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Reverse Integer
Reverse digits of an integer.Example1:x = 123, return 321Example2:x = -123, return -321click to show spoilers.Have you thought about this?Here are som...
分类:其他好文   时间:2014-08-19 23:54:05    阅读次数:245
【Math】GCD XOR 证明
题目:Given an integer N, and how many pairs (A;B) are there such that: gcd(A;B) = A xor B where 1=2)是不同位数的。和同学讨论后得出如下证明:设最大公约数为 j, 假设这两个数是b 和 b+k*j,(k>....
分类:其他好文   时间:2014-08-19 23:50:55    阅读次数:233
Palindrome Number
Determine whether an integer is a palindrome. Do this without extra space.click to show spoilers.Some hints:Could negative integers be palindromes? (i...
分类:其他好文   时间:2014-08-19 23:42:35    阅读次数:227
倒序的种子数据结构 - Reversed
package?com.victor.sort.seeds; import?java.util.ArrayList; /** ?*?倒序 ?*?@author?黑妹妹牙膏 ?* ?*/ public?class?Reversed?extends?Seeds?{ @Override protected?ArrayList<Integer>?...
分类:其他好文   时间:2014-08-19 19:19:35    阅读次数:190
Cracking the Coding Interview 5.7
An array A[1...n] contains all the integers from 0 to n except for one number which is missing.In this problem, we cannot access an entire integer in ...
分类:其他好文   时间:2014-08-19 18:49:45    阅读次数:251
BNUOJ 2528 Mayor's posters
Mayor's postersTime Limit: 3000msMemory Limit: 131072KBThis problem will be judged onUVA. Original ID:1058764-bit integer IO format:%lld Java class na...
分类:其他好文   时间:2014-08-19 18:10:15    阅读次数:400
【瑞星系统】修改会员积分处理
SET QUOTED_IDENTIFIER ON GOSET ANSI_NULLS ON GOALTER procedure GetVIPMark(@SLDAT datetime, @PNO integer, @SNO integer, @VIPNO varchar(20), @PLUInfo...
分类:其他好文   时间:2014-08-19 12:25:34    阅读次数:228
UVA 10791 Minimum Sum LCM (数论)
LCM (Least Common Multiple) of a set of integers is defined as the minimum number, which is a multiple of all integers of that set. It is interesting to note that any positive integer can be expressed...
分类:其他好文   时间:2014-08-18 23:38:13    阅读次数:275
Digital Roots
Digital RootsTime Limit: 1000ms Memory limit: 65536K有疑问?点这里^_^题目描述The digital root of a positive integer is found by summing the digits of the integer...
分类:其他好文   时间:2014-08-18 22:00:43    阅读次数:228
1-2+3-4+........+M方法一;方法二
1-2+3-4+......+m=(1-2)+(3-4)+...+[(m-2)-(m-1)]+m=-1+(-1)+(-1)+...+(-1)+m,一共有(m-1)/2个-1相加,再加上m,所以上式等于:(-1)*(m-1)/2+m代码:function sum(m as integer)sum=(-...
分类:其他好文   时间:2014-08-18 22:00:12    阅读次数:232
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