看看这里

A - Tenki

模拟

#include <bits/stdc++.h>
using namespace std;



int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    string s, t;
    cin >> s >> t;
    int res = 0;
    for(int i = 0; i < 3; i++) if(s[i] == t[i]) res++;
    cout << res << endl;
    return 0;
}

B - Power Socket

模拟

#include <bits/stdc++.h>
using namespace std;



int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int a, b;
    cin >> a >> b;
    int t = 1;
    int res = 0;
    while(t < b){
        res++;
        t += a - 1;
    }
    cout << res << endl;
    return 0;
}

C - Lower

对每个$i$都做移动肯定不行，想办法找点性质，发现在一段不递增子序列里最大值肯定是序列长度
中间的点完全可以不用管，维护当前非递增序列长度，当递增时更新

#include <bits/stdc++.h>
using namespace std;



int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int n;
    cin >> n;
    vector<int> a(n);
    int res = 0, t = 0;
    for(int i = 0; i < n; i++) cin >> a[i];
    for(int i = 0, j = 0; i < n - 1; i++)
        if(a[i] >= a[i + 1]) t++;
        else 
        {
            res = max(res, t);
            t = 0;
        }
    res = max(res, t);
    cout << res << endl;
    return 0;
}

D - ModSum

瞪眼出，这种只能算脑筋急转弯，但也不难猜

#include <bits/stdc++.h>
using namespace std;



int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    long long n;
    cin >> n;
    cout << (n - 1) * n / 2 << endl;
    return 0;
}

E - League

因为每个人的比赛都是有先后顺序的，很容易往图论想
如果能想到将比赛hash的话，那拓扑排序也是很容易想的
将每场比赛看作一个点，按照先后顺序连边，跑拓扑排序
找最大距离，没有拓扑序说明比赛有冲突，无解

#include <bits/stdc++.h>
using namespace std;

const int N = 3e6 + 10;
int h[N], e[N], ne[N], idx;
int d[N], n, dist[N];
void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

int getcode(int x, int y)
{
    if(x > y) swap(x, y);
    return (x - 1) * n + y;
}

int main(int argc, char const *argv[])
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin >> n;
    vector<vector<int>> a(n + 1, vector<int>(n + 1));
    memset(h, -1, sizeof h);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j < n; j++) cin >> a[i][j];
        for(int j = 1; j < n - 1; j++)
        {
            add(getcode(i, a[i][j]), getcode(i, a[i][j + 1]));
            d[(getcode(i, a[i][j + 1]))]++;
        }      
    }
    queue<int> q;
    int res = 0;
    for(int i = 1; i <= n * n; i++) if(!d[i]) {q.push(i);dist[i] = 1;}
    while(!q.empty())
    {
        int ver = q.front(); q.pop();
        res = max(res, dist[ver]);
        for(int i = h[ver]; ~i; i = ne[i])
        {
            int j = e[i];
            d[j]--;
            if(!d[j])
            {
                q.push(j);
                dist[j] = dist[ver] + 1;
            }
        }
    }
    for(int i = 1; i <=  n * n; i++)
        if(d[i]) res = -1;
    cout << res << endl;
    return 0;
}

F - Engines

F几何题，写不了一点