hdu-5587 Array（递归）

Array

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 592    Accepted Submission(s): 284

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

 

Sample Output

1
4
7

 

Source

BestCoder Round #64 (div.2)

 

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hujie

 

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规律题；

思路：可以写一下前几项的式子。

1. 1
2. 112
3. 1121223
4. 112122312232334
5. ......

那么求把每项的和的递推公式就是a[n]=2*a[n-1]+2^n-1.

每项有2ⁿ-1个数，然后打表。打到2ⁿ-1<=1e16;

然后递归求解。复杂度为log(n);

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<stdlib.h>
 6 const long long N=1e16;
 7 typedef long long ll;
 8 void  ss(ll n,ll k);
 9 ll a[60];
10 ll sum=0;
11 using namespace std;
12 int main(void)
13 {
14     ll i,j,k,p,q,M;
15     a[1]=1;
16     for(i=2; (ll)1<<i<=N; i++)
17     {
18         a[i]=a[i-1]*2+((ll)1<<(i-1));
19
20     }//打表（ll）1强制将1转为64位的。不然会爆
21     scanf("%lld",&k);//k值可求得为54；
22     while(k--)
23     {
24         sum=0;
25         scanf("%lld",&M);
26         ss(M,0);
27         printf("%lld\n",sum);
28     }
29     return 0;
30
31 }
32 void  ss(ll n,ll k)
33 {
34     ll i,j,p,q;
35     for(i=1; i<=54; i++)
36     {
37         if(n==((ll)1<<(i))-1)//如果所求的位数正好是上面打表中an的某一项就是对应那项加上项数*翻的倍数。
38         {
39             sum+=a[i]+k*(((ll)1<<i)-1);
40             return ;
41         }
42         else if(((ll)1<<(i))-1>n)//找前面的比n小的an
43         {
44             break;
45         }
46
47     }
48     ll pp=k*(((ll)1<<(i-1))-1)+a[i-1]+k+1;//前一项an和（k+1）中间多加的0那个点对应的此时的值。
49     sum+=pp;
50     ll qq=n-((ll)1<<(i-1));//剩下后面没求位数
51     if(qq==0)
52     {
53         return ;
54     }
55     ss(qq,k+1);
56 }