hdu 5587 Array 二分

最新推荐文章于 2025-12-17 18:40:36 发布

转载最新推荐文章于 2025-12-17 18:40:36 发布 · 93 阅读

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原文链接：http://www.cnblogs.com/jhz033/p/6627709.html

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#java

Vicky是一位热爱数学的魔法师，她使用一种特殊的复制与创建技巧处理数组。本篇介绍了如何通过这种技巧处理初始数组{1}

Array

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

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

BestCoder Round #64 (div.2)

题目链接：点击传送

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+30010,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=2147493647;
set<ll>s;
set<ll>::iterator it;
ll dfs(ll n)
{
    //cout<<n<<endl;
    if(n==0) return 0;
    it=s.find(n);
    if(it!=s.end())
       return 2*dfs(n/2)+n-n/2;
    it=s.upper_bound(n);
    it--;
    ll x=n-(*it);
    ll z=(*it);
    return dfs(z)+dfs(x-1)+n-z;
}
int main()
{
    for(int i=0;i<55;i++)
        s.insert((1LL<<i)-1);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll n;
        scanf("%lld",&n);
        ll ans=dfs(n);
        printf("%lld\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/jhz033/p/6627709.html