Hihocoder 1288

微软2016校园招聘4月在线笔试

水题

题目说明

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
2
1 10 4 3
10
2 10 4 3
10 10
样例输出
3
2

思路

主要是自己太久没有打算法题～ 搞得一个水题都要做很久。
从今天起没事做一下hihocoder的算法题练练手。

这个题很简单，先设定字体大小为宽和高中小的那一个，然后向下遍历求解即可。
注意每一段要换行。（不能用ceil）正解是 段的字数 + 每一行的字数 - 1 然后去除以每一行的字数

直接贴代码

#include 
using namespace std;

int main() {
    int num;
    int arr[1000];
    cin >> num;
    while (num--) {
        int n, p, w, h;
        cin >> n >> p >> w >> h;
        for (int i = 0; i < n; i++) cin >> arr[i];
        int size = min(w, h);
        int row;
        while (size >= 1) {
            row = 0;
            int r_word_num = (w / size);
            int c_word_num = (h / size);
            for (int i = 0; i < n; i++) {
                row += (arr[i] + r_word_num - 1) / r_word_num;
            }
            if (row <= p * c_word_num) {
                cout << size << endl;
                break;
            }
            else size--;
        }
    }
    return 0;
}