Dichotomous answer [TJOI2007] signs set

 

It is limited, at the problem is not the solution, a lot of understanding

 

Thank you for watching this konjac

 

Topic: https://www.luogu.org/problem/P3853

 

Because data n <= 10000000; can be derived by analyzing a sample which is one to two sub-themes.

Sample Analysis: The interval period is divided into (k + n) segments, identify (k + n) of the maximum section. Because k uncertain, so not sure how the points, we are to find out the maximum value of this compared with the rest of the period throughout several cases, these take the smallest inside .

This question is one half of the answer to the question of the template, you can just pay attention to details

code:

#include<bits/stdc++.h>
//#pragma GCC optimize(3)
int l,n,k,a[110000],dis[110000],ans=0,kk,mm,yzl;
using namespace std;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch<=' . 9 ' && CH> = ' 0 ' ) {X = (X << . 1 ) + (X << . 3 ) + (CH ^ 48 ); CH = getchar ();}
     return X * F; 
} 
BOOL Check ( int MID) // determines whether to continue bipartite go 
{
     int NUM = 0 ;
     for ( int I = . 1 ; I <n-; I ++ ) 
    { 
            NUM + = DIS [I] / MID;
         // since this problem is always a straight line that is: 0 ~ len mid half continue, eventually several statistical 1/2 
    }
     IF (NUM> K)return  0 ;
     return  . 1 ; 
} 
int main () 
{ 
    int LS, MID, RS; 
    L = Read (), n-= Read (), K = Read ();
     for ( int I = . 1 ; I <= n-; I ++ ) 
    a [I] = Read ();
     for ( int I = . 1 ; I <n-; I ++ ) { 
    DIS [I] = a [I + . 1 ] -a [I] - . 1 ;
     // must be decremented, since the this problem is a spacer, which ensures the start point and the end point are provided signs
     // need only be evenly distributed intermediate portion 
    } 
    LS = 0; RS = L;
     the while (LS <= RS) { 
        MID = LS + RS >> . 1 ;
         IF (Check (MID)) { 
            RS = mid- . 1 ; 
            YZL = MID; 
        } 
        the else { 
            LS = MID + . 1 ; 
        } 
    } 
    the printf ( " % D \ n- " , YZL); // At this mid == ls, may be directly output LS 
    return  0 ; 
} 
/ * 
. 9 2 2 
0. 8 

. 3 
* /