2021-05-20 21:44:08
badboy

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一般与就 jmeter 结合使用，可以直接录制脚本，并导出 jmeter 支持的 jmx 格式脚本。
我是打不开，你可以试试
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链接：https://pan.baidu.com/s/1u_7bKxu1w03CKdZDRQusPQ
提取码：SonG
（2）录制脚本

输入网址，进行操作

结束录制

（3）导出脚本
File-》Export to JMeter…保存到本地

直接用JMeter打开就可以，但录制脚本有很多没用的请求，可以做优化

（4）设置检查点
选择文本-》添加检查点-》添加成功-》回放？变成√号

（5）设置参数化-文本
添加参数文本

设置文本

添加成功显示位置

使用参数，找到需要设置参数的位置，设置参数
参数格式\${参数名}

设置运行策略

全部运行
报告


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2021-05-20 21:19:21
文章目录一、安装badboy脚本软件二、打开badboy软件三、badboy脚本录制四、添加验证点

傻瓜式安装即可

四、添加验证点
验证点的作用就是验证脚本是否按照我们测试的思路执行，判断脚本执行过程中是否存现问题


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2021-06-20 00:49:35
Bad Boy time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Riley is a very bad boy, but at the same time, he is a yo-yo master. So, he decided to...
B. Bad Boy
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Riley is a very bad boy, but at the same time, he is a yo-yo master. So, he decided to use his yo-yo skills to annoy his friend Anton.
Anton’s room can be represented as a grid with n rows and m columns. Let (i,j) denote the cell in row i and column j. Anton is currently standing at position (i,j) in his room. To annoy Anton, Riley decided to throw exactly two yo-yos in cells of the room (they can be in the same cell).
Because Anton doesn’t like yo-yos thrown on the floor, he has to pick up both of them and return back to the initial position. The distance travelled by Anton is the shortest path that goes through the positions of both yo-yos and returns back to (i,j) by travelling only to adjacent by side cells. That is, if he is in cell (x,y) then he can travel to the cells (x+1,y), (x−1,y), (x,y+1) and (x,y−1) in one step (if a cell with those coordinates exists).
Riley is wondering where he should throw these two yo-yos so that the distance travelled by Anton is maximized. But because he is very busy, he asked you to tell him.
Input
The first line contains a single integer t (1≤t≤104) — the number of test cases. Then t test cases follow.
The only line of each test case contains four integers n, m, i, j (1≤n,m≤109, 1≤i≤n, 1≤j≤m) — the dimensions of the room, and the cell at which Anton is currently standing.
Output
For each test case, print four integers x1, y1, x2, y2 (1≤x1,x2≤n, 1≤y1,y2≤m) — the coordinates of where the two yo-yos should be thrown. They will be thrown at coordinates (x1,y1) and (x2,y2).
If there are multiple answers, you may print any.
Example
inputCopy
7
2 3 1 1
4 4 1 2
3 5 2 2
5 1 2 1
3 1 3 1
1 1 1 1
1000000000 1000000000 1000000000 50
outputCopy
1 2 2 3
4 1 4 4
3 1 1 5
5 1 1 1
1 1 2 1
1 1 1 1
50 1 1 1000000000
Note
Here is a visualization of the first test case.
我的思路：
根据几何思想，那个yo-yo master只能放到矩形的四个端点，那么，可以选择的只有6种可能，枚举即可
我的错误，爆int了
明明测试案例里面的最后一个就爆int了，我太垃圾了。在计算路径长度的时候，需要算三个边长的和。
ac代码：
#include <bits/stdc++.h>
using namespace std;
struct dian{
int x;
int y;
//int len;
}s[4];
long long longway(int a,int b,int i, int j)
{
return abs(a-i)+abs(b-j);
}
int main() {
//std::cout << "Hello, World!" << std::endl;
int t,m,n,i,j,ans;
cin >> t;
vector<long long > le(6);
while(t--)
{
cin>>m>>n>>i>>j;
s[0].x=1;s[0].y =1;
s[1].x = 1;s[1].y = n;
s[2].x = m;s[2].y = n;
s[3].x = m;s[3].y = 1;
for (int k = 0; k < 4; ++k) {
le[k] = longway(s[k].x,s[k].y,i,j)+ longway(s[(k+1)%4].x,s[(k+1)%4].y,i,j)+longway(s[k].x,s[k].y,s[(k+1)%4].x,s[(k+1)%4].y);
}
le[4] = longway(s[0].x,s[0].y,i,j)+ longway(s[2].x,s[2].y,i,j)+ longway(s[0].x,s[0].y,s[2].x,s[2].y);
le[5] = longway(s[1].x,s[1].y,i,j)+ longway(s[3].x,s[3].y,i,j)+ longway(s[1].x,s[1].y,s[3].x,s[3].y);
ans = 0;
for (int k = 1; k < 6; ++k) {
if(le[k]>=le[ans]) ans = k;
}
if(ans <4)
cout<<s[ans].x<< ' '<<s[ans].y<<' '<<s[(ans+1)%4].x<< ' '<<s[(ans+1)%4].y<< ' '<<endl;
else if(ans == 4) cout<<s[0].x<< ' '<<s[0].y<< ' '<<s[2].x<< ' '<<s[2].y<<endl;
else cout<<s[1].x<< ' '<<s[1].y<< ' '<<s[3].x<< ' '<<s[3].y<<endl;
}
return 0;
}

在思考一下发现
其实对角线两个点的路径长度是最长的只需要直接输出即可
#include "bits/stdc++.h"
using namespace std;
int main ()
{
int T,n,m,i,j ;
cin >> T;
while(T--)
{
cin>>m>>n>>i>>j;
cout <<'1' <<' '<<'1'<<' '<<m<<' '<<n<<endl;

}
return 0;
}



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万次阅读 多人点赞 2019-07-17 17:16:08
badboy  现在好像维护了下     有需要的可以下载这个版本用下
链接：https://pan.baidu.com/s/1WustinShXDfQ4Sr08_L61g
提取码：1fmq


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千次下载 热门讨论 2011-12-07 10:30:11