找规律

题目

You are given a positive integer N(1≦N≦1e18). 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 xor b=u and a+b=v. Here, xor denotes the bitwise exclusive OR. Since it can be extremely large, compute the answer modulo 1e9+7.

Input

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

Output

u,v 对的数量 模1e9+7

Sample Input 1

3

Sample Output 1

5

Sample Input 2

1422

Sample Output 2

52277

Sample Input 3

1000000000000000000

Sample Output 3

787014179

题意

给出正整数N,求出整数对u和v(0≤u,v≤N)的数目，使得存在两个非负整数a和b满足a xo rb=u 和a+b=v。这里，xor表示按位异或。对答案取模1e9+7

分析

这个题就是找规律，然后递归找答案，当然，递归是需要记忆化搜索的，不然可能会爆掉。如果写递推也是可以的（当然我写的递归）；

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int Mod = 1e9+7;
map<ll, ll>ans;
ll f(ll x){
    if (ans[x])return ans[x] % Mod;
    return ans[x]=(f(x/2)+f((x-1)/2)+f((x-2)/2))%Mod;
}
int main(){
    ll n;
    scanf("%lld",&n);
    ans[0] = 1;
    ans[1] = 2;
    printf("%lld\n", f(n) % Mod);
}

 

 

PS：感谢zhl大佬提供的查找公式的网站。