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.

样例输入

2
1 10 4 3
10
2 10 4 3
10 10

样例输出

3
2

模拟题

题意：有n段文字可以调节字体，调节字体后每行所能显示的字会变少，屏幕的宽度为w长度为l，和字体的关系为一行能显示的字数为[w/s],一页能显示的行数为[l/s];s为字号，要求s最大且在p页能显示完，没啥好说，水题，模拟

#include<stdio.h>
#include<string.h>
#include<math.h>
#define INF 0x3fffffff
#include<iostream>
using namespace std;
int main()
{
    int N;
    scanf("%d",&N);
    while(N--)
    {
        int n,p,w,h;
        scanf("%d %d %d %d",&n,&p,&w,&h);
        int a[1010];
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        int l=0;
        for(int i=1;i<=min(w,h);i++)
        {
            int wi=w/i;///一行存几个字
            int hi=h/i;///一页存多少行
            int ans=0;
            for(int j=0;j<n;j++)
            {
                if(a[j]%wi==0)
                    ans+=a[j]/wi;
                else
                    ans+=(a[j]/wi+1);
            }
            if(ans>p*hi)
            {
                l=1;
                printf("%d\n",i-1);
                break;
            }
        }
        if(l==0)
            printf("%d\n",min(w,h));
    }
}