Xor_Sum 题解

题目

You are given a positive integer \(N(1≦N≦10^{18})\). 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 \(10^9+7\).

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\(a xor b = u\) 和\(a + b= v\)。这里，\(xor\)表示按位异或。对答案取模\(10^9 + 7\)

输入格式

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

一个整数\(N\)

输出格式

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

\(u,v\)对的数量模\(10^9+7\)

输入样例

3

输出样例

5

题解

把\(n=1,2,3,4,5...\)的答案手算出来, 是1, 2, 4, 5, 8, 10, 13, 14, 18, 21, 26, 28, 33, 36, 40, 41, 46, 50, 57, 60...然后找规律, 如果不好找, 可以在这个网站搜索.

用记忆化搜索优化效率, 如果开数组开不下, 用map即可

我怀疑这个不是正解

代码

#include <cstdio>
#include <map>
const long long MOD = 1e9 + 7;
std::map<long long, long long> dp;
long long f(long long x) {
    if (dp[x]) return dp[x] % MOD;
    return dp[x] = (f((x - 1) / 2) + f(x / 2) + f((x - 2) / 2)) % MOD;
}
int main() {
    long long n;
    scanf("%lld", &n);
    dp[0] = 1;
    dp[1] = 2;
    printf("%lld\n", f(n) % MOD);
}