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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 500 Accepted Submission(s): 189
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
#include <math.h>
#include <iostream>
#include <string>
#include <algorithm>
#define PI acos(-1.0)
#define INF 1e9
using namespace std;
int dd[100];
int main()
{
int t;
scanf("%d", &t);
int n;
memset(dd, 0, sizeof(dd));
int ans;
int e=1;
dd[e++]=1;
ans=1;
int i=1;
while(ans < INF )
{
ans=ans+pow(2, i);
i++;
dd[e++]=ans;
}
// printf("-----%d\n", i);
int cnt=1;
while(t--)
{
scanf("%d", &n);
int pos;
for(i=0; i<e; i++)
{
if(n>=dd[i] && n<dd[i+1])
{
pos=i; break;
}
}
printf("Case #%d: %d\n", cnt++, pos );
}
return 0;
}
hdu 2015校赛1002 Dual horsetail (思维题 )
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原文地址:http://www.cnblogs.com/yspworld/p/4377036.html