We have two rows. There are a dots on the top row and b dots on the bottom row. We draw line segments connecting every dot on the top row with every dot on the bottom row. The dots are arranged in such a way that the number of internal intersections among the line segments is maximized. To achieve this goal we must not allow more than two line segments to intersect in a point. The intersection points on the top row and the bottom are not included in our count; we can allow more than two line segments to intersect on those two rows. Given the value of a and b, your task is to compute P(a, b), the number of intersections in between the two rows. For example, in the following figure a = 2 and b = 3. This figure illustrates that P(2, 3) = 3.



Input

Each line in the input will contain two positive integers a (0 < a ≤ 20000) and b (0 < b ≤ 20000). Input is terminated by a line where both a and b are zero. This case should not be processed. You will need to process at most 1200 sets of inputs.

Output

For each line of input, print in a line the serial of output followed by the value of P(a, b). Look at the output for sample input for details. You can assume that the output for the test cases will fit in 64-bit signed integers.

Sample Input

2 2

2 3

3 3

0 0

Sample Output

Case 1: 1

Case 2: 3

Case 3: 9
问题链接：UVA10790 How Many Points of Intersection?

问题简述：（略）

问题分析：

    如图，找交点，不存在多于两条线相交于一点。

    每个交点都由两条线上的两个点得到，所以结果为C(a,2)*C(b,2)=ab(a-1)(b-1)/4。

程序说明：（略）

参考链接：（略）

题记：（略）
AC的C++语言程序如下：
/* UVA10790 How Many Points of Intersection? */

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int caseno = 0;
    long long a, b;
    while (scanf("%lld%lld", &a, &b) && (a || b))
        printf("Case %d: %lld\n", ++caseno, a * b * (a - 1) * (b - 1) / 4);

    return 0;
}

1.

画完图后，在空白处鼠标右键-Properties-Print界面修改Width和Height：

组合1（12:10）；组合2（10:8），单位：cm



 

2.

双击中心图画区域，在Size/Speed界面修改Left/Top使得图标位于画布中心区域

Width:75，Height:71.8



 

3.

修改坐标轴粗细和字体大小

左右上下左右的坐标轴都可修改，若要添加上边和右边的坐标轴，在Line and Ticks模块

修改Length为3，Thickness为1

所有的字体可在Tools-Options里修改，字体设置为12大小，点确定。也可在上面界面处调整



 

4.

线条和标号大小修改

双击线条，对线条和标号大小进行反反复复修改



 

5.

导出图片

tif格式

image settings设置参数，DPI根据要求设置，一般300

Export seetings中选择page

 

6.建议

6.1 一张图一个opj，文件名：20201225-章2-图5-简短解释

6.2 系列图合理利用 Tools-Theme Organizer，严格一图一opj

6.3 耐心再耐心，图跟着论文排版走

 

看题一看就知道估计得爆破，烦躁！！！



打开页面后出现，点击click



输入admin与admin



通过burp抓包发现是通过base64编码了



我一开始憨批了，没看到有个附件自己去搞了个字典没爆出来，记得下载附件
将请求发送给测试器



下一步


点击导入字典



但是发现前面没有admin：

写了个python来组合
import base64

with open(10_million_password_list_top_100.txt, r+) as f
    for line in f.readlines()
        line = line.strip('n')
        line = 'admin' + line
        print(line)
        line1 = base64.b64encode(line.encode('utf-8'))
        with open(final.txt, a+) as res
            res.write(str(line1, 'utf-8') + 'n');




再次导入字典

点击Attack开始攻击，发现个长度不一样的



双击点开得到flag