2020牛客多校第六场 C Combination of Physics and Maths

C Combination of Physics and Maths

题目链接：https://ac.nowcoder.com/acm/contest/5671/C

 
题目描述

Roundgod has an $n\times m$ matrix $A=[a_{i,j}]$. One day while she’s doing her physics homework, she wonders is it possible to define the physical quantity for matrices.
As we all know, the pressure $p$ satisfies a formula $p=\frac{F}{S}$, where $F$ is the compressive force and $S$ is the base area.
To describe it in maths, Roundgod puts forward that the compressive force of a matrix equals the sum of all its entries, while the base area of a matrix equals the sum of the entries in its last row. Then she can calculate the pressure for a matrix with the same formula.

Your goal is to find the submatrix of AA with maximum pressure.
A submatrix is obtained by taking nonempty subsets of its rows and columns. Formally, given a nonempty subsequence $S$ of $\{1,2,\dots ,n\}$ and a nonempty subsequence $T$ of $\{1,2,\dots ,m\}$, then

$\left[\begin{array}{cccc}{a}_{S_1,T_1}& a_{S_1,T_2}& \cdots & a_{S_1,T_{|T|}}\\ a_{S_2,T_1}& a_{S_2,T_2}& \cdots & a_{S_2,T_{|T|}}\\ ⋮& ⋮& \ddots & ⋮\\ a_{S_{|S|},T_1}& a_{S_{|S|},T_2}& \cdots & a_{S_{|S|},T_{|T|}}\end{array}\right]$

is a submatrix of $A$.

输入描述:
There are multiple test cases. The first line of input contains an integer $T(T\le 100)T(T\le 100)$, indicating the number of test cases. For each test case:
The first line contains two integers $n,m(1\le n,m\le 200)$, the number of rows and columns of the matrix, respectively.
Each of the next nn lines contains mm integers, specifying the matrix $(1\le a_{i,j}\le 5\cdot 10^4).$

输出描述:
For each test case, print the maximum pressure within an absolute or relative error of no more than $10^{-8}$ in one line.

示例1
输入

1
3 3
1 3 5
6 8 9
2 7 4

输出

4.50000000

说明

$\left[\begin{array}{cc}1& 5\\ 6& 9\\ 2& 4\end{array}\right]$is one of submatrices of $A$ with maximum pressure $\frac{1+5+6+9+2+4}{2+4}=\frac{27}{6}=4.5.$

 

题面

思路

题解：


题意就不再解释

假如我们对于每列都进行这样的计算：从第一行到终止行(任一行)的数之和除以终止行的数，那么我们可以得到以每一列的某些行作为矩阵的最大值，然后如果以多列作为选中的矩阵的话，合并后的值一定不会大于合并前中的一个矩阵的压强值，证明如下：

以两个矩阵来合并，设其中压强值较大的值为$\frac{a}{b}$,较小的为$\frac{c}{d}$，那么合并后的值为$\frac{a+c}{b+d}$。

则 $\frac{a}{b}$ - $\frac{a+c}{b+d}$ = $\frac{a\cdot d-c\cdot b}{b\cdot b+b\cdot d}$   又由$\frac{a}{b}$ > $\frac{c}{d}$ => $a\cdot d>b\cdot c$

故$\frac{a}{b}$ - $\frac{a+c}{b+d}$ $=$$\frac{a\cdot d-c\cdot b}{b\cdot b+b\cdot d}$ $>0$

即$\frac{a}{b}$$>$$\frac{a+c}{b+d}$ （如果$\frac{a}{b}=\frac{c}{d}$,就取等）

即所得到的结论：
$\frac{c}{d}\le \frac{a+c}{b+d}\le \frac{a}{b}$

故最大的结果就是只有一列的情况，找一列情况中最大的就可以了。

代码

#include 

const int N = 205;

int T, n, m, a[N][N];

int main() {
    for (scanf("%d", &T); T; T--) {
        scanf("%d%d", &n, &m);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                scanf("%d", &a[i][j]);
            }
        }
        double ans = 1; //只有一个数时就为1，不可能比这还小
        for (int j = 0; j < m; j++) {
            double sum = 0;
            for (int i = 0; i < n; i++) {
                sum += a[i][j];
                ans = std::max(ans, sum / a[i][j]);
            }
        }
        printf("%.8f\n", ans);
    }
    return 0;
}