微軟在線筆試題：FontSize

時間限制:10000ms
單點時限:1000ms
內存限制: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

直接貼代碼：

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int caseNum = 0;
        while(sc.hasNext()){
            caseNum = sc.nextInt();
            int N,P,W,H;
            int[] size = new int[caseNum];      // 總的字符數
            for(int i=0; i<caseNum; i++){
                N = sc.nextInt();
                P = sc.nextInt();
                W = sc.nextInt();
                H = sc.nextInt();
                int[] pChar = new int[N];   // 每一段的字符數
                for(int j=0; j<N; j++){
                    pChar[j] = sc.nextInt();    // 每一段的字符數
                }
                /////////此處將所有數據已經讀入完畢
                boolean flag = false;
                int s = 1;
                while(!flag){
                    int onePageRow = H / s; // 每頁可以有的行數
                    int onePageCol = W / s; // 每行可以有的列數
                    int allRow = onePageRow * P;    // 總共的行數
                    int[] col = new int[N]; //每一段可以有的行數
                    for(int k=0; k<N; k++){
                        col[k] = (pChar[k] - 1) / onePageCol+1; //每段需要的行數 
                    }
                    int sum = 0;
                    for(int num : col){
                        sum+=num;
                    }
                    if(allRow < sum){
                        flag = true;
                    }else{
                        s++;
                    }
                }
                size[i] = s-1;
            }
            for(int maxSize : size){
                System.out.println(maxSize);
            }
        }
        sc.close();
    }
}

懶得寫了，Game Over