[CQOI2015] 选数

Portal

重庆的题目质量还是不错的.

这个刚好为\(k\)就是一个暗示, 暗示我们枚举超集/子集进行容斥.

容斥做法:

咕咕咕

---

以下是反演做法.

设\(F(n)\)表示n == k的时候题目所求.

设\(G(n)\)表示k | n 的时候题目所求.

那么\([l, r]\)范围内有\(\frac{r}{k} - \frac{l - 1}{k}\)个k的倍数

那么G(n) = \((\frac{r}{k} - \frac{l - 1}{k}) ^ N\)

所以F(n) = \(\sum_{n | d} \mu(\frac{d}{n}) * (\frac{H}{d} - \frac{L - 1}{d}) ^ N\)

\[ = \sum_{s} \mu(s) (\frac{H}{sk} - \frac{L - 1}{sk}) ^ N ​\]

然后直接杜教筛就可以了

其实Mu反演也是一种容斥

Upd 20190109:

关于这类反演, 有几个可以注意的地方, 帮我们发现如何进行反演:

关于枚举子集/超集: 简单来说哦就是如何好算如何算.

发现这题是一个区间内的问题,一个区间内的倍数的数量可直接由除法原理求得.

如果要是枚举约数, 考虑这个式子:
\[ \sum_{i = 1}^{n} \sigma_0(n) = \sum_{i} \lfloor \frac{n}{i} \rfloor \]
如果i从1开始, 还是可以考虑枚举因数的.

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
using namespace std;
#define rep(i, a, b) for(int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for(int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
typedef long long LL;
typedef long double LD;
int read() {
    char ch = getchar();
    int x = 0, flag = 1;
    for (;!isdigit(ch); ch = getchar()) if (ch == '-') flag *= -1;
    for (;isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
    return x * flag;
}
void write(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) write(x / 10);
    putchar(x % 10 + 48);
}

const int MaxDta = 1e5 + 9, Threshold = 1e6 + 9, Mod = 1e9 + 7;
static int prime[Threshold], tot, mu[Threshold], pSumMu[Threshold];
static bool isnprime[Threshold];
__gnu_pbds :: gp_hash_table <int, int> prefixMu;

int fpm(int base, int tims) { 
    int r = 1;
    while (tims) {
        if (tims & 1) r = 1ll * r * base % Mod;
        base = 1ll * base * base % Mod;
        tims >>= 1;
    }
    return r;
}

void linearSieve() {
    mu[1] = 1;
    rep (i, 2, Threshold - 1) {
        if (!isnprime[i]) prime[++tot] = i, mu[i] = -1;
        for (int k, j = 1; j <= tot && (k = i * prime[j]) < Threshold; ++j) {
            isnprime[k] = 1;
            if (i % prime[j] == 0) {
                mu[k] = 0;
                break;
            } else mu[k] = -mu[i];
        }
    }

    rep (i, 1, Threshold - 1) pSumMu[i] = pSumMu[i - 1] + mu[i];
}

int muSum(int Bound) {
    if (Bound < Threshold) return pSumMu[Bound];
    if (prefixMu[Bound]) return prefixMu[Bound];

    int Sum = 1;
    for (int l = 2, r; l <= Bound; l = r + 1) {
        r = Bound / (Bound / l);
        Sum -= (r - l + 1ll) * muSum(Bound / l);
    }

    return prefixMu[Bound] = Sum;
}

static int N, K, L, H;

void init() {
    linearSieve();
    N = read(), K = read(), L = read() - 1, H = read();
    H /= K; L /= K;
}

void solve() {
    LL ans = 0;

    for (int l = 1, r; l <= H; l = r + 1) {
        if (L / l == 0) r = H / (H / l); else r = min(H / (H / l), L / (L / l));
        (ans += 1ll * (muSum(r) - muSum(l - 1) + Mod) % Mod * fpm(H / l - L / l, N) % Mod) %= Mod;
    }

    cout << ans << endl;
}

int main() {
//  freopen("LG3172.in", "r", stdin);
//  freopen("LG3172.out", "w", stdout);

    init();
    solve();

#ifdef Qrsikno
    debug("\nRunning time: %.3lf(s)\n", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
    return 0;
}