1000和1024的争论,其实是传输领域和存储领域概念不清引起的;在传输领域,1秒钟传输多少字位(即b,bit),肯定是用10进制表示,所以是1kb=1000b,即1秒钟传输1000个比特位;就好像:我走路的速度是1km/小时,即1000米/小时,根1024没半点关系;在存储领域,1kB=1024B,这里的B是大写,表示字节,即8个字位(bit);有人说因为1024约等于1000,这种说法是不对的.

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4229    Accepted Submission(s): 2205 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4178    Accepted Submission(s): 2174 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took pa

总时间限制:  1000ms 内存限制:  65536kB 描述 输入两个正整数 m 和 k,其中1 < m < 100000,1 < k < 5 ,判断 m 能否被19整除,且恰好含有k个3,如果满足条件,则输出YES,否则,输出NO. 例如,输入: 43833 3 满足条件,输出YES. 如果输入:39331 3 尽管有3个3,但不能被19整除,也不满足条件,应输出NO. 输入 m 和 k 的值,中间用单个空格间隔. 输出 满足条件时输出 YES,不满足时输出 NO. 样例输入

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4739    Accepted Submission(s): 2470 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took pa

实例三:将n的k位清0 方法: result= n &~(1<<k) 使第k为变成0,再与运算,0和任何数进行与运算都是0. 解释:  0000 0001 ---- 1 左移k位  0000 1000 取反操作  1111 0111 原数 1111 1111 取反后与原数进行与操作 结果 1111 0111 ---- k清零 int _tmain(int argc, _TCHAR* argv[]){ int n,k,nResult=0; cout << "请输入原

http://acm.hdu.edu.cn/showproblem.php?pid=2639       Problem Description The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it does

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2639 Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5817    Accepted Submission(s): 3067 Problem Description The title of this

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4824    Accepted Submission(s): 2514 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3718    Accepted Submission(s): 1903 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par

The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link: Here is the link: http://acm.hdu.edu.c

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3437    Accepted Submission(s): 1773 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par

解题思路:对于01背包的状态转移方程式f[v]=max(f[v],f[v-c[i]+w[i]]);其实01背包记录了每一个装法的背包值,但是在01背包中我们通常求的是最优解, 即为取的是f[v],f[v-c[i]]+w[i]中的最大值,但是现在要求第k大的值,我们就分别用两个数组保留f[v]的前k个值,f[v-c[i]]+w[i]的前k个值,再将这两个数组合并,取第k名. 即f的数组会增加一维. http://blog.csdn.net/lulipeng_cpp/article/details/

The title of this problem is familiar,isn't it?yeah,if you had took part in the "Rookie Cup" competition,you must have seem this title.If you haven't seen it before,it doesn't matter,I will give you a link: Here is the link: http://acm.hdu.edu.c

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7165    Accepted Submission(s): 3802 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3355    Accepted Submission(s): 1726 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took par

Anton has a positive integer n, however, it quite looks like a mess, so he wants to make it beautiful after k swaps of digits. Let the decimal representation of n as (x1x2⋯xm)10 satisfying that 1≤x1≤9, 0≤xi≤9 (2≤i≤m), which means n=∑mi=1xi10m−i. In e

附题目链接:Bone Collector II Input The first line contain a integer T , the number of cases.Followed by T cases , each case three lines , the first line contain two integer N , V, K(N <= 100 , V <= 1000 , K <= 30)representing the number of bones and t

Description 从键盘输入一个自然数K(K>1),若存在自然数M和N(M>N),使得K^M^和K^N^均大于或等于1000,且他们的末尾三位数相等,则称M和N是一对"K尾相等数".请编一个程序,输出M+N值最小的K尾相等数. Input 输入一个整数K Output 输出M+N的最小值 Sample Input 2 Sample Output 120 用到的知识点: 取模运算 \((a * b) \% mod = ((a \% mod) * (b \% mod))

//字节大小,K,M,G public static final long KB = 1024; public static final long MB = KB * 1024; public static final long GB = MB * 1024; /** * 文件字节大小显示成M,G和K * @param size * @return */ public static String displayFileSize(long size) { if (size >= GB) { ret