Coin Test
来源:互联网 发布:正装 知乎 编辑:程序博客网 时间:2024/06/11 01:50
Coin Test
- 描述
As is known to all,if you throw a coin up and let it droped on the desk there are usually three results. Yes,just believe what I say ~it can be the right side or the other side or standing on the desk, If you don't believe this,just try In the past there were some famous mathematicians working on this .They repeat the throwing job once again. But jacmy is a lazy boy.He is busy with dating or playing games.He have no time to throw a single coin for 100000 times. Here comes his idea,He just go bank and exchange thousands of dollars into coins and then throw then on the desk only once. The only job left for him is to count the number of coins with three conditions.
He will show you the coins on the desk to you one by one. Please tell him the possiblility of the coin on the right side as a fractional number if the possiblity between the result and 0.5 is no larger than 0.003. BE CAREFUL that even 1/2,50/100,33/66 are equal only 1/2 is accepted ! if the difference between the result and 0.5 is larger than 0.003,Please tell him "Fail".Or if you see one coin standing on the desk,just say "Bingo" any way.
- 输入
- Three will be two line as input.
The first line is a number N(1<N<65536)
telling you the number of coins on the desk.
The second line is the result with N litters.The letter are "U","D",or "S","U" means the coin is on the right side. "D" means the coin is on the other side ."S" means standing on the desk. - 输出
- If test successeded,just output the possibility of the coin on the right side.If the test failed please output "Fail",If there is one or more"S",please output "Bingo"
- 样例输入
6UUUDDD
- 样例输出
1/2
查看代码---运行号:252134----结果:Accepted
01.
#include <iostream>
02.
#include <cmath>
03.
04.
using
namespace
std;
05.
#define MAX_LEN 65536
06.
int
main()
07.
{
08.
int
nCoins;
//硬币
09.
char
str[MAX_LEN];
10.
11.
cin >> nCoins;
12.
cin >> str;
13.
14.
int
u = 0, d = 0, s = 0;
15.
for
(
int
i = 0; str[i] !=
'\0'
; i++)
16.
{
17.
if
(str[i] ==
'U'
)
18.
u++;
19.
else
if
(str[i] ==
'D'
)
20.
d++;
21.
else
if
(str[i] ==
'S'
)
22.
s++;
23.
}
24.
if
(s != 0)
25.
{
26.
cout <<
"Bingo\n"
;
27.
}
28.
else
if
(
fabs
(1.0*u / (
double
)nCoins - 1.0/2.0) > 0.003)
29.
{
30.
cout <<
"Fail\n"
;
31.
}
32.
else
33.
{
34.
int
r, m = u, n = nCoins;//保存原来的变量值
35.
while
(n)
36.
{
37.
r = m % n;
38.
m= n;
39.
n = r;
40.
}
41.
cout << u / m<<
"/"
<< nCoins / m << endl;
42.
}
43.
return
0;
44.
}
- Coin Test
- Coin Test
- Coin test
- Coin Test
- coin test
- Coin Test
- Coin Test
- Coin Test
- Coin Test
- Coin Test
- Coin Test
- NYOJ - Coin Test
- nyoj-204Coin Test
- nyoj 204-Coin Test
- NYOJ 204 Coin Test
- nyoj-204Coin Test
- 204 Coin Test
- nyoj 204 Coin Test
- 100个乒乓球,两个人轮流取,每人每次最多取5个,最少取1个,如果甲先取,怎样才能取到第100个乒乓球?
- HDU 1047 Integer Inquiry
- Java Map接口简介
- SQL知识
- 三,软件需求分析
- Coin Test
- SQL触发器实例
- 线程实用解析--------(二)创建调用有参函数的线程和线程池简介
- as3结构学习笔记:ENTER_FRAME 监听的函数
- winform窗体缩小到托盘
- union struct 内存对齐
- 求n!最右端非 0的数字!!hdu Last non-zero Digit in N!
- LSMW的流程及其注意事项
- 求一个字符串中连续出现次数最多的子串