Introduction to Algorithms learning-sub-arrays Max and problem

1#include <iostream>2#include <algorithm>3 using namespacestd;4 Const intMaxsize=101;5 intA[maxsize],sum[maxsize],n,inf= (1<< -);6 voidsolve () {7     if(n==0)return;8     intans=-inf;9sum[1]=a[1];Ten     intleft=1, right=1; One      for(intI=2; i<=n;i++){ A         if(sum[i-1]>0){ -sum[i]=sum[i-1]+A[i]; -right++;  the         } -         Else{ -sum[i]=A[i]; -left=i; +         } -         if(Sum[i]>ans) ans=Sum[i]; +     } Aprintf"Subarray from Indice%d->%d have the maximum sum:%d\n", Left,right,ans);  at } - intMain () { -      while(SCANF ("%d", &n)! =EOF) { -          for(intI=1; i<=n;i++) scanf ("%d",&a[i]); - solve (); -         /* in Maxisum-subarray problem, in fact, was try to find Subarray A[I...J] - of array a[1..i...j. N] (1<=i<=j<=n). and the problem is meaingful to Only if array A has both the positve value and negative elements. +              - considering the size of problem: the * If n==1, the problem is quite easy. A[1] is the right answer.  * * If array A has more than one value and then we can use DP method to $ solve it.Panax Notoginseng              - Considering the procedure have already checked the a[i-1], and is the going to enter the i th check. And we can assume that we have + already get the maximum subarray of the a[1...i-1].supposing the A maxisum Subarray of A[1..i-1] is a[p. I-1]. the              + we have aleady get procedure are going to the ith check - the solution of a[1..i], and we'll move to check a[i]. $ relationship between Sum[i-1] and Sum[i]? From our experience, $               - * If sum[i-1]<0, then whatever the A[i) is, the sum of Subarray - A[p...i-1,i] must is smaller than that of Subarray a[p. I-1]. So we the Modify the starting indice of Maxisum from ' P ' to ' I '. and at the - same time, we can know that the maxisum subarray of a[1...i] are notWuyi The a[p...i] but the a[p...i-1]. So the value of Maxisum remains the the same:the sum of a[p...i]. - * For another circumstance, if sum[i-1]>0, we can take both situations Wu int consideration, if a[i]>0, obviously, we should let - Sum[i]=sum[i-1]+a[i]. Then for a[i]<0? In fact, we also need to plus About A[i] onto sum[i-1]. As we know, in the (i-1) th calculation, acoording $ to the above assumption, we had already found the solution to the - Maxisum Subarray:a[p...i-1]. So although the sum of a[p...i-1] is - higher than that's a[p...i], in fact, with each iteration, we compare - The value of sum[i-1] and Sum[i] to the Maxinum and store the maxinum. A So we let the Sum[i-1]+a[i] have no matter to the maxinum. +              the (Sum[i] stores the max sum of Subarray end in A[i]).  -         */ $     } the     return 0; the}

Introduction to Algorithms learning-sub-arrays Max and problem