Dylans loves numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 569 Accepted Submission(s): 320
Problem Description
Who is Dylans?You can find his ID in UOJ and Codeforces.
His another ID is s1451900 in BestCoder.
And now today‘s problems are all about him.
Dylans is given a number N.
He wants to find out how many groups of "1" in its Binary representation.
If there are some "0"（at least one）that are between two "1",
then we call these two "1" are not in a group,otherwise they are in a group.
Input
In the first line there is a number T.
T is the test number.
In the next T lines there is a number N.
0≤N≤1018,T≤1000
Output
For each test case,output an answer.
Sample Input
1
5
Sample Output
2
Source
快要炸了..
tle成狗
因为是tle,看了下自己没有写cin cout,估计就是算法的问题...
我是先存了二进制的每一位到数组,然后扫一遍...
嗯,这都tle...
那我不存不扫,直接记录当前二进制位和之前二进制位..
logn的复杂度总可以了吧啊?
还TLE..........
嗯,其实已经发现 n是小于等于1e18的,没开long long
但是一位没开long long 会是wa...就没理...
之后实在黔驴技穷,改了下,竟然过了...
然后想明白了.
因为存二进制的时候有一个while
没开long long 的话就炸了,不知道读进去的是什么,while就出不来,于是就tle了.T T
果然太年轻.
/*************************************************************************
> File Name: code/bc/#45/1001.cpp
> Author: 111qqz
> Email: [email protected]
> Created Time: 2015年07月29日 星期三 13时25分01秒
************************************************************************/
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=1E3+5;
int a[N];
LL n,nn;
int main()
{
int T;
cin>>T;
while (T--)
{
scanf("%lld",&n);
nn = n ;
int k = 0;
while (nn)
{
k++;
a[k]=nn&1;
nn = nn >>1;
}
int ans = 0;
// for ( int i = 1 ; i <= k ; i++ )
// cout<<a[i]<<endl;
for ( int i = 1 ; i <= k ; i++ )
{
if (a[i]==1&&a[i-1]==0)
ans++;
}
printf("%d\n",ans);
}
return 0;
}