题目传送门

做法：注意行和列能整除p的情况，然后分块构造。赛场上写的太难看，码力不够QAQ

AC代码：

#include<bits/stdc++.h>
#define IO          ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x)       push_back(x)
#define sz(x)       (int)(x).size()
#define sc(x)       scanf("%d",&x)
#define pr(x)       printf("%d\n",x)
#define abs(x)      ((x)<0 ? -(x) : x)
#define all(x)      x.begin(),x.end()
#define mk(x,y)     make_pair(x,y)
#define debug       printf("!!!!!!\n")
#define fin         freopen("in.txt","r",stdin)
#define fout        freopen("out.txt","w",stdout)
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxm = 1e8+5;
const int maxn = 1e6+5;
const int INF = 0x3f3f3f3f;
int mp[105][105];
int main()
{
    //fin;
    IO;
    int n,m,p;
    cin>>n>>m>>p;
    int t = n*m;
    if(t%p!=0) cout<<"No"<<endl;
    else{
        if(p>n && p>m){
            cout<<"No"<<endl;
            return 0;
        }
        int x = t/p;
        if(x<=t)
        {
            cout<<"Yes"<<endl;
            if(n%p == 0)
            {
                int pos = 1;
                for(int k=0;k<n/p;k++)
                {
                    for(int i=0;i<p;i++)
                    {
                        for(int j=0;j<m;j++)
                        {
                            cout<<j+pos<<" ";
                        }
                        cout<<endl;
                    }
                    pos+=m;
                }
            }
            else
            {
                int pos = 1;
                int tmp = 0;
                for(int k=0;k<m/p;k++)
                {
                    for(int i=0;i<p;i++)
                    {
                        for(int j=0;j<n;j++)
                        {
                            mp[i+tmp][j] = pos+j;
                        }
                    }
                    pos+=n;
                    tmp+=p;
                }
                for(int i=0;i<n;i++)
                {
                    for(int j=0;j<m;j++)
                        cout<<mp[j][i]<<" ";
                    cout<<endl;
                }
            }
        }
        else cout<<"No"<<endl;
    }
    return 0;
}