C题为撒会卡cin呢？

运行超时代码

// Problem: Karashi的生日蛋糕
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/49244/C
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// Date: 2022-12-30 19:13:55
// 
// Powered by CP Editor (https://cpeditor.org)

/**
  * @author  : SDTBU_LY
  * @version : V1.0.0
  * @上联    : ac自动机fail树上dfs序建可持久化线段树
  * @下联    : 后缀自动机的next指针DAG图上求SG函数
**/

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

#define MAXN 1000010

using namespace std;

typedef long long ll;

int n,k;
vector<int> v[MAXN];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    cin >> n >> k;
    
    int pos=0;
    for(int i=1;i<=n;i++)
    {
        int num=i;
        int temp=num/k,d=num%k;
        for(int j=0;j<=d-1;j++)
            v[(pos+j)%k].push_back(temp+1);
        for(int j=d;j<=k-1;j++)
            v[(pos+j)%k].push_back(temp);
            
        pos=(pos+num)%k;
        //cout << pos << endl;
    }
    for(int i=0;i<=k-1;i++)
    {
        for(int j=0;j<n;j++)
            cout << v[i][j] << ' ';
        cout << endl;
    }       
    return 0;
}

AC代码

// Problem: Karashi的生日蛋糕
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/49244/C
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// Date: 2022-12-30 19:13:55
// 
// Powered by CP Editor (https://cpeditor.org)

/**
  * @author  : SDTBU_LY
  * @version : V1.0.0
  * @上联    : ac自动机fail树上dfs序建可持久化线段树
  * @下联    : 后缀自动机的next指针DAG图上求SG函数
**/

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

#define MAXN 1000010

using namespace std;

typedef long long ll;

int n,k;
vector<int> v[MAXN];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    scanf("%d %d",&n,&k);
    
    int pos=0;
    for(int i=1;i<=n;i++)
    {
        int num=i;
        int temp=num/k,d=num%k;
        for(int j=0;j<=d-1;j++)
            v[(pos+j)%k].push_back(temp+1);
        for(int j=d;j<=k-1;j++)
            v[(pos+j)%k].push_back(temp);
            
        pos=(pos+num)%k;
        //cout << pos << endl;
    }
    for(int i=0;i<=k-1;i++)
    {
        for(int j=0;j<n;j++)
            printf("%d ",v[i][j]);
        printf("\n");
    }       
    return 0;
}