hdu-4561 Continuous maximum product (water problem)

http://acm.hdu.edu.cn/showproblem.php?pid=4561

The maximum continuous product.

How the hell did the dog push the train of thought?

The idea is%&* () * (& )) *) *&) ...%......%** (* () () ()--+ (* * (... &*......%&* ... y%^&%%^*&& (&* (&* (*& () *& (**......&%......&

%&......%&......& (&......* ()) & (**&......%*&&* (--+--) & (**......&%......&......*&%￥s%^& $**&%%&^a *&^&*** (^u (*) () ^ (*&^&* () &^**&*$%YT^&^%& ^^ &%%^%&%^ &%&^

%&^& (^&& (*&_) (&^*^u&*&op) _ (+_{() *&*&%^&%$&^$^%$#%^&*^& )(**)_(_+................................................

You know ...

I will translate;

In fact, the idea is to find a breakpoint, breakpoint is 0, that is, the demarcation of 0, the number of the given string into the K-segment such as 2220-2-2-22022;

Can be divided into 222-2-2-22 223, because if the selection of 0 words is definitely 0, the title of the product is less than or equal to 0 results are 0, so with 0 demarcation.

So that is the number of 2 per paragraph, and 2 of the number, if the number of 2 in the paragraph is an even number, then the length of this paragraph is directly compared with Maxx, Update Maxx,

If it is an odd number, suppose that n is looped from the left side of the paragraph to the edge until it encounters the nth-2 jump, so the front is not an even number-2.

Then from the right end of the loop to the left side, until the nth-2 jump, compared to two times the size of the large is the continuous maximum product of this section.

Because to be continuous and only one more-2, so say maximum either left continuous, or right continuous.

The last Maxx is the biggest;

1#include <stdio.h>2#include <iostream>3#include <stdlib.h>4#include <string.h>5#include <math.h>6typedefstructpp7 {8     intx;9     inty;Ten     intX1; One     intY1; A } SS; - using namespacestd; - intMainvoid) the { -     inta[10005]; -SS cou[10005]; -     intn,i,j,k,p,q; +scanf"%d",&n); -      for(i=1; i<=n; i++) +     { Ascanf"%d",&k); at          for(j=0; j<k; J + +) -         { -scanf"%d",&a[j]); -cou[j].x=0; -cou[j].y=0; -cou[j].x1=0; incou[j].y1=0; -         } to         if(a[0]>0) +         { -cou[0].x++; the         } *         Else if(a[0]<0) $         {Panax Notoginsengcou[0].y++; -         } the         intmaxx=0; +          for(j=1; j<k; J + +)//2-2 More from breakpoint to this point (from left loop) A         { the             if(a[j]>0) +             { -cou[j].x=cou[j-1].x+1; $cou[j].y=cou[j-1].y; $             } -             Else if(a[j]<0) -             { thecou[j].y=cou[j-1].y+1; -cou[j].x=cou[j-1].x;Wuyi             } the  -         } Wu         if(a[k-1]>0) -         { Aboutcou[k-1].x1++; $         } -         Else if(a[k-1]<0) -         { -cou[k-1].y1++; A         } +          for(j=k-2; j>=0; j--)//2-2 More from breakpoint to this point (right loop) the         { -             if(a[j]>0) $             { thecou[j].x1=cou[j+1].x1+1; thecou[j].y1=cou[j+1].y1; the             } the             Else if(a[j]<0) -             { incou[j].y1=cou[j+1].y1+1; thecou[j].x1=cou[j+1].x1; the             } About  the         } the          for(j=0; j<k; J + +) the         { +             if(cou[j].y%2==0) -             { the                 if(maxx<cou[j].x+cou[j].y)Bayi                 { themaxx=cou[j].x+cou[j].y; the                 } -             } -             if(cou[j].y1%2==0) the             { the                 if(maxx<cou[j].x1+cou[j].y1) the                 { themaxx=cou[j].x1+cou[j].y1; -                 } the             } the         } theprintf"Case #%d:%d\n", I,maxx);94     } the     return 0; the}

hdu-4561 Continuous maximum product (water problem)