• In the second line, output “Collision will not happen.” (without quotes) if the collision will never happen. Otherwise, output two real numbers xc and yc, rounded to one decimal place, which ...
• There's a round medal fixed on an ideal smooth table, Fancy is trying to throw some coins and make them slip towards the medal to collide. There's also a round range which shares exact ...

There's a round medal fixed on an ideal smooth table, Fancy is trying to throw some coins and make them slip towards the medal to collide. There's also a round range which shares exact
the same center as the round medal, and radius of the medal is strictly less than radius of the round range. Since that the round medal is fixed and the coin is a piece of solid metal, we can assume that energy of the coin will not lose, the coin will collide
and then moving as reflect.
Now assume that the center of the round medal and the round range is origin ( Namely (0, 0) ) and the coin's initial position is strictly outside the round range. Given radius of the
medal Rm, radius of coin r, radius of the round range R, initial position (x, y) and initial speed vector (vx, vy) of the coin, please calculate the total time that any
part of the coin is inside the round range.
Please note that the coin might not even touch the medal or slip through the round range.
Input
There will be several test cases. Each test case contains 7 integers Rm, R, r, x, y, vx and vy in
one line. Here 1 ≤ Rm < R ≤ 2000, 1 ≤ r ≤ 1000, R + r < |(x, y)| ≤ 20000, 1 ≤ |(vx, vy)| ≤ 100.
Output
For each test case, please calculate the total time that any part of the coin is inside the round range. Please output the time in one line, an absolute error not more than 1e-3 is acceptable.
Sample Input
5 20 1 0 100 0 -1
5 20 1 30 15 -1 0
Sample Output
30.000
29.394
题意：光滑的桌面上放两个同心圆，一枚硬币以一定速度行驶，碰到小圆会反弹，求硬币部分在圆环中所花的总时间
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

int main()
{
double Rm,R,r,x,y,vx,vy;
while(scanf("%lf%lf%lf%lf%lf%lf%lf",&Rm,&R,&r,&x,&y,&vx,&vy)!=EOF)
{
double ans=0;
Rm+=r; R+=r;
double A=vx*vx+vy*vy;
double B=2*x*vx+2*y*vy;
double C1=x*x+y*y-R*R;
double C2=x*x+y*y-Rm*Rm;
double delta1=B*B-4*A*C1;
double delta2=B*B-4*A*C2;
if(delta1>0&&x*vx+y*vy<0)
{
if(delta2<=0) ans=fabs(sqrt(delta1)/A);
else ans=fabs((sqrt(delta1)-sqrt(delta2))/A);
}
printf("%.3lf\n",ans);
}
return 0;
}
思路：转化成直线与圆的问题
注意：几何中整形数据可用浮点型输入


展开全文
• 提示md5哈希碰撞（并没有用） 照旧ssh ls -al 和上一关差不多,一个可执行文件,一个源码,一个无法读取的flag文件 查看源码 #include <stdio.h> #include <...unsigned long che...

提示md5哈希碰撞（并没有用）
照旧ssh ls -al

和上一关差不多,一个可执行文件,一个源码,一个无法读取的flag文件
查看源码

#include <stdio.h>
#include <string.h>
unsigned long hashcode = 0x21DD09EC;
//转换并相加
int* ip = (int*)p;
int i;
int res=0;
for(i=0; i<5; i++){
res += ip[i];
}
return res;
}

int main(int argc, char* argv[]){
//至少有一个参数
if(argc<2){
printf("usage : %s [passcode]\n", argv[0]);
return 0;
}
//参数长度必须为20
if(strlen(argv[1]) != 20){
printf("passcode length should be 20 bytes\n");
return 0;
}

system("/bin/cat flag");
return 0;
}
else
printf("wrong passcode.\n");
return 0;
}

分析源码:
将长度为20byte的char型数组转换为int型数组.得到五个整型
(char型为1Byte，int型为4Byte，而且数据是从右向左存储,所以每四个char转换成一个int)
最终要求五个整型数字之和等于0x21DD09EC
解决方法:
理论上有无数种组合,方便起见这里我们设置四组都是相同的’0x02020202’，最后一组算出来是：
>>> hex(0x21dd09ec-0x02020202*4)'0x19d501e4'
所以反序输入应该为0xe4 0x01 0xd5 0x19
由于有非可见字符串，我们无法手动从终端输入，利用python进行输入：
python -c 'print 4*"\x02\x02\x02\x02"+"\xe4\x01\xd5\x19"' | xargs ./col
操作:

flag: daddy! I just managed to create a hash collision
完毕
附:
char型数组转换为int型数组:
例如我们构造了char型数组abcdefgh,即传入了四个char:0xab、0xcd、0xef、0xgh，转换成int后对应的数值变成0xghefcdab,(小端存储)
然后一共构造了5组这样的0xghefcdab进行求和最后的结果为0x21DD09EC

python -c (command):

Execute the Python code in command. command can be one or more statements separated by newlines, with significant leading whitespace as in normal module code.
If this option is given, the first element of sys.argv will be "-c" and the current directory will be added to the start of sys.path (allowing modules in that directory to be imported as top level modules).

xargs命令:
xargs是给命令传递参数的一个过滤器.通常情况下，xargs从管道或者stdin中读取数据.
xargs的默认命令是echo,这意味着数据将会包含换行和空白,不过通过xargs的处理，换行和空白将被空格取代。

转载于:https://www.cnblogs.com/Archimedes/p/8612659.html
展开全文
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