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lightOJ 1047   Neighbor House (DP) 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730#problem/C 题目: Description The people of Mohammadpur have decided to paint each of their houses red, green, or blue. They've also decided that no two n…
Neighbor House LightOJ - 1047 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int T,TT,n; ][],ans[][]; int main() { int i,j,k; scanf("%d",&T); ;TT<=T;TT++) { scanf("%d",&n); ;i&l…
Description In this problem, you are given an integer number s. You can transform any integer number A to another integer number B by adding x to A. This x is an integer number which is a prime factor of A (please note that 1 and A are not being cons…
C - C Time Limit:500MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu   Description The people of Mohammadpur have decided to paint each of their houses red, green, or blue. They've also decided that no two neighboring houses will be painte…
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1047 Integer Inquiry Description One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of t…
当我尝试在java中通过ProcessBuilder运行window的cmd命令时出现错误: public static void main(String [] args) throws IOException { ProcessBuilder builder = new ProcessBuilder(); Process process = builder.command("dir d:\\").start(); InputStream inputStream = process.g…
传送门:http://www.lightoj.com/volume_showproblem.php?problem=1030 Discovering Gold Time Limit: 2 second(s) Memory Limit: 32 MB Program Description You are in a cave, a long cave! The cave can be represented by a 1 x N grid. Each cell of the cave can con…
上一次接触 project euler 还是2011年的事情,做了前三道题,后来被第四题卡住了,前面几题的代码也没有保留下来. 今天试着暴力破解了一下,代码如下: (我大概是第 172,719 个解出这道题的人) program 4 A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.…
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题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1047 题意:见中文题面 思路:该题是求二维的子矩阵的最大值与最小值的差值尽量小.所以可以考虑求出每个子矩阵的最大值和最小值.考虑一维求子段的最小值/最大值的思路.滑动窗口+单调队列. 转换成二维.设minNum[i][j]表示右下角为(i,j)的子矩阵的最小值.先对矩阵每一行用一维的做法求出每一行的子段的最小值,然后同样的方法求列的最值.注意在求列的子段最小值时比较的元素不是原矩阵的元素…