kk Blog —— 通用基础


date [-d @int|str] [+%s|"+%F %T"]
netstat -ltunp
sar -n DEV 1

crash vs gdb work

贴自https://www.redhat.com/archives/crash-utility/2014-October/msg00002.html
Yes, sure. GDB works very differently from crash. There main conceptual
difference is that GDB only handles with VIRTUAL addresses, while the
crash utility first translates everything to PHYSICAL addresses.
Consequently, GDB ignores the PhysAddr field in ELF program headers,
and crash ignores the VirtAddr field.

I have looked at some of my ELF dump files, and it seems to me that
VirtAddr is not filled correctly, except for kernel text and static
data (address range 0xffffffff80000000-0xffffffff9fffffff). Your linked
list is most likely allocated in the direct mapping
(0xffff880000000000-0xffffc7ffffffffff). However, I found out that the
virtual addresses for the direct mapping segments are wrong, e.g. my
dump file specifies it at 0xffff810000000000 (hypervisor area). This is
most likely a bug in the kernel code that implements /proc/vmcore.

But that’s beside the point. Why? The Linux kernel maps many physical
pages more than once into the virtual address space. It would be waste
of space if you saved it multiple times (for each virtual address that
maps to it). The crash utility can translate each virtual address to
the physical address and map it onto ELF segments using PhysAddr.
Incidentally, the PhysAddr fields are correct in my dump files…

I’m glad you’re interested in using GDB to read kernel dump files,
especially if you’re willing to make it work for real. I have proposed
more than once that the crash utility be re-implemented in pure gdb.
Last time I looked (approx. 1.5 years ago) the main missing pieces were:

  1. Use of physical addresses (described above)
  2. Support for multiple virtual address spaces (for different process contexts)
  3. Ability to read compressed kdump files
  4. Ability to use 64-bit files on 32-bit platforms (to handle PAE)

HTH,
Petr Tesarik

静态编译crash + xbt + bt -H

要在centos6上编译，为了能在centos5用，用静态编译
有两个显示函数参数的patch，但是不一定能起作用
patch1:

https://github.com/jhammond/xbt https://www.redhat.com/archives/crash-utility/2013-September/msg00010.html

patch2:

https://github.com/hziSot/crash-stack-parser https://github.com/hziSot/crash-stack-parser/blob/master/crash-parse-stack-7.0.1.patch

一、依赖包：

yum install bison zlib zlib-static glibc-static elfutils-devel elfutils-devel-static elfutils-libelf-devel-static ncurses ncurses-static crash-devel

二、patch1: xbt 显示参数

patch: https://github.com/hziSot/crash-stack-parser
make CFLAGS+=–static LDFLAGS+=–static

三、patch2: bt -H 显示参数

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依赖：有些没有静态包，要自己编译安装：
liblzma.a: http://tukaani.org/xz/xz-5.0.7.tar.bz2
libbz2.a:  http://www.bzip.org/1.0.6/bzip2-1.0.6.tar.gz
下载代码：git clone https://github.com/jhammond/xbt.git xbt.git
把xbt.git/xbt_crash.c中函数xbt_func前的static删了
把xbt.git/xbt_crash.c中函数xmod_init的register_extension删了
把 xbt 命令加到global_data.c        函数x86_64_exception_frame已经在其他库中定义了，所以要换个名字
编译xbt代码：make   ==  rm -rf *.so
把 xbt.git/xbt_crash.o  xbt.git/xbt_dwarf.o  xbt.git/xbt_dwfl.o  xbt.git/xbt_eval.o  xbt.git/xbt_frame_print.o 加到 Makefile 的 OBJECT_FILES= 中
make CFLAGS+=--static LDFLAGS+="--static -lc  -lm -ldl -ldw -lebl -lelf -lbz2 -llzma"


注意:-lelf -lebl要放在-ldw后面。

数A到数B之间的统计

Problem 1896 神奇的魔法数

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Accept: 98    Submit: 307
Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem Description

John定义了一种“神奇的魔法数”。 不含前导零且相邻两个数字之差至少为m的正整数被称为“神奇的魔法数”。特别的，对于任意的m，数字1..9都是“神奇的魔法数”。
John想知道，对于给定的m，在正整数a和b之间，包括a和b，总共有多少个“神奇的魔法数”？

Input

第一行一个数字T（1<=T<=100），表示测试数据组数。
接下来T行，每行代表一组测试数据，包括三个整数a,b,m。（1<=a<=b<=2,000,000,000, 0<=m<=9）

Output

对于每组测试数据，输出一行表示“神奇的魔法数”的个数。

Sample Input

7 1 10 2 1 20 3 1 100 0 10 20 4 20 30 5 1 10 9 11 100 9

Sample Output

9 15 100 5 3 9 1

Source福州大学第七届程序设计竞赛
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#include <stdio.h>

int n,m,d,dp[13][13],sum[13],dn[13],dm[13];


// DFS的时候这两个地方根据不同要求写。
int dfs(int da[], int dep, int all)
{
	int i,j,ret=0;
	if (dep == 0) return 1;
	for (i=0;i<da[dep];i++)
	{
		if (all > 0 || i > 0) {
			if (all == 0 || i-da[dep+1]>=d || i-da[dep+1]<=-d)
				ret += dp[dep][i];
		} else
			ret += sum[dep-1];
	}
	if (all == 0 || da[dep]-da[dep+1]>=d || da[dep]-da[dep+1]<=-d)
		ret += dfs(da, dep-1, all+da[dep]);
	return ret;
}

int main()
{
	int i,j,k,l,T;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d %d %d", &m, &n, &d);
		for (i=0;i<13;i++)
			for (j=0;j<13;j++) dp[i][j] = 0;
		sum[0] = 0; sum[1] = 9;
		for (i=0;i<10;i++) dp[1][i] = 1;
		for (i=2;i<13;i++) {
			sum[i] = sum[i-1];
			for (j=0;j<10;j++) {
				for (k=0;k<10;k++)
					if (j-k>=d || j-k<=-d)
						dp[i][j] += dp[i-1][k];
				if (j > 0)
					sum[i] += dp[i][j];
			}
		}
//        for (i=0;i<=2;i++)
//            for (j=0;j<10;j++) printf("%d %d %d\n", i, j, dp[i][j]);
		i = 1; k = n;
		while (i < 13) {
			dn[i] = k % 10; k /= 10;
			i++;
		}
		i = 1; k = m-1;
		while (i < 13) {
			dm[i] = k % 10; k /= 10;
			i++;
		}
		n = dfs(dn, 11, 0);
		if (m == 1)
			m = 0;
		else
			m = dfs(dm, 11, 0);
		printf("%d\n", n-m);
	}
	return 0;
}

How many 0’s?

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Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 2997
Accepted: 1603

Description

A Benedict monk No.16 writes down the decimal representations of all natural numbers between and including m and n, m ≤ n. How many 0's will he write down?

Input

Input consists of a sequence of lines. Each line contains two unsigned 32-bit integers m and n, m ≤ n. The last line of input has the value of m negative and this line should not be processed.

Output

For each line of input print one line of output with one integer number giving the number of 0's written down by the monk.

Sample Input

10 11
100 200
0 500
1234567890 2345678901
0 4294967295
-1 -1

Sample Output

1
22
92
987654304
3825876150

Source

Waterloo Local Contest, 2006.5.27
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import java.util.*;
import java.math.*;
import java.io.*;

public class Main {
	static long val,n,m,dp[][]=new long[13][13],a[]=new long[13],dn[]=new long[13], dm[]=new long[13], sum[]=new long[13];
	static long dfs(long dnm[], int dep, long all)
	{
		int i, j, k;
		long ret=0;
		if (dep == 0) return 0;
		for (i=0;i<dnm[dep];i++) {
			if (all > 0 || i > 0)
				ret += dp[dep][i]; // 需要计算前导0
			else
				ret += sum[dep-1]; // 不需要计算前导0
		}
		if (all > 0 && dnm[dep] == 0)
			ret += val % a[dep] + 1;
		ret += dfs(dnm, dep-1, all+dnm[dep]);
		return ret;
	}

	public static void main(String[] args) {
		int i,j,k,l;
		Scanner cin = new Scanner(System.in);
		a[1] = 10;
		for (i=2;i<13;i++) a[i] = a[i-1]*10;
		for (i=0;i<13;i++)
			for (j=0;j<13;j++) dp[i][j] = 0;
		dp[1][0] = 1;
		sum[0] = sum[1] = 0;
		for (i=2;i<13;i++) {
			sum[i] = sum[i-1];
			for (j=0;j<10;j++) {
				for (k=0;k<10;k++)
					dp[i][j] += dp[i-1][k];
				dp[i][j] += j==0 ? a[i-1] : 0;
				if (j > 0)
					sum[i] += dp[i][j];
			}
		}
		while (true) {
			m = cin.nextLong();
			n = cin.nextLong();
			if (m == -1 || n == -1) break;
			for (i=0;i<13;i++) dn[i] = dm[i] = 0;
			i = 1;
			val = n;
			while (val > 0) {
				dn[i] = val % 10;
				val /= 10;
				i++;
			}
			i = 1;
			val = m-1;
			while (val > 0) {
				dm[i] = val % 10;
				val /= 10;
				i++;
			}
			val = n;
			n = dfs(dn, 12, 0) + 1; // 0 还有一个0
			val = m-1;
			m = dfs(dm, 12, 0) + 1;
			if (val < 0) m = 0;
			System.out.println(n-m);
		}
	}
}