示的是变量的相对关系,这种情况下在得到一组可行解后, 我们让所有可行解同时+c, 变量相对关系不变, 即可行解有无数个, 不存在最值.”是什么情况呢？
我有点懵……

题目链接 361. 观光奶牛

请问代码里这个循环队列是什么含义？

if (hh == N) hh = 0;

if (tt == N) tt = 0;

y老师代码

1141. 局域网

题目中哪里说存在环呢？

题目链接 1134. 最短路计数

else if (dist[j] == dist[t] + 1)
{
    cnt[j] = (cnt[j] + cnt[t]) % mod;
}

请问这表示的是什么情况？

#include <cstdio>

typedef long long LL;

LL qadd(LL a, LL b, LL p)
{
    LL res = 0;
    while (b)
    {
        if (b & 1) res = (res + a) % p;
        a = (a + a) % p;
        b >>= 1;
    }
    return res;
}

int main()
{
    LL a, b, p;
    scanf("%lld%lld%lld", &a, &b, &p);
    printf("%lld\n", qadd(a, b, p));

    return 0;
}

#include <iostream>
using namespace std;

typedef long long LL;
LL qmi(int a, int b, int p) {
    LL ans = 1 % p;
    while (b) {
        if (b & 1) ans = ans * a % p;
        a = a * (LL)a % p;
        b >>= 1;
    }
    return ans;
}

int main() {
    int n;
    scanf("%d", &n);
    while (n --) {
        int a, b, p;
        scanf("%d%d%d", &a, &b, &p);
        printf("%lld\n", qmi(a, b, p));
    }
}

/* 
    f[i][j] 表示以i开始的长度为2^j的区间最大值
*/ 
#include <bits/stdc++.h>
#define md(x, y)  ((x + y) >> 1)
#define ls(x)      (x << 1)
#define rs(x)      (x << 1 | 1)
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int MAXN = 2e5 + 10;
int f[MAXN][20], n, m, a[MAXN];

void init()
{
    for(int i = 0; i < 20; i++)
        for(int j = 1; j + (1 << i) - 1 <= n; j++)
            if(!i) f[j][i] = a[j];
            else   f[j][i] = max(f[j][i - 1], f[j + (1 << (i - 1))][i - 1]);
}
int query(int l, int r)
{
    int len = r - l + 1;
    int k = log2(len);
    return max(f[l][k], f[r - (1 << k) + 1][k]);
}
int main()
{
    cin >> n;
    for(int i = 1; i <= n; i++)
        cin >> a[i];
    init();
    cin >> m;
    while(m--)
    {
        int l, r;
        cin >> l >> r;
        cout << query(l, r) << endl;
    }
    return 0;
}

/* 

*/ 
#include <bits/stdc++.h>
#define md(x, y)  ((x + y) >> 1)
#define ls(x)      (x << 1)
#define rs(x)      (x << 1 | 1)
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int MAXN = 5e5 + 10;
ll a[MAXN], c[MAXN], n;
typedef pair<int, int> P;
P b[MAXN];
int lowbit(int x)
{
    return x & (-x);
}
ll query(int x)
{
    ll ret = 0;
    for(int i = x; i > 0; i -= lowbit(i))
        ret += c[i];
    return ret;
}
void update(int x, int k)
{
    for(int i = x; i <= n; i += lowbit(i))
        c[i] += k;
} 
int main()
{
    while(cin >> n, n)
    {
        for(int i = 1, x; i <= n; i++) cin >> x, b[i] = {x, i};
        sort(b + 1, b + 1 + n);
        memset(c, 0, sizeof c);
        ll res = 0;
        for(int i = n; i >= 1; i--)
        {
            int x = b[i].second;
            res += query(x);
            update(x, 1);
        }
        cout << res << endl;
    }
    return 0;
}

/* 
    down 存的是低部
*/ 
#include <bits/stdc++.h>
#define md(x, y)  ((x + y) >> 1)
#define ls(x)      (x << 1)
#define rs(x)      (x << 1 | 1)
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 10;
vector<int> vec;
priority_queue<int> down;
priority_queue<int, vector<int>, greater<int>> up;
void update()
{
    int a = up.size(), b = down.size();
    if(b > a + 1)
    {
        int u = down.top();
        down.pop();
        up.push(u);
    }
    else if(a > b)
    {
        int u = up.top();
        up.pop();
        down.push(u);
    }
}
int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        while(up.size()) up.pop();
        while(down.size()) down.pop();
        vec.clear();
        int kase, n;
        cin >> kase >> n;
        for(int i = 1; i <= n; i++)
        {
            int u;
            cin >> u;
            if(i == 1)
            {
                vec.push_back(u);
                down.push(u);
                continue;
            }
            int x = down.top();
            if(u >= x) up.push(u);
            else down.push(u);
            update();
            if(i % 2)
                vec.push_back(down.top());
        }
        cout << kase <<" "<< (n + 1) / 2 << endl;
        for(int i = 1; i <= vec.size(); i++)
        {
            printf("%d ", vec[i - 1]);
            if(i % 10 == 0)
                printf("\n");
        }
        if(vec.size() % 10)
            printf("\n");
    }
    return 0;
}