zhousilijames的专栏

#include #include#include#include#include#includeusing namespace std;int T,m,a[10],ans;vector mg[10];int g[10][10];bool used[10];void init(){ memset(g,0,sizeof(g)); g[1][3]=g[3][

#1680 : hiho字符串2时间限制:10000ms单点时限:1000ms内存限制:256MB描述我们定义第一代hiho字符串是"hiho"。  第N代hiho字符串是由第N-1代hiho字符串变化得到，规则是：h -> hio  i -> hi  o -> ho  例如第二代hiho字符串是： hiohihioho  第三代是： ...

#include using namespace std;const int mx=1000000001+100000;int n;vector g;int dfs(int num,int cnt,int limit){ if(num==0) return 0; int p=upper_bound(g.begin(),g.end(),num)-g.begin()-1;

时间限制:10000ms单点时限:1000ms内存限制:256MB描述Given an integer n, for all integers not larger than n, find the integer with the most divisors. If there is more than one integer with the same

时间限制:10000ms单点时限:1000ms内存限制:256MB描述Little Hi and Little Ho are playing a construction simulation game. They build N cities (numbered from 1 to N) in the game and connect them by N-

时间限制:1000ms单点时限:1000ms内存限制:256MB描述wuzhengkai是一个炉石传说玩家。有一天，他碰到对面上了个诺兹多姆并上了嘲讽和圣盾。诺兹多姆是一个有8点攻击力和8点生命值的随从，这意味着他能够承受总量小于8点的伤害而不死亡，且对于任何攻击他的单位造成8点伤害。嘲讽意味着当诺兹多姆死亡之前你不能攻击对方玩家。圣盾意味着诺兹

#include #include#include#include#include#include#include#includeusing namespace std;const int mx=50000+1000;int T,num,p;char name[mx][10] ,s[mx*10];bool used[mx];vector g[mx];struct no

DescriptionIn order to understand early civilizations, archaeologists often study texts written in ancient languages. One such language, used in Egypt more than 3000 years ago, is based on charact

刚开始我还很怀疑我的想法，后来百度没发现题解，看到discuss里面说数据很弱，于是我就大胆的写了个暴力,每次枚举一个点找到一次到合法终点的合法路径是，结束dfs，简单暴力代码如下#include #include#includeusing namespace std;const int mx=111;int mp[mx][mx],T,n,m,num,ans;bool use

A spreadsheet is a rectangular array of cells. Cells contain data or expressions that can be evaluated toobtain data. A “simple” spreadsheet is one in which data are integers and expressions are mixed

#include #include#include#includeusing namespace std;int n,op[40],sum[40];void init(){ memset(op,0,sizeof(op)); op[1]=sum[1]=1;op[0]=1;sum[0]=0; for(int i=2;i<20;i++) {

#include #include#include#includeusing namespace std;int n,m,q,mp[1100][1100];bool used[1100][1100];int dx[4]={0,0,-1,1},dy[4]={1,-1,0,0};int x1,x2,y1,y2;void dfs(int x,int y){ used[x]

#1689 : 推断大小关系时间限制:20000ms单点时限:2000ms内存限制:256MB描述有N个整数A1, A2, ... AN，现在我们知道M条关于这N个整数的信息。每条信息是：Ai &lt; Aj 或者 Ai = Aj  小Hi希望你能从第一条信息开始依次逐条处理这些信息。一旦能推断出A1和AN的大小关系就立即停止。  输出在处理第几条时第一次推断出A1...