To find the maximum number of sub-arrays 02

Title: The and of the largest sub-arrays in an integer array

Requirements:

1. Enter an array of shapes with positive and negative numbers in the array.
2. One or more consecutive integers in an array make up a sub-array, each of which has a and.
3. If the array a[0] ... A[j-1] next to each other, allowing a[i-1], ... A[n-1],a[0] ... A[J-1] and the largest.
4. Returns the position of the largest subarray at the same time. The maximum value for the and of all sub-arrays.

One more requirement for this mission is that the array is ring-shaped,

which can be from a[i-1], ... A[n-1],a[0], a[1] Such an array is also wide, then the first thought is the loop linked list, and then control the sum of the sub-array length, but, where the first element of the list is determined ... With DP words termination conditions how to set ...

So I feel that this method may not be very good to go, so, think of since it is a ring array, that sub-array must have requirements, if the array elements all the sum is positive,

So the sum of the sums of the subarray is not +∞, so it can be ...

, the length of the sub-array must be less than or equal to the original array, then a new array of two array arr is bound to contain all the sub-arrays.

And then the table to find all the sub-array of the and, and then Max, this is not extremely easy, hurriedly yards up

1#include <iostream>2#include <vector>3 using namespacestd;4 #defineMax_num 1005 intMainintargcChar*argv[])6 {7     inta[]={ the, the,-565,9465,- the, One, the,9400};8vector<int> Arr (a,a+ (int)(sizeof(a)/sizeof(int)));9 Arr.insert (Arr.end (), Arr.begin (), Arr.end ());Ten     intarrtemp[max_num][max_num]={0};//table arrtemp[i][j] for the sum of the sequence of records is represented as a subarray of the sum of [i,j] elements in arr One      for(inti =0; I < (int) (Arr.size ()/2); i++) A     { -          for(intj =0;j< (int) (Arr.size ()); J + +) -         { the             if((i <= J) && (J < i + (int) (Arr.size ()/2)))//I<=j Control Element order, J <i+length Control sub-series length -             { -                  for(intK = i; K <= J; k++) -                 { +ARRTEMP[I][J] + =Arr[k]; -                 } +             } A         } at     } -  -     intMaxtemp = arrtemp[0][0];//the temporary maximum value is set to Arr[0][0] that is arr[0] -     intStart =0; -     intEnd =0; -      for(inti =0; I < (int) (Arr.size ()/2); i++) in     { -          for(intj =0; J < (int) (Arr.size ()); J + +) to         { +             if((i <= J) && (J < (int) (I+arr.size ()/2)))//I<=j Control Element Order, J<i+length control sub-series length -             { the                 if(Arrtemp[i][j] >maxtemp) *                 { $Maxtemp =Arrtemp[i][j];Panax NotoginsengStart =i; -End =J; the                 } +             } A         } the     } +  -cout <<"sum:"<< maxtemp <<"Start:"<< Start <<"End:"<<end% (arr.size ()/2) <<Endl; $     return 0; $}

(i <= J) && (J < (int) (I+arr.size ()/2) is the sum of the control elements, such as an array with 4 elements, then the contents of the sum table can only be filled to

x o o o o x x x
x x o o o x x
x x x o o o x
Position in O, other locations cannot be the sum of sub-arrays.



also drunk.

Date	Lectures	Programming	Reading	Look at the code	Write a blog	Summarize
Monday	120	120		150		390
Tuesday		60		120		180
Wednesday				90		90
Thursday	120	120		60		300
Friday				120		120
Saturday		60				60
Sunday		240			90	330
Week Summary	240	600	0	540	90	1470

Defect Logging

Date	Number	Type	Introduction phase	Exclusion phase	Repair time	Fixing defects
3/27	1	Stack Overflow	Coding	Debugging	20min

My group members: Liu Wei

To find the maximum number of sub-arrays 02