Web服务器性能/压力测试工具http_load、webbench、ab、Siege使用教程

首先，我要对给予大众无私帮助和奉献的作者：Tanky Woo 表示感谢！同时提供他的文章链接：

http://www.wutianqi.com/?p=596

这题是典型的母函数（也叫做生成函数）题，有一定的难度，关键是理解母函数的概念和解法。关于这些我就不多说了，大家只要在百度上搜索就有了。

母函数这题，有很多要注意的地方，像我就是由于小的细节不注意，导致一错再错。比如那个 j 的步进问题，一开始是 j++, 后来是 j = j * 2，再后来是 j *= expr，最后才找到正确答案。

有些注释掉的代码，是我发现错误用的。在草稿上演算了许久，也没找到错误，便借助于观查过程来检查了。

我还把递归的解法也贴上来了，因为递归比较好做，同时理解递归的解题方法也比较有意义。可惜就是超时了。

// 母函数解法，终于成功了！

/* THE PROGRAM IS MADE BY PYY */
/*----------------------------------------------------------------------------// http://acm.hdu.edu.cn/showproblem.php?pid=1028 Ignatius and the Princess III

Date  : 2011/3/8 Thuesday
Result:
3574731 2011-03-08 15:56:26 Accepted 1028 15MS 324K 1635 B C++ pyy
//----------------------------------------------------------------------------*/

#include <iostream>
#include <stdio.h>
using namespace std;
void out(int *p, int n)
{
for (int i = 0; i <= n; i++)
printf("%d ", *(i+p));
putchar('/n');
}
int main()
{
int g[1001], t[1001], n;
while (cin >> n) {
int i, expr, j;
expr = 0;
// 初始化
for (i = expr; i <= n; i++) {
g[i] = 1;
t[i] = 1;
}
// 第 expr 个表达式。
for (expr = 2; expr <= n; expr++) {
// 前一个表达式与当前表达式第 j 个数相乘。
// 此处 j 增加的步长应该注意！
for (j = expr; j <= n; j += expr) {
// 前一个表达式与当前表达式的 第 i 幂次的数相乘后，
// 前一个表达式第 i 幂次的系数加 1.
for (i = j; i <= n; i++) {
// t 为前一个表达式，在与当前表达式
// 逐个相乘的过程中，始终不变。
g[i] += t[i - j];
}
//                                out(g, n);
}
// t 复制为当前表达式，充当下一个表达式的前表达式
for (j = 0; j <= n; j++)
t[j] = g[j];
//                        cout << "----/n";
//                        out(g, n);
//                        cout << "-------/n/n";
}

cout << g
<< endl;
}
return 0;
}

======================================递归解法======================================

// 递归解法，果断超时, 但能理解递归解法，很有意义。

/* THE PROGRAM IS MADE BY PYY */
/*----------------------------------------------------------------------------// http://acm.hdu.edu.cn/showproblem.php?pid=1028 Ignatius and the Princess III

Date  : 2011/2/27 Thursday
Begin : 15:50
End   : 17:35
//----------------------------------------------------------------------------*/

#include <iostream>
#include <stdio.h>
using namespace std;
const int flag = 0xFFFFFFFE;  // flag = 1111,1111,1111,1110

// the possibile forms of sequences with head of "head", example for Sample-1
/*------------------------------- Sample-1 -----------------------------------//
such as: (head = 2, total = 4)
4 = 2 + 2;
4 = 2 + 1 + 1;
//----------------------------------------------------------------------------*/
// 以 head 为开头的序列的可能形式, 示例见 Sample-1
int cal(int head, int total)
{
// the synonym of "if (head == 1 || head == 0)"
// 相当于 "if (head == 1 || head == 0)"
if (!(head & flag))
return 1;

int tail, min, sum = 0, i;
// the number behind head
// 在 head 后面的数字
tail = total - head;

// current trailing sequence forms after head could be diverse
/*----------------------------- Sample-2 -----------------------------//
such as : 3 = 3;
3 = 2 + 1;
3 = 1 + 1 + 1;
//--------------------------------------------------------------------*/
// 当前 head 后面的尾序列的组成形式可能是多样性的, 比如上例。
if (tail > 1) {
/*------------------------------------------------------------//
the beggest element in trailing sequence must be the smaller
one between tail and head
//------------------------------------------------------------*/
/*------------------------------------------------------------//
尾序列中的最大元素必须是 head 和 tail 中的较小者。
//------------------------------------------------------------*/
min = tail > head ? head : tail;
// calculate the possibile forms of tailing sequence, see
// Sample-2
// 计算尾序列的可能形式 见 Sample-2
sum += cal(min, tail);
}
// there is only one possibility
else
sum += 1;
// the possibile forms of sequences with head of "head = head - 1",
// example for Sample-1
// 以 head = head - 1 为开头的序列的形式，示例见 Sample-1
sum += cal(head - 1, total);
return sum;
}
int main()
{
int n;
while (scanf("%d", &n)) {
printf("%d/n", cal(n, n));
}
return 0;
}

======================================原题如下======================================

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4144Accepted Submission(s): 2890


Problem Description
"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.

"The second problem is, given an positive integer N, we define an equation like this:
N=a[1]+a[2]+a[3]+...+a[m];
a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"

Input
The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.

Output
For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

Sample Input

4
10
20

Sample Output

5
42
627

Author
Ignatius.L