題目1 : Font Size

時間限制:10000ms
單點時限:1000ms
記憶體限制:256MB

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.
輸入
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.
輸出
For each testcase, output a line with an integer Ans, indicating the maximum font size Steven can set.
樣例輸入
2
1 10 4 3
10
2 10 4 3
10 10
樣例輸出
3
2

以下是我寫的程式碼：

#include<iostream>
#include<math.h>
#include<iomanip>
using namespace std;

int main(){
    int ta;cin>>ta;
    for(int i=0;i<ta;i++){
        int n,p,w,h;    
        
        cin>>n>>p>>w>>h;
        int *a=new int[n+1];//new
        for(int j=0;j<n;j++){
            cin>>a[j];
        }
        a[n]=` `;
        int s;
        int wh= w>h ? h:w;
        //cout<<wh<<"zuixiaode "<<endl;
        s=wh;//chushihua為最大
        int *b=new int[n+1];//每個段落需要的行數
        while(1){
            int wn=w/s;
            int hn=h/s;
            int B=0;//所需總行數
            for(int j=0;j<n;j++){
                double t=ceil(double(a[j])/double(wn));
                b[j]=(int)t;
                B=B+b[j];
            }
            b[n]=` `;

            double te=ceil(double(B)/double(hn));
            int P=(int)te;//所需總頁數
            if(P>p)
                s--;
            else
                break;
        }
        delete []b;
        delete []a;
        cout<<s<<endl;
        

    }
    return 0;
}