Array

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

Total Submission(s): 339 Accepted Submission(s): 167

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.

Output

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

Sample Input

      3 1 3 5 

Sample Output

      1 4 7 

Source

题目大意：输入数字M，输出前M项和。

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;long long sum[1000];long long num[1000];void Init(){    sum[1]=1;    num[1]=1;    for(int i=2;i<=60;i++)    {        num[i]=num[i-1]*2+1;        sum[i]=sum[i-1]*2+num[i-1]+1;    }}long long Find(long long M){    if(M<=0)        return 0;    int k=lower_bound(num+1,num+55,M)-num;    if(num[k]==M)        return sum[k];    else        return Find(M-num[k-1]-1)+sum[k-1]+M-num[k-1];}int main(){    int t;    long long M;    Init();    scanf("%d",&t);    while(t--)    {        scanf("%lld",&M);        long long ans=Find(M);        printf("%lld\n",ans);    }    return 0;}