Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 20995 Accepted: 8150 Description You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled wit…
Mad scientist Mike has constructed a rooted tree, which consists of n vertices. Each vertex is a reservoir which can be either empty or filled with water. The vertices of the tree are numbered from 1 to n with the root at vertex 1. For each vertex, t…
Beautiful People Time Limit: 10000/5000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) Special Judge SubmitStatus Problem Description The most prestigious sports club in one city has exactly N members. Each of its members is strong an…
FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 42953 Accepted Submission(s): 14336 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…
#include <stdio.h> int main(){ int num[101]={0}; int result[21]={0}; int sum[101][21]={0}; int n,i,j,m,k; for(i=1;i<101;++i){ num[i]=i;} for(i=2;i<101;++i){ for (j=1;j<i-1;++j){ if (num[i]%num[j]==0){ num[i]/=num[j];}}} result[1]=1; m=1; fo…
首先介绍一下 Case ..When...Then..End 的用法: CASEJiXiaoFind_RowID WHEN '1' THENJiXiao_Money1 WHEN '2' THEN JiXiao_Money2 WHEN '3' THEN JiXiao_Money3 WHEN '4' THEN JiXiao_Money4 ELSE JiXiao_Money5 END 举例说明: 上图是一个视图的截图,我想在 表B 中添加一个字段 JiXiao_Money(绩效金额):…
首先登陆https://developer.apple.com(99美元账号) 选择iOS Developer program 板块下的 Certificates,Identifiers & Profiles 点击进入 b.进入 c.选择证书类型 distribution,选择添加 d.点击+后,选择distribution->App Store and Ad Hoc,continue PS:现在的itunes已经将这两项分开,要上传至AppStore请选择App Store,如果是要安装到…
一.SolrJ基础 1.相关资料 API:http://lucene.apache.org/solr/4_9_0/solr-solrj/ apache_solr_ref_guide_4.9.pdf:Client APIs---Using SolrJ http://wiki.apache.org/solr/Solrj solr in action:Using the SolrJ client library to add documents from Java, Using SolrJ from…
#include <bits/stdc++.h> using namespace std; const int MAXN = 100; const int X = 3; long long f[1005]; void init() { f[0] = 1; for(int i = 1;i < MAXN; i++) { f[i] = f[i-1]*X; } } int hash(string num) { int ans = 0; for (int i = 0; i < num.len…
#include<stdio.h> #include<stdlib.h> //快速幂算法,数论二分 long long powermod(int a,int b, int c) //不用longlong就报错,题目中那个取值范围不就在2的31次方内 { long long t; if(b==0) return 1%c; if(b==1) return a%c; t=powermod(a,b/2,c);//递归调用,采用二分递归算法,,注意这里n/2会带来奇偶性问题 t=t*t%c;…
C - Rightmost Digit Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1061 Description Given a positive integer N, you should output the most right digit of N^N. Input The input contains several test case…
