Font Size
Time Limit: 1 Sec Memory Limit: 256 MB
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.

传送门:HIHO1288

Input

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. 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.

Output

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

Sample Input

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

Sample Output

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2

题解

字体大小不会超过w和h的最小值,一一枚举找最大的size就好了。
## AC code:(不包含输入类)
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import java.io.*;  
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
int o=sc.nextInt();
for(int k=1;k<=o;k++){
int n=sc.nextInt();
int p=sc.nextInt();
int w=sc.nextInt();
int h=sc.nextInt();
int[]para=new int[n];
for(int i=0;i<n;i++){
para[i]=sc.nextInt();
}
int result=0;
int size;
for(size=1;size<=Math.min(w, h);size++){
int line=0;
for(int i=0;i<n;i++){
int gs=w/size;
if(para[i]%gs==0){
line+=para[i]/gs;
}
else
line=line+para[i]/gs+1;
}
int curP;
int hs=h/size;
if(line%hs==0){
curP=line/hs;
}
else
curP=line/hs+1;
result=curP;
if(result>p){
break;
}
}
System.out.println(size-1);
}
}
}
}