if((k1/(m-1))%2==1) cout<<n-k1/(m-1)<<" "<<m-k1%(m-1)<<endl;
		else cout<<n-k1/(m-1)<<" "<<k1%(m-1)+2<<endl;

注意到底是往左走还是往右走以及判断条件的奇偶性，本蒟蒻调了1h才搞清楚