摘要： 题意：在N*N的图中，找出孤立存在的十字架的个数。十字架要求为正十字，孤立表示组成十字架的‘#的周围的一格再无’#‘。dfs找出在中心的‘#’（周围四格也为‘#'），则缩小了搜索范围，再bfs找出是否是符合要求。#include #include #include #include #include using namespace std;char map[55][55];int n,cnt,head,tail,vis[55][55],center[55][55];int dirx[4] = {1,-1,0,0};int diry[4] = {0,0,1,-1};struct Queu 阅读全文

摘要： import java.beans.BeanInfo; import java.beans.IntrospectionException; import java.beans.Introspector; import java.beans.PropertyDescriptor; import java.lang.reflect.InvocationTargetException; import java.lang.reflect.Method; import java.util.Map;import org.apache.commons.beanutils.BeanUtils; import 阅读全文

摘要： 问题：求1~r中有多少个数与n互素。 对于这个问题由容斥原理，我们有3种写法，其实效率差不多。分别是：dfs，队列数组，位运算。 先说说位运算吧：用二进制1,0来表示第几个素因子是否被用到,如m=3，三个因子是2,3,5，则i=3时二进制是011，表示第2、3个因子被用到 LL Solve(LL n,LL r){ vector p; for(LL i=2; i*i1) p.push_back(n); LL ans=0; for(LL msk=1; msk1) p.push_back(n);}void dfs(LL k,LL t,LL s,LL... 阅读全文

摘要： Problem Description The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-to 阅读全文

摘要： OverflowWrite a program that reads an expression consisting of twonon-negative integer and an operator. Determine if either integer orthe result of the expression is too large to be represented as a``normal'' signed integer (typeinteger if you are workingPascal, type int if you are working i 阅读全文

摘要： Problem Description Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M #include#includeusing namespace std;typedef struct nn{ int x,y; int time; friend bool operator Q; node q,p; q.time=0;q.x=sj;q.y=si; Q.push(q); while(!Q... 阅读全文

摘要： 点此连接到UVA10494思路: 采取一种, 边取余边取整的方法, 让这题变的简单许多~ AC代码:#include#includeint main() { long long mod; long long k, tmp; int len; int ans[10010]; char num[10010], ch[2]; while(scanf("%s%s%lld", num, ch, &mod) != EOF) { len = strlen(num); k = 0; tmp = 0; memset(ans, 0, sizeof(ans)); for(int i = 阅读全文

摘要： 题意：有n个人，n个房子，一个人每走一步（可上、下、左、右取一个方向）花费1美元，问让这n个人走到n个房子里最少需要多少美元(n >在学最小费用最大流，找模版题找到了这道。建图：1到n为人的编号，n+1到2*n为房子的编号，另加上源点0和汇点2*n+1；源点到每个人各建立一条容量为1的流，费用为0；每个人到每个房子各建立一条容量为1的流，费用按题意计算；（注意：反向费用！！！）每个房子到汇点建立一条容量为1的流，费用为0。当满足最大流时，一定是源点发出n，每个人接收1并发出1到一个房子，n个房子各发出1汇成n到汇点，所以，真是最小费用最大流的模版题。#include #include # 阅读全文

摘要： LVM有扩容功能，无容错功能 物理卷： [root@localhost ~]# pvscan PV /dev/sda2 VG VolGroup lvm2 [19.51 GiB / 0 free] Total: 1 [19.51 GiB] / in use: 1 [19.51 GiB] / in no VG: 0 [0 ] [root@localhost ~]# pvcreate /dev/sd[bcd] 把bcd磁盘都设置为... 阅读全文

摘要： Cheapest PalindromeTime Limit: 2000MSMemory Limit: 65536KTotal Submissions: 4592Accepted: 2236DescriptionKeeping track of all the cows can be a tricky task so Farmer John has installed a system to automate it. He has installed on each cow an electronic ID tag that the system will read as the cows pa 阅读全文