多校1 A Add More Zero

Add More Zero

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0


[align=left]Problem Description[/align]
There is a youngster known for amateur propositions concerning several mathematical hard problems.

Nowadays, he is preparing a thought-provoking problem on a specific type of supercomputer which has ability to support calculations of integers between0
and (2m−1)
(inclusive).

As a young man born with ten fingers, he loves the powers of
10
so much, which results in his eccentricity that he always ranges integers he would like to use from1
to 10k
(inclusive).

For the sake of processing, all integers he would use possibly in this interesting problem ought to be as computable as this supercomputer could.

Given the positive integer m,
your task is to determine maximum possible integer
k
that is suitable for the specific supercomputer.
 

[align=left]Input[/align]
The input contains multiple test cases. Each test case in one line contains only one positive integerm,
satisfying 1≤m≤105.
 

[align=left]Output[/align]
For each test case, output "Case #x:y"
in one line (without quotes), where x
indicates the case number starting from 1
and y
denotes the answer of corresponding case.
 

[align=left]Sample Input[/align]

1
64

 

[align=left]Sample Output[/align]

Case #1: 0
Case #2: 19

 

问给一个n,问2^n-1用科学计数法表示，10上面的的数。一开始和队友打表找规律，找半天头都懵了。结果别人说代码就10行，静下来一想就是对10取对数嘛，用一下log函数就行了。。。。。感觉自己真是很菜。
码是我们4级都没过的队长打的

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
int a[11111];
int main()
{
int cas=0;
long long int n,k;
long long int cmp,res;
while(cin>>n>>k)
{
cas++;
cmp=2*(n-1);
res=(k-n)%cmp;
if(n>=k)
printf("Case #%d: %d\n",cas,k);
else if(res==0)
printf("Case #%d: %d\n",cas,n);
else if(res<=cmp/2)
printf("Case #%d: %d\n",cas,res);
else
printf("Case #%d: %d\n",cas,res-cmp/2);
}
return 0;
}