(Bestcoder Round #64 (Div.2)) Array--Bestcoder Round #64 (Div.2)

Bestcoder Round #64 (Div.2)

Array

Problem Description

Vicky is a math-loving magician with the ability to replicate and create. At first he had a sequence of {1}. Every day, he copies the sequence of his day, puts it at the end of the series, and divides it by 0 between the two series. Vicky wanted to make some changes, so he added all the new figures for the day (including 0) to 1. Vicky now want to test you, after 100 days, this series of the first m and how much?

Enter a description

Enter more than one set of data. The first line contains an integer t, which represents the number of data groups. T.\left (1 \leq T \leq 2 * {10}^{3} \right)(1≤T≤2∗10?3??The first row of each group of data contains an integer m.\left (1\leq M \leq {10}^{16} \right)< Span class= "Minner textstyle uncramped" > (1≤m≤< Span class= "Mord" >10 ?16 ?? )    /span>                

Output Description

For each set of data output a row of answers.

Input Sample

3135

Output Sample

147

Hint

The first item is always a number11, so sample 1 output11 The next day first copy, separated by 0, get {1,0,1}, and then add the resulting number 1, get {1,1,2}, so the sample 2 output the first 3 items and1+1+2=41+ 1+2=4. The third day first get { 1,1,2,0,1,1,2}, then get {1,1,2,1,2,2,3}, so sample 3 outputs the first 5 items and 1+1+2+1+2=71+1+2 +1+2= 7               

#include <stdio.h>#include<string.h>#defineN 110000#defineOO 0xFFFFFLong Longsum[ the], a[ the];Long LongFind (Long LongNintnn) {    if(nn<=0)        return 0; if(n<A[nn])returnFind (N, nn-1); Else if(n==A[nn])returnSum[nn]; Else    returnSUM[NN] + Find (n-a[nn]-1, nn) + N-a[nn];}intMain () {intI, nn; Long LongW =1e17; sum[1] =1, a[1] =1;  for(i=2; i< the; i++) {A[i]= a[i-1]*2+1; Sum[i]= sum[i-1]*2+ a[i-1] +1; if(a[i]>w) {nn=i;  Break; }    }    intT; scanf ("%d", &T);  while(t--)    {        Long LongN, ans; scanf ("%i64d", &N); Ans=Find (n, nn); printf ("%i64d\n", ans); }    return 0;}

(Bestcoder Round #64 (Div.2)) Array--Bestcoder Round #64 (Div.2)