BestCoder Round #64 (div.2) HDOJ5587 Array(dfs)

Array

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

Total Submission(s): 490    Accepted Submission(s): 235


Problem Description

Vicky is a magician who loves math. She has great power in copying and creating.

One day she gets an array {1}。 After that, every day she copies all the numbers in the arrays she has, and puts them into the tail of the array, with a signle '0' to separat.

Vicky wants to make difference. So every number which is made today (include the 0) will be plused by one.

Vicky wonders after 100 days, what is the sum of the first M numbers.

 

Input

There are multiple test cases.

First line contains a single integer T, means the number of test cases.(1≤T≤2∗103)

Next T line contains, each line contains one interger M. (1≤M≤1016)

 

Output

For each test case,output the answer in a line.

 

Sample Input

3
1
3
5

 

题目链接:点击打开链接

a[i]为i的二进制表示下1的个数, 预处理得到f, g数组, 对每次读入的m进行dfs即可.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 105;
ll m, f[MAXN], g[MAXN];
void magic()
{
f[1] = 1, g[1] = 1;
for(int i = 2; i <= 63; ++i)
f[i] = f[i - 1] * 2 + 1;
for(int i = 2; i <= 63; ++i)
g[i] = g[i - 1] * 2 + f[i - 1] + 1;
}
ll dfs(ll x)
{
if(x <= 1) return x;
int y = lower_bound(f + 1, f + 63, x) - f;
if(f[y] == x) return g[y];
return g[y - 1] + dfs(x - f[y - 1] - 1) - f[y - 1] + x;
}
int main(int argc, char const *argv[])
{
magic();
int t;
scanf("%d", &t);
while(t--) {
scanf("%lld", &m);
printf("%lld\n", dfs(m));
}
return 0;
}