微软2016实习生笔试--第一题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

21 10 4 3102 10 4 310 10

样例输出

32

代码如下：

#include <iostream>#include <vector>#include <cstring>using namespace std;int main(){int N;  //保存要处理的任务数while(cin >> N){int *parNum = new int[N];           // N组任务的段落数int *pageNum = new int[N];          // N组任务的页数int *width = new int[N];            // N组任务的屏幕宽度int *height = new int[N];           // N组任务的屏幕高度vector<vector<int> > parWord(N);    // N组任务的每一段落的字数for(int i = 0; i < N; i++){cin >> parNum[i] >> pageNum[i] >> width[i] >> height[i];parWord[i].resize(parNum[i]);for(int j = 0; j < parNum[i]; j++)cin >> parWord[i][j];}int *result = new int[N];           // 保存最终结果memset(result, 0, N*sizeof(int));for(int i = 0; i < N; i++){int sumWord = 0;for(int j = 0; j < parNum[i]; j++)sumWord += parWord[i][j];while((width[i]/(result[i]+1)) * (height[i]/(result[i]+1)) * pageNum[i] >= sumWord)  // 如果当前字的大小为result[i]+1,计算这些页数能存的总字数，result[i]++;                                                                     // 当总字数刚好小于sumWord时，result[i]存放的为正确结果}for(int i = 0; i < N; i++)                                                               // 输出最终答案{cout << result[i] <<endl;}delete[] parNum;delete[] pageNum;delete[] width;delete[] height;delete[] result;}return 0;}