hdu 3354 Probability One
Probability One
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 379 Accepted Submission(s): 293
Then:
1. Ask your friend to compute n1 = 3 * n0 and to tell you if n1 is even or odd.
2. If n1 is even, ask your friend to compute n2 = n1/2. If, otherwise, n1 was odd then let your friend compute n2 = (n1 + 1)/2.
3. Now ask your friend to calculate n3 = 3 * n2.
4. Ask your friend to tell tell you the result of n4 = n3/9. (n4 is the quotient of the division operation. In computer lingo, ’/’ is the integer-division operator.)
5. Now you can simply reveal the original number by calculating n0 = 2 * n4 if n1 was even, or n0 = 2 * n4 + 1 otherwise.
Here’s an example that you can follow: If n0 = 37, then n1 = 111 which is odd. Now we can calculate n2 = 56, n3= 168, and n4 = 18, which is what your friend will tell you. Doing the calculation 2 × n4 + 1 = 37 reveals n0.
The last line of the input file has a single zero (which is not part of the test cases.)
k. B Q
Where k is the test case number (starting at one,) B is either ’even’ or ’odd’ (without the quotes) depending on your friend’s answer in step 1. Q is your friend’s answer to step 4.
Note: There is a blank space before B.
38
0
2. even 19
//0MS 228K 366 B G++
/* 题意:
题目看似很难.其实是水题.直接按它的过程计算一遍即可得出解 */
#include<stdio.h>
int main(void)
{
int n;
int k=;
while(scanf("%d",&n),n)
{
int odd=;
n*=;
if(n%){
odd=;
n=(n+)/;
}else n/=;
n*=;
n/=;
if(odd) printf("%d. odd %d\n",k++,n);
else printf("%d. even %d\n",k++,n);
}
return ;
}
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