[atARC140E]Not Equal Rectangle

取$P=23$，构造$A_{i,j}=(\lfloor\frac{i}{P}\rfloor\lfloor\frac{j}{P}\rfloor+i+j)mod\ P+1$（$i$和$j$均从0开始）

关于正确性，证明如下——

考虑同列的两数，代入得$A_{i_{1},j}=A_{i_{2},j}$当且仅当$(\lfloor\frac{i_{2}}{P}\rfloor-\lfloor\frac{i_{1}}{P}\rfloor)\lfloor\frac{j}{P}\rfloor\equiv i_{2}-i_{1}(mod\ P)$

结合$P^{2}\ge n$，当$\lfloor\frac{i_{1}}{P}\rfloor\equiv \lfloor\frac{i_{2}}{P}\rfloor(mod\ P)$时，简单分析可得$i_{1}=i_{2}$，矛盾

结合$P$是素数，$\lfloor\frac{i_{2}}{P}\rfloor-\lfloor\frac{i_{1}}{P}\rfloor$存在模$P$的逆元，进而条件也即$\lfloor\frac{j}{P}\rfloor\equiv f(i_{1},i_{2})(mod\ P)$

关于限制，即存在$j_{1}$和$j_{2}$均满足上式，进而$\lfloor\frac{j_{1}}{P}\rfloor\equiv \lfloor\frac{j_{2}}{P}\rfloor$，同理可得$j_{1}=j_{2}$，矛盾

综上，即得证

#include<bits/stdc++.h>
using namespace std;
#define P 23
int n,m;
int main(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++)printf("%d ",((i/P)*(j/P)+i+j)%P+1);
        printf("\n");
    }
    return 0;
}