http://acm.hdu.edu.cn/showproblem.php?pid=5524

 

Subtrees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 184    Accepted Submission(s): 99


Problem Description
There is a complete binary tree with N nodes.The subtree of the node i has Ai nodes.How many distinct numbers are there of Ai?
 

 

Input
There are multiple test cases, no more than 1000 cases.
For each case contains a single integer N on a line.(1N1018)
 

 

Output
The output of each case will be a single integer on a line:the number of subtrees that contain different nodes.
 

 

Sample Input
5
6
7
8
 

 

Sample Output
3
4
3
5
 

 

Source
 
 
在Hack的时候截别人的代码, 然而我并不懂什么意思, 但相信自己以后会会的!!!

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

typedef long long ll;

using namespace std;

ll ans=0 , Max=0, n;

void search(ll x)
{
    ll lc=x, rc=x, dep=0;
    while(lc*2<=n) lc *= 2, dep++;
    while(rc*2+1<=n) rc = rc*2 + 1;

    if(lc<=rc) Max = max(Max, dep);
    else
    {
        search(x*2);
        search(x*2+1);
        ans++;
    }
}

int main()
{
    while(~scanf("%I64d", &n))
    {
        ans = 0;
        Max = 0;
        search(1);
        printf("%I64d\n", ans+Max+1);
    }
    return 0;
}
代码1:
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<set>

using namespace std;
typedef long long ll;
set<ll> st;
set<ll>::iterator it;

void deal(ll n)
{
    ll k, t;
    if((it=st.find(n)) !=st.end())
        return ;
    st.insert(n);
    for(k=62; k>=0; k--)
    {
        if((1ll<<k)&(n+1))break;
    }
    t = n - (1ll<<k) + 1;
    deal((1ll<<(k-1)) +min(t, 1ll<<(k-1))-1);
    deal((1ll<<(k-1)) +max(0ll, t-(1ll<<(k-1)))-1);
}

int main()
{
    ll n;

    while(~scanf("%I64d", &n))
    {
        st.clear();
        st.insert(0);
        st.insert(1);
        deal(n);
        printf("%d\n", st.size()-1);
    }
    return 0;
}
代码2:

 

posted on 2015-11-03 15:09  栀蓝  阅读(232)  评论(0编辑  收藏  举报

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