Two-dimensional array-end-to-end concatenated maximum sub-array

The computer broke down and forgot to send it.

Title: Returns the number of the largest sub-arrays in a two-dimensional integer array.

Requirements: Enter a two-dimensional shape array with positive and negative numbers in the array.

A two-dimensional array is connected to the end of the line, like a belt.

One or more consecutive integers in an array make up a sub-array, each of which has a and.

The maximum value for the and of all sub-arrays.

Ideas:

The maximum subarray of two-dimensional array and the one-dimensional array are combined to find the maximum subarray.

Code:

#include <iostream>#include<cstring>#include<assert.h>Const intN = -;Const intINF =-9999;using namespacestd;voidOutputintA[][n],intNintHeadintFootintBeginintLast )//head, foot represents the upper and lower bounds, begin, last represents the left and right bounds {inti,j; cout<<"the sub-arrays are:"<<Endl;  for(I=head; i<=foot; i++)    {        if(Last <begin) {             for(J=begin; j<n; J + +) {cout<<a[i][j]<<" "; }             for(j=0; j<=last; J + +) {cout<<a[i][j]<<" "; } cout<<Endl; }        Else        {             for(J=begin; j<=last; J + +) {cout<<a[i][j]<<" "; } cout<<Endl; }} cout<<Endl;}intMaxsubarray (intA[],intNint&begin,int&Last ) {Assert (a!=null && n>0); intMax =INF; intsum; inti,j,k;  for(i=0; i<n; i++) {sum=0;  for(j=i; j<n; J + +) {sum+=A[j]; if(Sum >max) {Max=sum; Begin=i; Last=J; }        }         for(k=0; k<i; k++)//starting from the first {sum+=A[k]; if(Sum >max) {Max=sum; Begin=i; Last=K; }        }     }        returnMax;}intFindmaxsubmatrix (intA[][n],intN) {intTmpsum[n]; intMax =INF; intbegin, last, Begin1, Last1, head, foot; //enumerate possible combinations of all rows     for(intI=0; i<n; i++)    {        //the tmpsum will be cleared 0memset (Tmpsum,0,sizeof(tmpsum));  for(intJ=i; j<n; J + +)             {              //plus the elements of the current row             for(intk=0; k<n; k++)//each column{Tmpsum[k]+=A[j][k]; }            intTmpmax =Maxsubarray (Tmpsum, N, Begin1, last1); if(Tmpmax >max) {Begin=begin1; Last=Last1; Head=i; Foot=J; Max=Tmpmax;        }}} output (A, n, head, foot, begin, last); returnMax;}intMain () {intA[n][n]; intN//the size of the arraycout<<"Please enter the size of n in the array n*n:"<<Endl;  while(Cin>>n &&N) { for(intI=0; i<n; i++)        {             for(intj=0; j<n; J + +)            {                intk=rand (); A[I][J]= k%2==0? rand ()% -+1:(-rand ()% -+1); }        }         for(intI=0; i<n; i++)        {             for(intj=0; j<n; J + +) {cout<<a[i][j]<<" "; } cout<<Endl; } cout<<Endl; cout<<"the number of the largest sub-arrays is:"<<findmaxsubmatrix (A, N) <<Endl; }        return 0;}

Test:

1.n=4

2.n=5

Sentiment:

The program is still a little cumbersome, you have to define a K to start from the beginning.

Two-dimensional array-end-to-end concatenated maximum sub-array