HDU 5587 Array

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.\left( 1 \leq T \leq 2 * {10}^{3} \right)(1≤T≤2∗10​3​​) Next T line contains, each line contains one interger M. \left( 1\leq M \leq {10}^{16} \right)(1≤M≤10​16​​)

Output

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

Sample Input

3
1
3
5

Sample Output

1
4
7
转换为2进制做类似数位dp就好了
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 5;
int n;
long long m;
long long a[maxn], b[maxn];

void init()
{
    for (int i = a[b[0] = 0] = 1; i <= 100; i++) a[i] = a[i - 1] << 1;
    for (int i = 1; i <= 100; i++) b[i] = (b[i - 1] << 1) + a[i - 1];
    for (int i = 0; i <= 100; i++) a[i]--;
}

long long solve(LL x,LL y)
{
    for (int i = 0; i <= 100; i++) 
    if (x > a[i] && x <= a[i + 1])
    {
        if (x == a[i + 1])    return b[i + 1] + y*a[i + 1];
        else return b[i] + y*(a[i] + 1) + 1 + solve(x - a[i] - 1, y + 1);
    }
}

int main()
{
    init();
    while (scanf("%d", &n) != EOF)
    {
        while (n--)
        {
            scanf("%I64d", &m);
            printf("%I64d\n", solve(m, 0));
        }
    }
    return 0;
}