Vida says she's part Elf: that at least one of her ancestors was an Elf. But she doesn't know if it was a parent (1 generation ago), a grandparent (2 generations ago), or someone from even more generations ago. Help her out!
Being part Elf works the way you probably expect. People who are Elves, Humans and part-Elves are all created in the same way: two parents get together and have a baby. If one parent is?monospace; font-size: 13px;">A/B?Elf, and the other parent is?C/D?Elf, then their baby will be(A/B?+?C/D)?/?2?Elf. For example, if someone who is?0/1?Elf and someone who is?1/2Elf have a baby, that baby will be?1/4?Elf.
Vida is certain about one thing: 40 generations ago, she had?240?different ancestors, and each one of them was?1/1?Elf or?0/1?Elf.
Vida says she's?P/Q?Elf. Tell her what is the minimum number of generations ago that there could have been a?1/1?Elf in her family. If it is not possible for her to be?P/Q?Elf, tell her that she must be wrong!
The first line of the input gives the number of test cases,?T.?T?lines follow. Each contains a fraction of the form?P/Q, where?P?and?Q?are integers.
For each test case, output one line containing "Case #x: y", where x is the test case number (starting from 1) and y is the minimum number of generations ago a 1/1 Elf in her family could have been if she is?P/Q?Elf. If it's impossible that Vida could be?P/Q?Elf, y should be the string "impossible" (without the quotes).
1 ≤?T?≤ 100.
1 ≤?P?<?Q?≤ 1000.
P?and?Q?have no common factors. That means?P/Q?is a fraction in?lowest terms.
1 ≤?P?<?Q?≤ 1012.
P?and?Q?may?have common factors.?P/Q?is?not?guaranteed to be a fraction in lowest terms.
5 1/2 3/4 1/4 2/23 123/31488
Case #1: 1 Case #2: 1 Case #3: 2 Case #4: impossible Case #5: 8
Note that the fifth sample case does not meet the limits for the Small input. Even if you don't solve it correctly, you might still have solved the Small input correctly.
In the first sample case, Vida could have a?1/1?Elf parent and a?0/1?Elf parent. That means she could have had a 1/1 Elf one generation ago, so the answer is 1.
In the second sample case, Vida could have had a?1/1?Elf parent and a?1/2?Elf parent. That means she could have had a?1/1?Elf one generation ago, so the answer is 1.
In the third sample case, Vida could have had a?0/1?Elf parent and a?1/2?Elf parent. The1/2?Elf parent could have had a?1/1?Elf parent and a?0/1?Elf parent. That means she could have had a?1/1?Elf two generations ago, so the answer is 2.
In the fourth sample case, it's impossible to be exactly?2/23?Elf if your ancestors 40 generations ago were all?0/1?Elf or?1/1?Elf.
Yes, Vida has a lot of ancestors. If that is the part of the problem that seems the most unrealistic to you, please re-read the part about Elves.
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时间搞错了，没来得及去做google 的2014 jam，不过没啥，因为做题的速度远跟不上。。。。，人老了，智商也不够，也就只能做个A。开始还把题目意思搞错了。。。要求的是最近的一代。我当时求得是，最少要经过多少代，才能得到P/Q,感觉我理解的更难。
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#include <stdio.h>
#include <iostream>
long long gcd(long long P,long long Q){
long long r =Q;
do{
r = Q;
Q = P%Q;
P = r;
}while(Q!=0);
return r;
}
bool simple(long long P,long long Q,long long &op,long long &oq)
{
if(P>Q)
return false;
long long u = gcd(Q,P);
op = P/u;
oq = Q/u;
return true;
}
int main(int argc ,char *argv[])
{
int N=0;
scanf("%d",&N);
long long P,Q;
long long op,oq;
for(int i=0;i<N;i++){
scanf("%I64d/%I64d",&P,&Q);
simple(P,Q,op,oq);
long long j=1;
int count =0;
for(;j<oq;j<<=1){count++;}
if(j != oq)
{
printf("Case #%d: impossible\n",i+1);
continue;
}
j=1;
int count2=0;
while(j<=op) {j<<=1;}
long long v = oq/j;
while(v!=0){
v >>=1;
count2++;
}
printf("Case #%d: %d\n",i+1,count2);
}
}
?
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思路：
?
?
如果改成计算，至少要经过多少代，才能得到P/Q.那么上面的需要修改下。
结果为：
代数 = op 中1的个数+ log(oq)-1;
?
记1{x}为二进制x1的个数
?
如果op = 5，那么1{op}=2，如果op=16；那么需要经过2+4-1=5代才能得到5/16;
?
1{x}
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