翻译 给定一个整型数,写一个函数决定它是否是3的幂(翻译可能不太合适-- 跟进: 你能否够不用不论什么循环或递归来完毕. 原文 Given an integer, write a function to determine if it is a power of three. Follow up: Could you do it without using any loop / recursion? 分析 题意我事实上不是满懂,比方说12究竟可不能够呢?还是说仅仅有:3.9.27.81这样的才行
题目链接:Log函数问题 2 / 49 Problem G FZU 2032 Log函数问题 不知道为什么...比赛时高精度难倒了一票人...成功搞出大新闻... 试了一下直接double相加超时,然后放弃(汗),然后double 有效数字最多16位的话,确实需要高精度了呢... 然后,机智的代码: #include <stdio.h> #include <string.h> #include <iostream> using namespace std; int an
main函数的定义形式 main函数能够不带參数,也能够带參数,这个參数能够觉得是 main函数的形式參数.C语言规定main函数的參数仅仅能有两个,习惯上这两个參数写为argc和argv.所以C99标准中规定仅仅有下面两种定义方式是正确的: int main(void) //无參形式 int main(int argc, char *argv[]) //有參形式 当然有參形式能够进行演变,所以以下的写法也是正确的(同一时候变量名是能够更换的): int
众所周知,静态SQL的输出结构必须也是静态的.对于经典的行转列问题,如果行数不定导致输出的列数不定,标准的答案就是使用动态SQL, 到11G里面则有XML结果的PIVOT. 但是 oracle 10G 没有 PIVOT 函数怎么办,自己写一个不久有了.上代码 直接点. CREATE OR REPLACEtype PivotImpl_shx as object( ret_type anytype, -- The return type of the table function stmt varc
//编写函数,将一个数的指定位置置0或置1 #include <stdio.h> unsigned int set_bit(unsigned int num, int pos, int flag) { int n = 1; n = n << (pos - 1); //将n的第pos位置1,其它全为0 if (flag == 0) { num = num&(~n); } else if (flag == 1) { num = num | n; } else printf(&q
Problem Description In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of
kiki's game Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/1000 K (Java/Others) Total Submission(s): 4972 Accepted Submission(s): 2908 Problem Description Recently kiki has nothing to do. While she is bored, an idea appears in his m
#include <iostream> using namespace std; int main(){ //求两数的和? int a,b,s; cout<<"请你输入两个整型的数字:"<<endl; cin>>a>>b; int sum(int x ,int y); s=sum(a,b);//实际参数 ,代表具体数值,在()当中 cout<<"The sum of a and b is:"&l