hdu 5587 Array

Array

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 255    Accepted Submission(s): 133

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

3135

 

Sample Output

147

 

Source

BestCoder Round #64 (div.2)

 

题目大意：

Vicky是个热爱数学的魔法师，拥有复制创造的能力。一开始他拥有一个数列{1}。每过一天，他将他当天的数列复制一遍，放在数列尾，并在两个数列间用0隔开。Vicky想做些改变，于是他将当天新产生的所有数字（包括0）全加1。Vicky现在想考考你，经过100天后，这个数列的前M项和是多少？。

解题思路：

先打个表，不可能达到 100天，因为最大是10^16次方的和。。。

然后推一下每个0也就是 1 的值就行了。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <algorithm>#include <set>using namespace std;#define MM(a) memset(a,0,sizeof(a))typedef long long LL;typedef unsigned long long ULL;const int maxn = 65;const int mod = 1e9+7;const double eps = 1e-10;const int INF = 0x3f3f3f3f;LL gcd(LL a, LL b){    if(b == 0)        return a;    return gcd(b, a%b);}LL sum[maxn];void Init()///每一个0的值{    sum[0] = 1;    LL tmp = 1;    for(int i=1; i<maxn; i++)///10^16次方最多不超过60天    {        sum[i] = (sum[i-1]-1)*2+tmp+1;        tmp *= 2;    }}int main(){    Init();//    for(int i=0; i<63; i++)//        cout<<sum[i]<<" ";    int T;    cin>>T;    while(T--)    {        LL m;        cin>>m;        LL ret, tmp, x, y;        ret = x = tmp = 0;        y = 1;        while(m)        {            if(m%2)            {                ret += sum[tmp]+x;                x += y;                ///cout<<ret<<endl;            }            y<<=1;            tmp++;            m>>=1;        }        cout<<ret<<endl;    }    return 0;}