HDU 3486 RMQ + 二分

算法：

1.RMQ 求区间最值

2.二分枚举段的数量，注意是段的数量和其和成正比，也就是枚举段的数量。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<vector>
#include<string>
#include<math.h>
#include<map>
#include<set>
#include<algorithm>
using namespace std;

int v[200010];
int f[200010][21];
int n,k;

void pre( )
{
  memset(f,0,sizeof(f));
  for( int i = 1; i <= n; i++)
     f[i][0] = v[i];
  int x = (int) log2(n);
  for( int j = 1; j <= x; j++)
  {
      for( int i = 1; i <= n + 1-(1<<j); i++)
      {
         f[i][j] = max(f[i][j-1],f[i + (1<<(j-1))][j-1]);   
             
      }     
       
  }      
     
}

int query( int L, int R )
{
  int x = (int) log2(R - L + 1);
  return max(f[L][x],f[R+1-(1<<x)][x]);
}

bool jugde( int x )
{ 
   int sum = 0;
   int t = n / x;
   for( int i = 1, k = 1; i <= x; i++, k += t)
   {  
       sum += query( k, k + t - 1);   
       //printf("a = %d b = %d %d\n",i, i + x - 1, sum);
   }
   if( sum > k )
     return true;  
   return false; 
}

//分的段数跟和成正比。 
int find( int l, int r)
{
   int ans = -1;
   while( l <= r )
   {
      int mid = (l + r) / 2;
      if( jugde(mid) )    
      {
        ans = mid;  
        r = mid - 1;
      }
      else
      {
        l = mid + 1; 
      }
          
   }    
   return ans; 
}


int main( )
{
  int a,b;
  while( scanf("%d%d",&n,&k) != EOF )
  {
    if( n == -1 &&  k == -1 )
        break;
    for( int i = 1; i <= n; i++)
     scanf("%d",&v[i]);
    pre( );       
    printf("%d\n",find(1,n));
  } 

}