以下是在开发过程中遇到的一些细节点: 1)called after throwing an instance of std::bad_alloc 此问题是由于publish(data),当中data赋值的问题导致的. 2)在用GDB调试CodeBlocks的时候,遇到权限不允许的情况. warning: GDB: Failed to set controlling terminal: Operation not permitted 此问题需要通过设置来解决: 参考资料: http://blog.…
uva12546. LCM Pair Sum One of your friends desperately needs your help. He is working with a secret agency and doing some encoding stuffs. As the mission is confidential he does not tell you much about that, he just want you to help him with a specia…
My system is CentOS7 x64, Maya2015 x64 for Linux. - Make sure that your project is built with flag -g - Start Maya, and find out the process ID of maya.bin. e.g. you can find the process ID in System Monitor. This is my screenshot, and my maya.bin's…
- For now, CodeBlocks doesn't privide binary packages for CentOS7.(http://www.codeblocks.org/downloads/26#linux64) - In this thread(http://forums.codeblocks.org/index.php/topic,19588.0.html), Jens told me that he provided the CodeBlocks binary packag…
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=109329#problem/B 全题在文末. 题意:在a,b中(a,b<=n)(1 ≤ n ≤ 1014),有多少组(a,b) (a<b)满足lcm(a,b)==n; 先来看个知识点: 素因子分解:n = p1 ^ e1 * p2 ^ e2 *..........*pn ^ en for i in range(1,n): ei 从0取到ei的所有组合 必能包含所有n的因子. 现…
组合数学 GCD and LCM Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 451 Accepted Submission(s): 216 Problem Description Given two positive integers G and L, could you tell me how many solutions…
google test 是google的c++开源单元测试工具,chrome的开发团队就是使用它. Code::Blocks 12.11(MinGW 4.7.1) (Windows版)Google Test 1.7.0 一 编译Google Test 运行cmake-gui在弹出来的对话框中Where is the source code之后填写解压的gtest,例如:D:/software/gtest-1.7.0Where to build the binaries之后填写生成的工程路径,例如…
B - Pairs Forming LCM Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1236 Description Find the result of the following code: long long pairsFormLCM( int n ) { long long res = 0; for( in…
题目链接:http://lightoj.com/volume_showproblem.php?problem=1215 题意:已知三个数a b c 的最小公倍数是 L ,现在告诉你 a b L 求最小的 c ; 其实就是告诉你(最小公倍数LCM)LCM(x, y) = L 已知 x 和 L 求 最小的 y ; L = LCM(x, y)=x*y/gcd(x, y);如果把x,y,L写成素因子之积的方式会很容易发现 L 就是 x 和 y 中素因子指数较大的那些数之积; 例如LCM(24, y)…