[zoj3629]找规律

题意：a[n] = ([n/1] + [n/2] + ... + [n/n]) & 1 == false，找出a数组的规律来就ok了。

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep(a, b) for (int a = 0; a < b; a++)
#define rrep(a, b) for (int a = (b - 1); a >= 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pc(a) putchar(a)
#define ps(a) printf("%s", a)
#define pd(a) printf("%d", a)
#define sd(a) scanf("%d", &a)

typedef double db;
typedef long long LL;
typedef unsigned long long uLL;
typedef pair<int, int> pii;
typedef multiset<int> msi;
typedef set<int> si;
typedef vector<int> vi;
typedef map<int, int> mii;

const int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
const int maxn = 5 * 1e4 + 7;
const int maxm = 1e5 + 7;
const int minv = 1e7 + 7;
const int max_val = 1e6 + 7;
const int MD = 1e9 +7;
const LL INF = 1e15;
const double PI = acos(-1.0);
const double eps = 1e-10;

template<class T> T gcd(T a, T b) { return b == 0? a : gcd(b, a % b); }

uLL a, b;

uLL g(LL n) { return n * (2 * n - 1); }

uLL f(uLL x) {
    uLL l = 1, r = 0xFFFFFFFF;
    while (l < r) {
        uLL m = (l + r) >> 1;
        if (m * m < x) l = m + 1;
        else r = m;
    }
    uLL n = l, rest = x - (n - 1) * (n - 1), ans = 0;
    if (n & 1) ans += rest;
    ans += g(n >> 1);
    return ans;
}

int main() {
    //freopen("in.txt", "r", stdin);
    while (cin >> a >> b) {
        a++; b++;
        cout << f(b) - f(a - 1) << endl;
    }
    return 0;
}