hdu6275 Mod, Xor and Everything（分块，类欧几里得）

时间限制: 10 Sec  内存限制: 128 MB
提交: 198  解决: 10
[提交] [状态] [命题人:admin]

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”.

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.

输出

For each test case, print the answer in one line.

样例输入

复制样例数据

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;
}