hiho一下 第148周

题目1 : 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

题目很简单，就是找最大符合条件的p。把题目看懂，枚举似乎也可以过。

这里，我用二分省了一部分时间，也许有更好的优化。。菜鸟我就没有去想了。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
ll T, n, w, h, pp;
ll p[maxn];

bool ok(ll s) {
    ll c_num = w/s, l_num = h/s, cnt = 0;
    if( c_num<=0 || l_num<=0 ) return false;
    for(int i=0; i<n; i++) {
        if( c_num == 1 ) cnt += p[i];
        else cnt += (p[i]+c_num-1)/c_num;
    }
    return ( cnt + l_num - 1 ) / l_num <= pp;
}

int main(){
    cin >> T;
    while( T -- ) {
        cin >> n >> pp >>  w >> h;
        for(int i=0; i<n; i++) cin >> p[i];
        ll l = 0, r = min(w, h), m;
        while( l < r ) {
            m = (l+r)/2;
            if( ok(m) ) l = m+1;
            else  r = m-1;
        }
        while( !ok(l) ) l--;
        cout << l << endl;
    }
    return 0;
}