時間限制:10000ms 單點時限:1000ms 內存限制: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 a_i 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.

輸入

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, ... a_N, indicating the number of characters in each paragraph.

For all test cases,

1 <= N <= 10³,

1 <= W, H, a_i <= 10³,

1 <= P <= 10⁶,

There is always a way to control the number of pages no more than P.

輸出

For each testcase, output a line with an integer Ans, indicating the maximum font size Steven can set.

樣例輸入

2 1 10 4 3 10 2 10 4 3 10 10

樣例輸出

3 2

題目鏈接：http://hihocoder.com/problemset/problem/1288

題目大意：n段文章，每段文章ai個字，手機尺寸是w*h的，現在要放大字體，字體為s時每行可以顯示w/s向下取整個字，每頁可以顯示h/s向下取整個行，求分頁不超過p時字體最大可以為多少。

題目分析：數據很小，直接枚舉，算出每次一共要多少行，然后再算要多少頁。

#include <cstdio> #include <algorithm> #define ll long long using namespace std; int const MAX = 1e3 + 5; ll a[MAX];int main() {int T;scanf("%d", &T);while(T --){ll n, p, w, h;ll sum = 0;scanf("%lld %lld %lld %lld", &n, &p, &w, &h);for(int i = 0; i < n; i++)scanf("%lld", &a[i]);ll mi = min(w, h);ll ans = 0, tot, perline, line, perpg, pg;for(ll i = mi; i >= 1; i--){tot = 0;perline = w / i;for(int j = 0; j < n; j++){line = a[j] / perline + (a[j] % perline != 0);tot += line;}perpg = h / i;pg = tot / perpg + (tot % perpg != 0);if(pg <= p){ans = i;break;}}printf("%lld\n", ans);} }

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