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题目链接
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.
, 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
题意：有n个目录名字符串，长度为a[1]~a[n] 屏幕宽为w ，现在要按照已经给的目录循序一列一列的放，每一列放x个，最后一列放<=x个 要求每一列目录名左端对整齐 ，形成一个长方形的块 ，且块与块之间空一格，且不能超过屏幕的宽度，求最小的行数；
思路：先对输入长度处理，用ST算出每个区间的最大值，然后枚举行数x 从1 ~ n；
代码如下：
#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; }
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原文地址：http://www.cnblogs.com/chen9510/p/5926174.html
