#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;
typedef vector<vi> vvi;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcount __builtin_popcount


int cnt;
map<int, int> mapp;
const int N = 2e5 + 10;
int discrete(int x){
    if (!mapp.count(x)) mapp[x] = cnt++;
    return mapp[x];
}

 //int a[200341] = {0};
int main(){
    //ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n, m;
    cin >> n;
    vi a(N, 0);
    mapp.clear();
    cnt = 0;
    for (int i = 0, t; i < n; ++i){
        cin >> t;t = discrete(t);
        a[t] += 1;
    }
    cin >> m;
    int happy = 0, ok = 0, ans = 1;
    set<int> sett;
    vvi b(N, vi(2, 0));
    for (int i = 0; i < m; ++i) cin >> b[i][0];
    for (int i = 0; i < m; ++i) cin >> b[i][1];
    for (int i = 0; i < m; ++i){
        int t = discrete(b[i][0]);
        if (a[t] > happy){
            happy = a[t];
            ans = i + 1;
            ok = a[discrete(b[i][1])];
        }
        else if (happy == a[t] && ok < a[discrete(b[i][1])]){
            ok = a[discrete(b[i][1])];
            ans = i + 1;
        }
    }

    cout << ans << endl;
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;
typedef vector<vi> vvi;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcount __builtin_popcount


int cnt;
map<int, int> mapp;
const int N = 2e5 + 10;
int discrete(int x){
    if (!mapp.count(x)) mapp[x] = cnt++;
    return mapp[x];
}

 //int a[200341] = {0};
int main(){
    //ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n, m;
    cin >> n;
    vi a(N, 0);
    mapp.clear();
    cnt = 0;
    for (int i = 0, t; i < n; ++i){
        cin >> t;t = discrete(t);
        a[t] += 1;
    }
    cin >> m;
    int happy = 0, ok = 0, ans = -1;
    set<int> sett;
    vvi b(N, vi(2, 0));
    for (int i = 0; i < m; ++i) cin >> b[i][0];
    for (int i = 0; i < m; ++i) cin >> b[i][1];
    for (int i = 0; i < m; ++i){
        //if (i > 1000) cout << b[i][0] << " \n"; 重点是这一句，数据有2000个，但是读到省略号以后的数据全是0
        int t = discrete(b[i][0]);
        if (a[t] > happy){
            happy = a[t];
            ans = i + 1;
            ok = a[discrete(b[i][1])];
        }
        else if (happy == a[t] && ok < a[discrete(b[i][1])]){
            ok = a[discrete(b[i][1])];
            ans = i + 1;
        }
    }

    cout << ans << endl;
    return 0;
}



//下面这种写法是第2000个数据第一千多就读到了，根本没法跑代码

#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;
typedef vector<vi> vvi;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcount __builtin_popcount


int cnt;
map<int, int> mapp;
int discrete(int x){
    if (!mapp.count(x)) mapp[x] = cnt++;
    return mapp[x];
}

 //int a[200341] = {0};
int main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n, m;
    cin >> n;
    vi a(2e5, 0);
    mapp.clear();
    cnt = 0;
    for (int i = 0, t; i < n; ++i){
        cin >> t;t = discrete(t);
        a[t] += 1;
    }
    cin >> m;
    int happy = 0, ok = 0, ans = -1;
    set<int> sett;
    for (int i = 1; i <= m; ++i){
        int t;  cin >> t; t = discrete(t);
        if (a[t] > happy){
            happy = a[t];
            sett.clear();
            sett.insert(i);
            ans = i;
        }
        else if (a[t] == happy){
            sett.insert(i);
        }
    }
    for (int i = 1; i <= m; ++i) {
        int t; cin >> t; t = discrete(t);
        if (sett.count(i) && ok < a[t]){
            ok = a[t];
            ans = i;
        }
    }
    cout << ans << endl;
    return 0;
}

要么省略号后数据读出来全是0，要么数据读出来顺序不对，第2000个数据第1300多就读到了

题目链接 https://www.acwing.com/problem/content/105/

我实在是受不了了，代码写的逻辑一点问题都没有，换了两种方法了，就是输入数据读的有问题

错误的代码：

#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;
typedef vector<vi> vvi;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcount __builtin_popcount


int cnt;
map<int, int> mapp;
const int N = 2e5 + 10;
int discrete(int x){
    if (!mapp.count(x)) mapp[x] = cnt++;
    return mapp[x];
}

 //int a[200341] = {0};
int main(){
    //ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n, m;
    cin >> n;
    vi a(N, 0);
    mapp.clear();
    cnt = 0;
    for (int i = 0, t; i < n; ++i){
        cin >> t;t = discrete(t);
        a[t] += 1;
    }
    cin >> m;
    int happy = 0, ok = 0, ans = 1;
    set<int> sett;
    vvi b(N, vi(2, 0));
    for (int i = 0; i < m; ++i) cin >> b[i][0];
    for (int i = 0; i < m; ++i) cin >> b[i][1];
    for (int i = 0; i < m; ++i){
        //if (i > 1000) cout << b[i][0] << " \n"; 重点是这一句，数据有2000个，但是读到省略号以后的数据全是0
        int t = discrete(b[i][0]);
        if (a[t] > happy){
            happy = a[t];
            ans = i + 1;
            ok = a[discrete(b[i][1])];
        }
        else if (happy == a[t] && ok < a[discrete(b[i][1])]){
            ok = a[discrete(b[i][1])];
            ans = i + 1;
        }
    }

    cout << ans << endl;
    return 0;
}



//下面这种写法是第2000个数据第一千多就读到了，根本没法跑代码

#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;
typedef vector<vi> vvi;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcount __builtin_popcount


int cnt;
map<int, int> mapp;
int discrete(int x){
    if (!mapp.count(x)) mapp[x] = cnt++;
    return mapp[x];
}

 //int a[200341] = {0};
int main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n, m;
    cin >> n;
    vi a(2e5, 0);
    mapp.clear();
    cnt = 0;
    for (int i = 0, t; i < n; ++i){
        cin >> t;t = discrete(t);
        a[t] += 1;
    }
    cin >> m;
    int happy = 0, ok = 0, ans = 1;
    set<int> sett;
    for (int i = 1; i <= m; ++i){
        int t;  cin >> t; t = discrete(t);
        if (a[t] > happy){
            happy = a[t];
            sett.clear();
            sett.insert(i);
            ans = i;
        }
        else if (a[t] == happy){
            sett.insert(i);
        }
    }
    for (int i = 1; i <= m; ++i) {
        int t; cin >> t; t = discrete(t);
        if (sett.count(i) && ok < a[t]){
            ok = a[t];
            ans = i;
        }
    }
    cout << ans << endl;
    return 0;
}

要么数据读出来全是0，要么数据读出来顺序不对，第2000个数据第1300多就读到了

什么情况啊，vector，数组都试过了，输入一直采用的cin>>输入，始终解决不了的问题

CF已AC，我是真服了..debug两个小时原来是网站有问题，以后再也不相信网站了

#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;
typedef vector<vi> vvi;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcount __builtin_popcount




class FenwickTree{
private:
    vll ft;
    void build(const vll& f){
        int m = sz(f) - 1;
        ft.assign(m + 1, 0);
        for (int i = 1; i <= m; ++i){
            ft[i] += f[i];
            if (i + lowbit(i) <= m) ft[i + lowbit(i)] += f[i];
        }
    }
public:
    FenwickTree(int m){ft.assign(m + 1, 0);}
    FenwickTree(const vll& f){
        int m = sz(f) - 1;
        ft.assign(m + 1, 0);
        build(f);
    }
    void update(int i, ll val){
        while (i < sz(ft)){
            ft[i] += val;
            i += lowbit(i);
        }
    }
    ll rsq(int i){
      ll sum = 0;
      while (i){sum += ft[i]; i -= lowbit(i);}
    }
    ll rsq(int i, int j){return rsq(j) - rsq(i - 1);}

};
int numFactors(int x){
    int result = 2;
    for (int i = 2; i <= x / i; ++i){
        if (x % i == 0){
            if (x / i != i) result += 1;
            result += 1;
        }
    }
    return result;
}


vi sieveFactors(int x){
    vi result(x + 1, 0);
    for (int i = 1; i <= x; ++i){
        for (int j = i; j <= x / i; ++j){
            result[j * i] += 1;
        }
    }
    return result;
}

int main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    /*
    int n = 2e9;
    vi a= sieveFactors(n);
    vi result;
    cout << sz(a) << endl;
    int maxx = 1;
    for (int i = 1; i <= n; ++i){
        if (maxx < a[i]){
            result.pb(i);
            maxx = a[i];
        }
    }
    cout << sz(result) << endl;
    for (auto x : result) cout << x << ",";*/
    set<int> sett{1, 2, 4, 6, 12,24,36,60,120,180,240,360,720,840,1260,1680,2520,5040,7560,10080,15120,20160,25200,27720,45360,50400,55440,83160,110880,166320,221760,277200,332640,
    498960,554400,665280,720720,1081080,1441440,2162160,2882880,3603600,4324320,6486480,7207200,8648640,10810800,14414400,17297280,21621600,32432400,36756720,43243200,
    61261200,73513440,110270160,122522400,147026880,183783600,245044800,294053760,367567200,551350800,698377680,735134400,1102701600,1396755360};
    int n;
    cin >> n;
    auto t = sett.lower_bound(n);
    if (t == sett.end() || *t > n) --t;
    cout << *t << endl;
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;
typedef vector<vi> vvi;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcount __builtin_popcount



void dfs(const int& n, bitset<10>& bs, vi& b, int k){
    if (k == n + 1){
        for (int i = 1; i <= n; ++i) cout << b[i] << " \n"[i == n];
        return;
    }
    for (int i = 1; i <= n; ++i){
        if (bs[i]) continue;
        bs[i] = 1;
        b[k] = i;
        dfs(n, bs, b, k + 1);
        bs[i] = 0;
    }
}



int main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    vi b(n + 1);
    bitset<10> bs;
    dfs(n, bs, b, 1);
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;
typedef vector<vi> vvi;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcount __builtin_popcount



//小失误，在编程的时候忘记把计数放到循环里面，而是放到了外面，重复计数了。
//先枚举第一行的所有按法，然后从第二行开始根据上面的灯的情况向下传递
//最后判断最后一行即可

void change(vvi& a, int i, int j){
    a[i][j] ^= 1;
    if (i) a[i - 1][j] ^= 1;
    if (i < 4) a[i + 1][j] ^= 1;
    if (j) a[i][j - 1] ^= 1;
    if (j < 4) a[i][j + 1] ^= 1;
}


int main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int TC;
    cin >> TC;
    while (TC--){
        vvi a(5, vi(5));
        int result = 1e7;
        for (int i = 0; i < 5; ++i) {
            string s;
            cin >> s;
            for (int j = 0; j < 5; ++j) {a[i][j] = s[j] - '0';}
        }
        auto b = a;
        for (int i = 0; i < (1 << 5); ++i){
            int j = i, k = 0;
            a = b;
            int cnt = 0;
            while (k < 5){
                if (j & (1 << k)) {change(a, 0, k);cnt += 1;}
                k += 1;
            }
            for (int j = 1; j < 5; ++j){
                for (int k = 0; k < 5; ++k) if (a[j - 1][k] == 0){
                    change(a, j, k);
                    cnt += 1;
                }
            }
            bool ok = true;
            for (int j = 0; j < 5; ++j) {
                if (a[4][j] == 0) {ok = false; break;}
            }
            if (ok) result = min(result, cnt);
        }
        cout << (result <= 6 ? result : -1) << endl;
    }
    return 0;
}

//关键在于如何保证序列递增且不重复，那么考虑接下来的位置时，就必须从当前位置的下一个数字开始搜索（这个序列必须有序）
//这样可以保证递增，但是一定保证不重复吗？答案是正确的，因为重复的原因都是因为大数在前面，小数在后面，排序后重复
#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;
typedef vector<vi> vvi;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcnt __builtin_popcount





void dfs(const int& n, const int& m, vi& b, bitset<100>& bs, int k, int start){
    if (k == m){
        for (int i = 0; i < m; ++i) cout << b[i] << " \n"[i + 1 == m];
    }
    for (int i = start; i <= n; ++i){
        if (bs[i]) continue;
        bs[i] = 1;
        b[k] = i;
        dfs(n, m, b, bs, k + 1, i + 1);
        bs[i] = 0;
    }
}


int main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n, m;
    cin >> n >> m;
    if(!m) return 0;
    vi b(m + 10);
    bitset<100> bs;
    dfs(n, m, b, bs, 0, 1);
    return 0;
}

//第一思路居然是写两个并查集，一个维护都相等的集合，一个维护不相等的集合，但是有一些测试数据跑不过去
//比如 1 7 1， 7 9 0， 13 9 1， 1 13 1，然后用了al维护，结果还是不行
//思路应该是：用一个uf维护相等的集合，在保存所有的不相等的点，在结束后判断一下不相等的点有没有在一个set中的即可
//或者维护所有不相等的点，然后存储相等的点对，最后看相等的点对有没有在一个集合中的即可  //不行的，只能维护相等的点
#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;
typedef vector<vi> vvi;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcnt __builtin_popcount



class UnionFind{
private:
    vi fa;
public:
    UnionFind(int m){
        fa.assign(m, 0); for (int i = 0 ; i < m; ++i) fa[i] = i;
    }
    int findSet(int x){return fa[x] == x ? x : fa[x] = findSet(fa[x]);}
    bool isSameset(int x, int y){return findSet(x) == findSet(y);}
    void unionSet(int x, int y){
        x = findSet(x), y = findSet(y);
        if (x == y) return;
        fa[x] = y;
    }
};



map<int, int> mapp;
int cnt;
int discrete(int x){
    if (mapp.count(x) == 0) mapp[x] = cnt++;
    return mapp[x];
}
int main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int tc;
    cin >> tc;
    while (tc--){
        int n;
        cin >> n;
        UnionFind uf(2 * n + 10);
        mapp.clear();
        cnt = 0;
        bool ok = true;
        vii a;
        for (int i = 0; i < n; ++i){
            int u, v, p;
            cin >> u >> v >> p;
            u = discrete(u);
            v = discrete(v);
            if (ok == false) continue;
            if (u == v) {ok = p == 1; continue;}
            if (p == 1){
                uf.unionSet(u, v);
            }
            else{
                a.pb({u, v});
            }
        }
        for (auto& x : a) if (uf.isSameset(x.first, x.second)){
            ok = false;
            break;
        }
        cout << (ok ? "YES" : "NO") << endl;
    }
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcnt __builtin_popcount


//题目大意：想办法排列，满足行和列的单调性。
//已知的信息：行数和每行的人数，且行数范围最大是5，总人数不会超过30个

//爆搜：感觉有点像背包问题，这个状态可以由5个子状态推过来，但是只有满足条件的子状态才能往这个状态推
//对于子状态的判断，只有行人数的判断和跟相邻行关系的判断，有个疑问就是为什么没有当前放置同学的编号的判断呢？
//初始条件，1只能放在1号位


//总的疑问：为什么可以直接按人数进行转移呢 不需要对状态的最后一个同学的编号进行判定吗

ll f[31][31][31][31][31] = {0};//vector写不出五维的东西..



ll dfs(int a, int b, int c, int d, int e){
    ll& ans = f[a][b][c][d][e];cout << a << b << c << d << e << endl;
    if (ans) return ans;

    if (a - 1 >= b && a >= 1)ans += dfs(a - 1, b, c, d, e);
    if (b - 1 >= c && b >= 1)ans += dfs(a, b - 1, c, d, e);
    if (c - 1 >= d && c >= 1)ans += dfs(a, b, c - 1, d, e);
    if (d - 1 >= e && d >= 1)ans += dfs(a, b, c, d - 1, e);
    if (e >= 1)ans += dfs(a, b, c, d, e - 1);
    return ans;
}





int main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int k;
    f[1][0][0][0][0] = 1;       //边界条件
    while (cin >> k && k){
        vi A(5, 0);
        for (int i = 0; i < k; ++i) cin >> A[i];
        cout << dfs(A[0], A[1], A[2], A[3], A[4]) << endl;
    }
    return 0;
}

#include <bits/stdc++.h>

using namespace std;

typedef vector<int> vi;
typedef long long int ll;
typedef vector<ll> vll;
typedef pair<int, int> ii;
typedef vector<bool> vb;
typedef vector<ii> vii;
typedef unsigned long long ull;



#define lowbit(S) ((S)& -(S))
#define pb push_back
#define sz(x) ((int)(x.size()))
#define bg begin()
#define ed end()
#define rbg rbegin()
#define red rend()
#define bitcnt __builtin_popcount


//求限定了最小长度连续区间的最大平均值，第一想法是枚举所有区间， tle
//平均值没有递增效应， 可以知道平均值的范围，对范围内的平均值进行判定，看看是否符合
//对平均值进行判定，可以将每个数都减去平均值，如果存在长度大于f的连续区间和>=0,满足条件
bool check(const vi& a, int f, double ave){
    int n = sz(a) - 1;
    double minn = 1e3, maxx = -1e3;
    vector<double> s(n + 1, 0);
    for (int i = 1; i <= n; ++i){
        s[i] = s[i - 1] + (double)a[i] - ave;
    }
    for (int r = f; r <= n; ++r){
        minn = min(minn, s[r - f]);
        maxx = max(maxx, s[r] - minn);
    }
    return maxx >= 0;
}

int main(){
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n, f, result = 0;
    double l = 0, rr = 0;
    cin >> n >> f;
    vi a (n + 1);
    for (int i = 1; i <= n; ++i) {cin >> a[i];rr = max(rr, double(a[i]));}
    while (rr - l > 1e-4){
        double mid = (l + rr) / 2;
        if (check(a, f, mid)) {l = mid;}
        else rr = mid;
    }
    cout << int(rr * 1000) << endl;
    return 0;
}