微软笔试题1:Font Size

描述

Steven loves reading book on his phone. The book he reads now consists of N paragraphs and the i-th paragraph contains ai characters.

Steven wants to make the characters easier to read, so he decides to increase the font size of characters. But the size of Steven's phone screen is limited. Its width is W and height is H. As a result, if the font size of characters is S then it can only show ⌊W / S⌋ characters in a line and ⌊H / S⌋ lines in a page. (⌊x⌋ is the largest integer no more than x)  

So here's the question, if Steven wants to control the number of pages no more than P, what's the maximum font size he can set? Note that paragraphs must start in a new line and there is no empty line between paragraphs.

输入

Input may contain multiple test cases.

The first line is an integer TASKS, representing the number of test cases.

For each test case, the first line contains four integers N, P, W and H, as described above.

The second line contains N integers a1, a2, ... aN, indicating the number of characters in each paragraph.

For all test cases,

1 <= N <= 103,

1 <= W, H, ai <= 103,

1 <= P <= 106,

There is always a way to control the number of pages no more than P.

输出

For each testcase, output a line with an integer Ans, indicating the maximum font size Steven can set.

思路:本题应该算本次笔试的唯一的一个水题吧！就是判断一共有多少行（h/j）*p,然后按照宽度算，需要的行数比不比这个多，注意本题的一个条件，页数不能超过p，根据此思想，可以很容易得到下面的代码

代码如下，笔试的时候是AC的

#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<time.h>
#include<math.h>
#include<memory>
#include<vector>
#include<bitset>
#include<fstream>
#include<stdio.h>
#include<utility>
#include<sstream>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int a[1005];
int main()
{
    #ifdef absi2011
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    #endif
    int zu;
    scanf("%d",&zu);
    int t;
    for (t=0;t<zu;t++)
    {
        int n,p,w,h;
        scanf("%d%d%d%d",&n,&p,&w,&h);
        int i;
        for (i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        int j;
        int ans=0;
        for (j=1;j<=min(w,h);j++)
        {
            int k=h/j*p;
            int sum=0;
            for (i=0;i<n;i++)
            {
                sum+=a[i]/(w/j);
                if (a[i]%(w/j)!=0) sum++;
            }
            if (sum<=k) ans=j;
        }
        printf("%d\n",ans);
    }
    return 0;
}