微软在线笔试题：FontSize

最新推荐文章于 2017-09-19 20:57:43 发布

原创最新推荐文章于 2017-09-19 20:57:43 发布 · 959 阅读

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本文介绍了一种算法，用于计算在限定屏幕尺寸和段落数量的情况下，如何确定书籍显示的最大字体大小，使得总页数不超过指定数值。该算法通过逐步增加字体大小并检查是否满足页面限制条件来找到最优解。

时间限制:10000ms
单点时限:1000ms
内存限制: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.

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

样例输出

3
2

直接贴代码：

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];      // 总的字符数
            for(int i=0; i<caseNum; i++){
                N = sc.nextInt();
                P = sc.nextInt();
                W = sc.nextInt();
                H = sc.nextInt();
                int[] pChar = new int[N];   // 每一段的字符数
                for(int j=0; j<N; j++){
                    pChar[j] = sc.nextInt();    // 每一段的字符数
                }
                /////////此处将所有数据已经读入完毕
                boolean flag = false;
                int s = 1;
                while(!flag){
                    int onePageRow = H / s; // 每页可以有的行数
                    int onePageCol = W / s; // 每行可以有的列数
                    int allRow = onePageRow * P;    // 总共的行数
                    int[] col = new int[N]; //每一段可以有的行数
                    for(int k=0; k<N; k++){
                        col[k] = (pChar[k] - 1) / onePageCol+1; //每段需要的行数 
                    }
                    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();
    }
}

懒得写了，Game Over