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原题链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1424 逐渐找到做这种题的感觉了. 二分法.g[i][j]存储坐标(i, j)的值,s[i][j]存储的值为左上角为起始点(1,1),右下角为(i, j)的矩形区域内所有值的和,那么: s[i][j] = g[i][j] + s[i-1][j] + s[i][j-1] - s[i-1][j-1] 扫描整个矩形,遇到为“1”的点就将其作为起点,开始二分边长,利用数组s在O(1)的时间复杂度…
To master any programming languages, you need to definitely solve/practice the below-listed problems. These problems range from easy to advanced difficulty level. I have collected these questions from various websites. For solutions refer this - http…
对抗网络是14年Goodfellow Ian在论文Generative Adversarial Nets中提出来的. 原理方面,对抗网络可以简单归纳为一个生成器(generator)和一个判断器(discriminator)之间博弈的过程.整个网络训练的过程中, 两个模块的分工 判断网络,直观来看就是一个简单的神经网络结构,输入就是一副图像,输出就是一个概率值,用于判断真假使用(概率值大于0.5那就是真,小于0.5那就是假) 生成网络,同样也可以看成是一个神经网络模型,输入是一组随机数Z,输出是…
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E - Maximum Square Time Limit:4500MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVALive 4867 Description Given an NxM matrix of all 1s and 0s, find the largest submatrix which is a square containing all 1s. Input Ther…
A. Maximum Square Ujan decided to make a new wooden roof for the house. He has…
题目如下: Given a m x n matrix mat and an integer threshold. Return the maximum side-length of a square with a sum less than or equal to threshold or return 0 if there is no such square. Example 1: Input: mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4…
题目链接 Given a m x n matrix mat and an integer threshold. Return the maximum side-length of a square with a sum less than or equal to threshold or return 0 if there is no such square. Example 1: Input: mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,…
Ujan decided to make a new wooden roof for the house. He has nn rectangular planks numbered from 11 to nn. The ii-th plank has size ai×1ai×1 (that is, the width is 11 and the height is aiai). Now, Ujan wants to make a square roof. He will first choos…
题目 我是按照边进行二分的 class Solution { public: int sum[100005]; int a[305][305]; int maxSideLength(vector<vector<int>>& mat, int threshold) { if(threshold==0) return 0; int n = mat.size(); int m = mat[0].size(); int len = min(n,m); for(int i=0;i&l…