First consider how to determine a r * c matrix eligibility, easy to find the upper left corner of the matrix and can not be hit anything else, requires the upper left corner of the matrix r * c can not exceed the top left corner of the element, after empathy constantly medals for the most & leftmost non-zero points can be used to optimize the differential, complexity is $ o (n ^ {4} ) $ (n and m are the same order)
(plus some divisible then, descending to enumeration pruning can live a)
positive solutions is such that the maximum enumeration 1 * c matrix, enumeration of the largest matrix r * 1, r * c matrix is the answer (the difference with maintenance)
in order to prove the correctness of this, divided consider two parts:
1. the necessity that r * c matrix can launch legal matrix 1 * c and r * 1 legitimate, apparently set up (with a matrix of c times r * 1 / r times c * matrix to 1 )
2. sufficiency, namely the matrix 1 * c and r * 1 can launch legal r * c matrix legitimate, considering the current top-left corner, knocked on end x number of times, indicating that each column in excess of x, then the inevitable each row can cause knock at least x, i.e. a matrix of r * c x times knock
problem to obtain certificates, then the complexity of the final solution was positive $ o (n ^ {3} ) $ (n and m are the same order )
1 #include<bits/stdc++.h> 2 using namespace std; 3 int n,m,s,ans,a[105][105],b[105][105]; 4 bool pd(int r,int c){ 5 memcpy(b,a,sizeof(a)); 6 for(int i=1;i<=n;i++) 7 for(int j=1;j<=m;j++) 8 if (b[i][j]){ 9 if ((i+r-1>n)||(j+c-1>m))return 0; 10 int p=b[i][j]; 11 for(int x=i;x<i+r;x++) 12 for(int y=j;y<j+c;y++) 13 if ((b[x][y]-=p)<0)return 0; 14 } 15 return 1; 16 } 17 int main(){ 18 scanf("%d%d",&n,&m); 19 for(int i=1;i<=n;i++) 20 for(int j=1;j<=m;j++){ 21 scanf("%d",&a[i][j]); 22 s+=a[i][j]; 23 } 24 for(int i=n;i;i--) 25 for(int j=m;j;j--) 26 if ((s%(i*j)==0)&&(i*j>ans)&&(pd(i,j)))ans=i*j; 27 printf("%d",s/ans); 28 }