Intermediate.
Put your skills to the test.

Problem summary

A side-adjacent top-left to bottom-right path visits exactly N+M−1 cells. Decide whether all such paths have equal sums.

Constraints

N,M≥1; NM≤30000; entries −33000…33000.

Worked example

Input and output for the worked example discussed below.

2 3
1 2 3
2 3 4
3 3
2 1 3
6 1 3
4 4 5
0 0

YN

Solution analysis

1. A path of the specified minimum length uses only right and down moves. Compare right-down and down-right alternatives around any 2×2 square: their shared endpoints cancel from the sum.
2. The alternatives agree exactly when the upper-right and lower-left entries are equal. Check this local condition for every 2×2 square. When all checks pass, every anti-diagonal is constant.
3. Every shortest path visits one cell from each anti-diagonal, so those local equalities suffice. The first sample table has anti-diagonal values 1,2,3,4 and qualifies; the second has upper-right 1 unequal to lower-left 6 and fails.

Why it works

Any local square can be embedded in a shortest prefix and suffix, so equal global sums force the middle alternatives to match. Conversely, the local equalities make every anti-diagonal constant, and every shortest path takes one value from each. This proves equivalence.

Watch out. Only minimum-length paths count. A one-row or one-column table always qualifies. Compare upper-right with lower-left, not opposite corners.

An independently written complete program with standard headers.

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main(){
    int rows;
    int columns;
    string answers;
    while(cin>>rows>>columns&&(rows!=0||columns!=0)){
        vector<vector<int>>grid(rows,vector<int>(columns));
        for(auto&row:grid){
            for(int&value:row){
                cin>>value;
            }
        }
        bool valid=true;
        for(int r=1;r<rows;r++){
            for(int c=1;c<columns;c++){
                // Swapping right-down with down-right must preserve the sum.
                if(grid[r-1][c]!=grid[r][c-1]){
                    valid=false;
                }
            }
        }
        answers+=valid?'Y':'N';
    }
    cout<<answers<<'\n';
}