Gym 100917J---dir，gym100917j---dir

Gym 100917J---dir，gym100917j---dir

題目鏈接

http://codeforces.com/gym/100917/problem/D

problem description

Famous Berland coder and IT manager Linus Gates announced his next proprietary open-source system "Winux 10.04 LTS"

In this system command "dir -C" prints list of all files in the current catalog in multicolumn mode.

Lets define the multicolumn mode for number of lines l. Assume that filenames are already sorted lexicographically.

• We split list of filenames into several continuous blocks such as all blocks except for maybe last one consist of l filenames, and last block consists of no more than l filenames, then blocks are printed as columns.
• Width of each column wi is defined as maximal length of the filename in appropriate block.
• Columns are separated by 1 × l column of spaces.
• So, width of the output is calculated as , i.e. sum of widths of each column plus number of columns minus one.

Example of multi-column output:

a       accd e t
aba     b    f wtrt
abacaba db   k

In the example above width of output is equal to 19.

"dir -C" command selects minimal l, such that width of the output does not exceed width of screen w.

Given information about filename lengths and width of screen, calculate number of lines l printed by "dir -C" command.

First line of the input contains two integers n and w — number of files in the list and width of screen (1 ≤ n ≤ 105, 1 ≤ w ≤ 109).

Second line contains n integers fi — lengths of filenames. i-th of those integers represents length of i-th filename in the lexicographically ordered list (1 ≤ fi ≤ w).

Print one integer — number of lines l, printed by "dir -C" command.

11 20
1 3 7 4 1 2 1 1 1 1 4

3

題意：有n個目錄名字符串，長度為a[1]~a[n] 屏幕寬為w  ，現在要按照已經給的目錄循序一列一列的放，每一列放x個，最後一列放<=x個  要求每一列目錄名左端對整齊 ，形成一個長方形的塊 ，且塊與塊之間空一格，且不能超過屏幕的寬度，求最小的行數；

思路：先對輸入長度處理，用ST算出每個區間的最大值，然後枚舉行數x 從1 ~ n；

代碼如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <map>
#include <cmath>
using namespace std;
typedef long long LL;
const int MAXN = 1e5+5;
int a[MAXN],m[30][MAXN];
int n;
LL w;

int main()
{
    while(scanf("%d%I64d",&n,&w)!=EOF)
    {
         memset(m,0,sizeof(m));
         for(int i=1;i<=n;i++)
         {
             scanf("%d",&a[i]);
             m[0][i]=a[i];
         }
         for(int i=1;i<=(int)log(n)/log(2);i++)
         {
             for(int j=1;j+(1<<i)-1<=n;j++)
                m[i][j]=max(m[i-1][j],m[i-1][j+(1<<(i-1))]);
         }

         for(int i=1;i<=n;i++)
         {
             int k=(int)log(i);
             long long sum=0;
             for(int j=0;j<n/i;j++)
             {
                 sum+=(long long)max(m[k][j*i+1],m[k][i*(j+1)-(1<<k)+1])+1;
             }
             if(n%i!=0) {
                k=(int)log(n%i);
                sum+=(long long)max(m[k][n-n%i+1],m[k][n-(1<<k)+1])+1;
             }
             if(sum-1<=w){
                printf("%d\n",i);
                break;
             }
         }
    }
    return 0;
}