题目1 : Font Size

最新推荐文章于 2022-11-12 22:35:17 发布

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本文介绍了一种算法，用于确定手机阅读时的最佳字体大小，确保在限定页面内完整显示文章，考虑到屏幕尺寸及每段文字长度。

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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

题目就是找到最大的字体，使N段文字能在P页之内放下。
我的思路就是暴力枚举求解，从最大的s开始判断，满足就退出。
因为没参加比赛，所以还没有提交，不知道能否AC，等题目挂出来再提交看看吧。

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        int a[3333],n,p,w,h;
        scanf("%d%d%d%d",&n,&p,&w,&h);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int s=min(w,h);s>=1;s--){
            int tot=0;//计算N段需要多少行
            int l=w/s;//每一行的字符个数
            for(int i=0;i<n;i++){
                    tot+=a[i]/l;
                if(a[i]%l!=0)tot++;
            }
            if(tot<=p*(h/s)){
                printf("%d\n",s);
                break;
            }
        }
    }
    return 0;
}