题目大意:有一个$n\times m$的方格图,求其中所有的格点正方形完整包含的小方格个数,多组询问.$n,m\leqslant 10^6$ 题解:令$n\leqslant m$.有一个显然的式子:$$ans=\sum\limits_{i=1}^n(n-i+1)(m-i+1)f(i)$$$f(i)$表示可以完整包含在$i\times i$的正方形中且顶点在这个正方形边上的正方形所包含的小方格总数.可以分选的正方形和$i\times i$正方形重合和边转动$j$格来计算$$f(n)=n^2+\su…

分析 https://www.cnblogs.com/Memory-of-winter/p/11209128.html 代码 #include<bits/stdc++.h> using namespace std; #define int long long ; ; ],],a[N+],b[N+],c[N+],n,m,t; signed main(){ ,y=; ;i<=N;i++)phi[i]=i; ;i<=N;i++) if(!is[i])for(j=i;j<=N;j+=…

A .Jzzhu and Chocolate pro:现在给定一个大小为N*M的巧克力,让你横着或者竖着切K刀,都是切的整数大小,而且不能切在相同的地方,求最大化其中最小的块. (N,M,K<1e9) sol:如果横着切X刀,竖着切Y刀,那么最小的面积=(N/(X+1))*(M/(Y+1)):一看这个公式知道是整数分块了,相同的部分合并. 需要保证X<N,Y<M: #include<bits/stdc++.h> #define ll long long using names…

主题链接:http://codeforces.com/problemset/problem/449/A ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943…

C. Jzzhu and Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k time…

Jzzhu and Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times.…

C. Jzzhu and Chocolate time limit per test: 1 seconds memory limit per test: 256 megabytes input: standard input output: standard output Jzzhu has a big rectangular chocolate bar that consists of \(n × m\) unit squares. He wants to cut this bar exact…

Given a set of words (without duplicates), find all word squares you can build from them. A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤ k < max(numRows, numColumns). For example, the wor…

本文介绍一种利用移动最小二乘法来实现图像变形的方法,该方法由用户指定图像中的控制点,并通过拖拽控制点来驱动图像变形.假设p为原图像中控制点的位置,q为拖拽后控制点的位置,我们利用移动最小二乘法来为原图像上的每个像素点v构建相应的仿射变换lv(x),并通过该变换来计算得到图像变形后的位置: 其中权重wi的表达式为wi = 1/|pi - v|2α. 仿射变换lv(x)由两部分组成lv(x) = xM + T,其中M为线性转换矩阵,T为平移量.事实上将最小化表达式对变量T求偏导后可以得到T的表达式T…

Given a set of words (without duplicates), find all word squares you can build from them. A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤ k < max(numRows, numColumns). For example, the wor…