zoj 2818 Root of the Problem】的更多相关文章

Root of the Problem Time Limit: 2 Seconds      Memory Limit: 65536 KB Given positive integers B and N, find an integer A such that AN is as close as possible to B. (The result A is an approximation to the Nth root of B.) Note that AN may be less than…
题目链接: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2818 题目描述: Given positive integers B and N, find an integer A such that AN is as close as possible to B. (The result A is an approximation to the Nth root of B.) Note that AN may be l…
水题三题: 1.给你B和N,求个整数A使得A^n最接近B 2. 输出第N个能被3或者5整除的数 3.给你整数n和k,让你求组合数c(n,k) 1.poj 3100 (zoj 2818) Root of the Problem: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2818 http://poj.org/problem?id=3100 #include<cstdio> #include<cmath>…
Root of the Problem Time Limit: 2 Seconds      Memory Limit: 65536 KB Given positive integers B and N, find an integer A such that AN is as close as possible to B. (The result A is an approximation to the Nth root of B.) Note that AN may be less than…
Root of the Problem Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 12060   Accepted: 6469 Description Given positive integers B and N, find an integer A such that AN is as close as possible to B. (The result A is an approximation to the…
水两发去建模,晚饭吃跟没吃似的,吃完没感觉啊. ---------------------------分割线"水过....."------------------------------- POJ 3100 Root of the Problem http://poj.org/problem?id=3100 大意: 给定B和N,求一个数A使得A^N最接近B 太水了..... #include<cstdio> #include<cmath> int main()…
题目链接: POJ:id=3100" style="font-size:18px">http://poj.org/problem? id=3100 ZOJ:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1818 HDU:pid=2740">http://acm.hdu.edu.cn/showproblem.php?pid=2740 Description Given positive…
A Simple Tree Problem Time Limit: 3 Seconds      Memory Limit: 65536 KB Given a rooted tree, each node has a boolean (0 or 1) labeled on it. Initially, all the labels are 0. We define this kind of operation: given a subtree, negate all its labels. An…
Description Given a rooted tree, each node has a boolean (0 or 1) labeled on it. Initially, all the labels are 0. We define this kind of operation: given a subtree, negate all its labels. And we want to query the numbers of 1's of a subtree. Input Mu…
Solution: 根据树的遍历道的时间给树的节点编号,记录下进入节点和退出节点的时间.这个时间区间覆盖了这个节点的所有子树,可以当做连续的区间利用线段树进行操作. /* 线段树 */ #pragma comment(linker, "/STACK:102400000,102400000") #include <iostream> #include <cstdio> #include <cstring> #include <cmath>…