ARC 066D Xor Sum AtCoder - 2272 （打表找规律）

Problem Statement

You are given a positive integer N. Find the number of the pairs of integers u and v(0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xorb=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 10^9+7.

Constraints

• 1≦N≦10^{18}

Input

The input is given from Standard Input in the following format:

N

Output

Print the number of the possible pairs of integers u and v, modulo 10^9+7.

Sample Input 1

3

Sample Output 1

5

The five possible pairs of u and v are:

• u=0,v=0 (Let a=0,b=0, then 0 xor 0=0, 0+0=0.)

• u=0,v=2 (Let a=1,b=1, then 1 xor 1=0, 1+1=2.）

• u=1,v=1 (Let a=1,b=0, then 1 xor 0=1, 1+0=1.）

• u=2,v=2 (Let a=2,b=0, then 2 xor 0=2, 2+0=2.）

• u=3,v=3 (Let a=3,b=0, then 3 xor 0=3, 3+0=3.）

Sample Input 2

1422

Sample Output 2

52277

Sample Input 3

1000000000000000000

Sample Output 3

787014179
 
思路：
打出前面一些项的答案：
1, 2, 4, 5, 8, 10, 13, 14, 18, 21, 26, 28, 33, 36, 40, 41, 46, 50, 57, 60, 68, 73, 80, 82, 89, 94, 102, 105, 112, 116, 121, 122, 128, 133, 142, 146, 157, 164, 174, 177, 188, 196, 209, 214, 226, 233, 242, 244, 253, 260, 272, 277, 290, 298, 309, 312, 322, 329,
可以发现这样的规律： 
d

a(2k) = 2*a(k-1) + a(k);
a(2k+1) = 2*a(k) + a(k-1).

然后用数组记录前面一些项目，然后开一个map记录中间递归过程访问过的变量（即，记忆话搜索）

然后直接递归处理上面发现的递归式即可了。

这也是OEIS中的一个数列。

http://oeis.org/A007729

A007729

6th binary partition function.

4

1, 2, 4, 5, 8, 10, 13, 14, 18, 21, 26, 28, 33, 36, 40, 41, 46, 50, 57, 60, 68, 73, 80, 82, 89, 94, 102, 105, 112, 116, 121, 122, 128, 133, 142, 146, 157, 164, 174, 177, 188, 196, 209, 214, 226, 233, 242, 244, 253, 260, 272, 277, 290, 298, 309, 312, 322, 329, 340, 344 (list; graph; refs; listen; history; text; internal format)

	OFFSET	0,2
	COMMENTS	From Gary W. Adamson, Aug 31 2016: (Start) The sequence is the left-shifted vector of the production matrix M, with lim_{k->inf} M^k. M =   1, 0, 0, 0, 0, ...   2, 0, 0, 0, 0, ...   2, 1, 0, 0, 0, ...   1, 2, 0, 0, 0, ...   0, 2, 1, 0, 0, ...   0, 1, 2, 0, 0, ...   0, 0, 2, 1, 0, ...   0, 0, 1, 2, 0, ...   ... The sequence is equal to the product of its aerated variant by (1,2,2,1): (1, 2, 2, 1) * (1, 0, 2, 0, 4, 0, 5, 0, 8, ...) = (1, 2, 4, 5, 8, 10, ...). Term a((2^n) - 1) = A007051: (1, 2, 5, 14, 41, 122, ...). (End) a(n) is the number of ways to represent 2n (or 2n+1) as a sum e_0 + 2*e_1 + ... + (2^k)*e_k with each e_i in {0,1,2,3,4,5}. - Michael J. Collins, Dec 25 2018
	LINKS	Alois P. Heinz, Table of n, a(n) for n = 0..10000 Michael J. Collins, David Wilson, Equivalence of OEIS A007729 and A174868, arXiv:1812.11174 [math.CO], 2018. B. Reznick, Some binary partition functions, in "Analytic number theory" (Conf. in honor P. T. Bateman, Allerton Park, IL, 1989), 451-477, Progr. Math., 85, Birkhäuser Boston, Boston, MA, 1990.
	FORMULA	G.f.: (r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) = (1 + 2x + 2x^2 + x^3 + 0 + 0 + 0 + ...). - Gary W. Adamson, Sep 01 2016 a(2k) = 2*a(k-1) + a(k); a(2k+1) = 2*a(k) + a(k-1). - Michael J. Collins, Dec 25 2018
	MAPLE	b:= proc(n) option remember;       `if`(n<2, n, `if`(irem(n, 2)=0, b(n/2), b((n-1)/2) +b((n+1)/2)))     end: a:= proc(n) option remember;       b(n+1) +`if`(n>0, a(n-1), 0)     end: seq(a(n), n=0..70);  # Alois P. Heinz, Jun 21 2012
	MATHEMATICA	b[n_] := b[n] = If[n<2, n, If[Mod[n, 2] == 0, b[n/2], b[(n-1)/2]+b[(n+1)/2]]]; a[n_] := a[n] = b[n+1] + If[n>0, a[n-1], 0]; Table[a[n], {n, 0, 70}] (* Jean-François Alcover, Mar 17 2014, after Alois P. Heinz *)
	CROSSREFS	A column of A072170. Cf. A002487, A007051. Apart from an initial zero, coincides with A174868. Sequence in context: A179509 A157007 A173509 * A174868 A268381 A186349 Adjacent sequences:  A007726 A007727 A007728 * A007730 A007731 A007732
	KEYWORD	nonn
	AUTHOR	N. J. A. Sloane
	EXTENSIONS	More terms from Vladeta Jovovic, May 06 2004
	STATUS	approved

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=;while(b){if(b%)ans=ans*a%MOD;a=a*a%MOD;b/=;}return ans;}
inline void getInt(int* p);
const int maxn=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
const ll mod=1e9+7ll;
ll a[]={, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , };
map<ll,ll> m;
ll f(ll k)
{
//    cout<<k<<endl;
    if(k<=)
    {
        return a[k];
    }
    if(m[k])
    {
        return m[k];
    }
    if(k&)
    {
        return m[k]=(2ll*f(k/2ll)%mod+f(k/2ll-1ll)%mod)%mod;
    }else
    {
        return m[k]=(2ll*f(k/2ll-1ll)%mod+f(k/2ll)%mod)%mod;
    }
}
int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    ll n;
    cin>>n;
    cout<<f(n)<<endl;
 
    return ;
}
 
inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '');
        while ((ch = getchar()) >= '' && ch <= '') {
            *p = *p *  - ch + '';
        }
    }
    else {
        *p = ch - '';
        while ((ch = getchar()) >= '' && ch <= '') {
            *p = *p *  + ch - '';
        }
    }
}