HDU 6033 Add More Zero (数学)

Add More Zero

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

Total Submission(s): 302    Accepted Submission(s): 223


Problem Description

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 between 0 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 from 1to 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.

 

Input

The input contains multiple test cases. Each test case in one line contains only one positive integer m,
satisfying 1≤m≤105.

 

Output

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.

 

Sample Input

1
64

 

Sample Output

Case #1: 0
Case #2: 19

 

Source

2017 Multi-University Training Contest - Team 1 

 

题意：
求出最大k，使10^k>=2^m-1，给出m。

point：
你敢相信这种签到题我们这种残疾队做了1个半小时？？？？？根本没往数学方面想，光想着打表暴力求解了。
一个简单的答案 k=m*(log2/log10).向下取整。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <map>
#include <algorithm>
#include <math.h>
using namespace std;
#define rt x<<1|1
#define lt x<<1
#define  LL long long

int main()
{
int m;
int p=0;
while(~scanf("%d",&m))
{
printf("Case #%d: ",++p);
printf("%.0f\n",floor(m*1.0*(log(2.0)/log(10.0))));
}
}

现场AC的代码，因为精度问题WA了5发也是辛苦我了。//赛中知道正规的解题方案很想一头撞死了，高中数学老师都要提刀砍我了。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <map>
#include <algorithm>
#include <math.h>
using namespace std;
#define rt x<<1|1
#define lt x<<1
#define  LL long long
int ans[100000+5];
int main()
{
LL now;
for(int i=1;i<=59;i++)
{
now=(LL)pow(2,i);
int l=0;
while(now)
{
now/=10;
l++;
}
ans[i]=l-1;
}
now=(LL)pow(2,59);
for(int i=60;i<=100000;i++)
{
now=now*2;
if(now>=1e18)
{
ans[i]=ans[i-1]+1;
now=now/10;
}
else
{
ans[i]=ans[i-1];
}

}
int n;
int p=0;
while(~scanf("%d",&n))
{
printf("Case #%d: %d\n",++p,ans
);
}
return 0;
}