Depth study of the largest sub-columns and the

For maximum child column and problems, we should be very familiar. With DP algorithm can very quickly and easily find the problem SUM, but for the beginning and end of a column, or a problem worthy of serious consideration.

   A not required at the beginning, ending sequence, the situation only requires values

To test points on the PTA as an example:

   In fact, the difficulty is that the number of columns negative, 0 mixed with three conditions are all negative 0, all for.

  If all 0, the output is 000;

  When the whole is negative, then the output is 0 a [0] a [n -1]

  If negative mix 0, then the output is 000.

  So it is natural to think of, so that the two variables results left = right = 0, then set a flag when detecting the full negative, left = a [0], right = a [n-1].

  Then the code is easy to write out:

#include<stdio.h>
    int f4(int a[],int n){
    int i,max=0,now=0,l=0,r=0,flag=0;
    a[10001]=0;a[10002]=0;
       
    for(i= 0;i<n;i++){
        if(a[i]>=0)
            flag=1;

        if(now<0){
            l=i;
            r=i;        
            now=a[i];   //刚吸收的数可能非常大，所以必须更新l与r
        }
        else {
            r=i;
            now+=a[i];
        }

        if(now>max){
             max=now;
             a[10002]=a[r];
             a[10001]=a[l];
        }

    }
        if(!flag){
            a[10001]=a[0];
            a[10002]=a[n-1];
           }

        return max;
}
      int main(){
      int i,n;
      scanf("%d",&n);
      int a[10002];
      for(i=0;i<n;i++)
      scanf("%d",&a[i]);

      int sum=f4(a,n);
      printf("%d %d %d",sum,a[10001],a[10002]);

    return 0;
}