Binary Tree Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6355   Accepted: 2922 Description Background Binary trees are a common data structure in computer science. In this problem we will look at an infinite binary tree where the node…

Snowflake Snow Snowflakes Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 34762 Accepted: 9126 Description You may have heard that no two snowflakes are alike. Your task is to write a program to determine whether this is really true. Your…

#include<iostream> #include<cstring> #include<cstdio> using namespace std; int n, m; int num[100005]; int front(int x) { return x&(-x); } int update(int x,int k) { while(x<=n) { num[x]+=k; x+=front(x); } return 1; } int sum(int x)…

#include<iostream> #include<stack> #include<stdio.h> using namespace std; struct node { __int64 num,pre,next; }; int main() { int n; freopen("in.txt","r",stdin); while(scanf("%d",&n)>0&&n) { s…

hadoop集群性能低下的常见原因 (一)硬件环境 1.CPU/内存不足,或未充分利用 2.网络原因 3.磁盘原因 (二)map任务原因 1.输入文件中小文件过多,导致多次启动和停止JVM进程.可以设置JVM重用. 2.数据倾斜:大文件且不可分割,导致处理这些文件的map需要很长时间. 3.数据本地化效果差. (三)reduce任务的原因 1.reduce任务数量过大或过小 2.数据倾斜:一部分key的记录数量太大,导致某些reduce执行过慢 3.缓慢的shuffle和排序 (四)hadoop…

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…

题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=119 解题思路: RMQ算法. 不会的可以去看看我总结的RMQ算法. http://blog.csdn.net/niushuai666/article/details/6624672 代码如下: #include<cstdio> #include<algorithm> #include<cmath> using namespace std; const int…

Help Me with the Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3706   Accepted: 2371 Description Your task is to read a picture of a chessboard position and print it in the chess notation. Input The input consists of an ASCII-art…

这里先只考虑x,y都大于0的情况 如果x^2+y^2=r^2,则(r-x)(r+x)=y*y 令d=gcd(r-x,r+x),r-x=d*u^2,r+x=d*v^2,显然有gcd(u,v)=1且u<v 有2r=d*(u^2+v^2),y=d*u*v,x=d(v^2-u^2)/2 枚举2r的约数d,再花费sqrt(2r/d)的时间枚举u,求出v=sqrt(2r/d-u^2)然后判断gcd(u,v)=1 最后结果乘以4(四个象限)+4(坐标轴上)即可 /***********************…

#include<iostream> #include<cstdio> using namespace std; struct node { int l, r, m; int max; }num[800005]; int val[200005]; int n, m; int init(int l, int r, int k) { num[k].l = l; num[k].r = r; if(l==r) { num[k].m = l; num[k].max=val[l]; retur…