提交: 198 解决: 10
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题目描述
You are given an integer n.
You are required to calculate (n mod 1) xor (n mod 2) xor ... xor (n mod (n - 1)) xor (n mod n).
The “xor” operation means “exclusive OR”.
You are required to calculate (n mod 1) xor (n mod 2) xor ... xor (n mod (n - 1)) xor (n mod n).
The “xor” operation means “exclusive OR”.
输入
The first line contains an integer T (1≤T≤5) representing the number of test cases.
For each test case, there is an integer n (1≤n≤1012) in one line.
For each test case, there is an integer n (1≤n≤1012) in one line.
输出
For each test case, print the answer in one line.
样例输入
5
1
2
3
4
5
样例输出
0 0 1 1 2
按位求出每一位的值,从左到右第k位为
然后用类欧算分块的每一块(类欧几里得算法小结)
$sum_{i=1}^{n} lfloor frac{n \% i}{2^k}
floor =
sum_{i=1}^{n} lfloor frac{n - lfloor frac{n}{i}
floor i}{2^k}
floor$
#include "bits/stdc++.h" using namespace std; typedef long long ll; bool f(ll a, ll b, ll c, ll n) { if (!a) return (((n + 1) & (b / c)) & 1ll) > 0; if (a >= c || b >= c) { ll temp = (n & 1ll) ? (n + 1) / 2 * n : n / 2 * (n + 1);//先除后乘防止溢出 return ((a / c * temp + (b / c) * (n + 1) + f(a % c, b % c, c, n)) & 1ll) > 0; } else { ll m = (a * n + b) / c; return (((n * m) ^ f(c, c - b - 1, a, m - 1)) & 1) > 0; } } int main() { int _; scanf("%d", &_); while (_--) { ll n; scanf("%lld", &n); ll ans = 0, to = min(30000000ll, n); for (ll i = 1; i < to; i++) { ans = ans ^ (n % i); } for (ll i = to, j; i <= n; i = j + 1) { j = n / (n / i); ll c = 1, ans1 = 0; for (int k = 1; k <= 50; k++) { ans1 += f(n / i, n % j, c, j - i) * c; c <<= 1; } ans ^= ans1; } printf("%lld ", ans); } return 0; }