hiho #1288 微软2016.4校招笔试题 Font Size

#1288 : Font Size

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

解题思路：
字体的最大值是min（H,W），依次枚举出所有的可能，算出每段字需要要多少行，让后算出总共需要多少行，再计算需要多少页。


AC代码

#include "iostream"
#include "algorithm"
#define MAX 1000

using namespace std;

int main(){

    int a[MAX];
    int T;
    
    cin >> T;
    while (T--){
        int N, P, W, H;    //总共N段，不超过P页，WH长宽，T组数据
        
        cin >> N >> P >> W >> H;
        
        for (int i = 0; i < N; i++){
            cin >> a[i];
        }

        int ans = min(H, W);
        int perline, allline, perpg,lineper,page;  //perpg每页行数，perline 每段行数，lineper每行字数

        for (int i = ans; ans >= 1; i--)
        {
            allline = 0;
            lineper = W / i;            //每行字数
            for (int j = 0; j < N; j++){
                perline = a[j] / lineper;
                if (a[j] % lineper!=0)
                    perline ++;
                allline += perline;
            }
            perpg = H / i;        //perpg每页行数

            page = allline / perpg;        //当前字体大小时需要的页数

            if (allline % perpg != 0)
                page++;

            if (page <= P){
                cout << i << endl;
                break;
            }
        }
    }
    return 0;
}

 

终于AC了，之前怎么都过不了，显示RE，返回去又把题目看了一遍，原来他MAX是10的3方，我设的100，数据集限定小了，所以一直RE