A

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=1e6+10;
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        int n,x=0;
        cin>>n;
        for(int i=1;;i++)
        {
            if(i%2)x-=2*i-1;
            else x+=2*i-1;
            if(abs(x)>n)
            {
                if(i%2)cout<<"Sakurako"<<'\n';
                else cout<<"Kosuke"<<'\n';
                break;
            }
        }
    } 
}

B
计算每条对角线的最小值和0取min依次累加即可

# include <bits/stdc++.h>
using namespace std;
#define int long long
int n;
const int N=510; 
int a[N][N];
signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
    int n;
    cin>>n;
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            cin>>a[i][j];
        }
    }
    for(int i=1;i<=n;i++)
    {
    int mn=1e18;
    int x=1,y=i;
    while(x<=n&&y<=n)
    { 
           //cout<<x<<" "<<y<<endl;
           mn=min(mn,a[x][y]);
           x++;
           y++;
    }
    if(mn<0)
    sum+=abs(mn);
    }
    for(int i=2;i<=n;i++)
    {
    int mn=1e18;
    int x=i,y=1;
    while(x<=n&&y<=n)
    { //cout<<x<<" "<<y<<endl;
           mn=min(mn,a[x][y]);
           x++;
           y++;
    }
    if(mn<0)
    sum+=abs(mn);
    }
    cout<<sum<<endl;
    }
}

C
计算操作的两种情况每次去取最优

# include <bits/stdc++.h>
using namespace std;
#define int long long 
void solve()
{
    int n;
    cin >> n;
    vector<int>a(n+1);
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    for (int i = 2; i <= n / 2; i++)
    {
        int pre = (a[i] == a[i - 1]) + (a[n - i + 1] == a[n - i + 2]);
        int nex = (a[i] == a[n - i + 2]) + (a[n - i + 1] == a[i - 1]);
        if (nex < pre)
            swap(a[i], a[n - i + 1]);
    }
    int ans = 0;
    for (int i = 2; i <= n; i++)
    {
        if (a[i] == a[i - 1])
            ans++;
    }
    cout << ans << "\n";
    return ;
}

signed main()
{
    int T;
    cin>>T;
    while(T--)
    {
        solve();
    }
}

D 当两个前缀和相等时区间和为零，贪心考虑区间越小越好 用map维护当前出现过的所有前缀和

# include <bits/stdc++.h>
using namespace std;
#define int long long 
int n;
const int N=1e5+10;
int a[N];
map<int,int>ma;
signed main()
{
        int T;
        cin>>T;
        while(T--)
        {
        ma.clear();
        cin>>n;
        for(int i=1;i<=n;i++)
        {
        cin>>a[i];
        }
        ma[0]=1;
        int sum=0,res=0;
        for(int i=1;i<=n;i++)
        {
        sum+=a[i];
        res+=ma[sum];
        if(ma[sum])
        ma.clear();
        ma[sum]=1;
        }
        cout<<res<<endl;
        }

}

E置换环问题

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

using ll=long long;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int N=1000005;
int a[N];
bool vis[N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
            vis[i]=0;
        }
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            if(vis[i])continue;
            int now=i,cnt=0;
            while(!vis[now])
            {
                vis[now]=1;
                now=a[now];
                cnt++;
            }
            ans+=cnt-1>>1;
        }
        cout<<ans<<'\n';
    }
    return 0;
}