Pair Development 6----end-to-end matrix

Pair member: Van der Zhao Yonghen

I. Topics and Requirements

Title, returns the and of the largest subarray in a two-dimensional integer array

requirements, 1, input a two-dimensional shaping array, there are positive numbers in the array and negative numbers.

2, two-dimensional array of the end-to-end phase, like a first-and-last tape.

3, a contiguous array of one or more integers constitute a sub-array, each sub-array has a and

Two. Design ideas

On the basis of the last thought of the loop array, and the two-dimensional array to find the maximum number of sub-arrays combined, the function is merged to complete the topic requirements.

• The first step, the array of each row as the loop array to find the largest sub-array
• The second step, by enumerating each case and depositing it into a new two-dimensional array
• The third, by the method of calculating the maximum sub-array by column, the largest and most of all sub-matrices

Three. Source code

#include "stdafx.h" #include "iostream" #include <vector> using namespace std; const int N = 101;    int a[n][n], p[n][n],b[n][n];         int maxrecsum (int n) {for (int i = 0; I <= N; ++i) {p[i][0] = 0;     P[0][i] = 0; } for (int i = 1, i <= N; ++i) {for (int j = 1; j <= N; ++j) p[i][j] = p[i-1][j] + P I     [J-1]-p[i-1][j-1] +a[i][j];     } int max = Int_min;             for (int i = 1, i <= N; ++i) {for (int j = i; j <= N; ++j) {int sum = 0; for (int k = 1; k <= N; ++k) {int temp = P[j][k]-p[j][k-1]-p[i-1][k] + p[i-1][k-1]                 ;                 if (sum > 0) sum + = temp;                 else sum = temp;             if (Sum > max) max = sum; }}}} return max;     } int main () {int n = 3;     int num; cout<< "Matrix with a size of three rows and three columns, enter a value:" <<endl;             for (int i = 1, i <= N; ++i) {for (int j = 1; j <= N; ++j) {cin >> num;         A[I][J] = num;   }} int b[3][6];   for (int t=0;t<3;t++) {b[t][0]=a[t][0];   B[T][1]=A[T][1];   B[T][2]=A[T][2];   B[T][3]=B[T][0];   B[T][4]=B[T][1];   B[T][5]=B[T][2];          } cout << "Maximum matrix of the and is:" << maxrecsum (n) << Endl; for (int i = 1, i <= N; ++i) {for (int j = 1; j <= N; ++j) {cout &LT;&LT;A[I][J]&L        t;< "";    } cout<<endl; } return 0; }

Four. Running

Five. Experimental impressions

The implementation of this experiment, time is a little tight, so there are many shortcomings in the function, the optimization of the program is enough to improve, I believe that there is more time, we can do better.

Six. Pair member group photo

Pair Development 6----end-to-end matrix