Font Size

一、题目大意

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

二、AC code

我直接暴力解决，但是这题应该是可以计算出来的。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define INF 0x3f3f3f3f

#define PRINT(x) cout<

using namespace std;

template <class Type>
Type stringToNum(const string& str)
{
    istringstream iss(str);
    Type num;
    iss >> num;
    return num;    
}

//======================================================
#define  MAXN 1002

int a[MAXN]; //每个段落字数

inline int upperCnt(int s,int t) {
    return s%t == 0?(s/t):(s/t+1);
}

int main()
{
    //freopen("input.txt","r",stdin);

    int T;
    cin>>T;

    while (T--) {

        int N,P,W,H;
        cin>>N>>P>>W>>H;

        for (int i=0;icin>>a[i];

        //start
        //枚举S
        int S;
        for(S=1;W>=S && H>=S;++S){
            int pageCnt = 0;
            int rowCapacity = H/S; //行容量
            int curRow = 0;
            for(int j=0;j//算出一个段落多少行

                while(curRow>=rowCapacity) {
                    curRow -= rowCapacity;
                    pageCnt++;
                }
            }
            if(pageCnt>P)
                break;
        }

        PRINT(S-1);
    }

    return 0;
}