HDU 5587 Array

Array

                                                              Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                  Total Submission(s): 411    Accepted Submission(s): 207

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)

 首先先把每一次变化后的字符个数与每一次变化后的和分别存放在a[i], sum[i], 然后搜索前n项，如何a[i] == n 说明这个长度恰好是n，sum[n]就是前n想和，否则就找剩下的部分,由于剩下的部分是a[i]一样，并且每一项+1，所以将这些提取出来，转化成前a[i]个，所以可以用之前的继续搜索

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <vector>#include <map>#include <list>#include <cmath>#include <queue>#include <stack>#include <set>#include <algorithm>#define LL long long#define INF 0x3f3f3f3f#define RR freopen("in.txt","r",stdin)#define WW freopen("out.txt","w",stdout)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;const LL maxn = 1e16+10;LL a[1010] = {0, 1}, sum[1010] = {0, 1};LL DFS(LL n){    if(n == 0)        return 0;    int pos;    for(int i=1; i<=100; i++)    {        if(a[i+1] > n)        {            pos = i;            break;        }    }    if(a[pos] == n)        return sum[pos];    else        return sum[pos] + DFS(n-a[pos]-1) + n - a[pos];}int main(){    for(int i=2; i<=110; i++)    {        a[i] = a[i-1] * 2 + 1;        sum[i] = sum[i-1] * 2 + a[i-1] + 1;        if(a[i] > maxn)            break;    }    int T;    scanf("%d",&T);    while(T--)    {        LL n;        scanf("%lld",&n);        printf("%lld\n",DFS(n));    }    return 0;}