LOJ#2095 选数

给定n，k，l，r

问从[l, r]中选出n个数gcd为k的方案数。

解：稍微一想就能想到反演，F(x)就是[l, r]中x的倍数个数的n次方。

后面那个莫比乌斯函数随便怎么搞都行，当然因为这是杜教筛的题就杜教筛了。

然后写一写，交上去80分......

然后枚举一下d是k的多少倍，我们发现F(x)的计算式[(r / x) - ((l - 1) / x)]ⁿ可以整除分块...

然后就做完了。

#include <cstdio>
#include <map>
#include <algorithm>

typedef long long LL;
const int N = 1000010, T = 1000008;
const LL MO = 1000000007;

std::map<LL, LL> mp;
int p[N], top, miu[N];
LL n, k, l, r, Miu[N], L, R;
bool vis[N];

inline void getp(int n) {
    miu[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]) {
            p[++top] = i;
            miu[i] = -1;
        }
        for(int j = 1; j <= top && i * p[j] <= n; j++) {
            vis[i * p[j]] = 1;
            if(i % p[j] == 0) {
                break;
            }
            miu[i * p[j]] = -miu[i];
        }
    }
    for(int i = 1; i <= n; i++) {
        Miu[i] = Miu[i - 1] + miu[i];
    }
    return;
}

inline LL qpow(LL a, LL b) {
    LL ans = 1;
    a %= MO;
    while(b) {
        if(b & 1) ans = ans * a % MO;
        a = a * a % MO;
        b = b >> 1;
    }
    return ans;
}

LL getMiu(LL x) {
    if(x <= 0) return 0;
    if(x <= T) return Miu[x];
    if(mp.count(x)) return mp[x];
    LL ans = 1;
    for(LL i = 2, j; i <= x; i = j + 1) {
        j = x / (x / i);
        ans -= (j - i + 1) % MO * getMiu(x / i) % MO;
        ans %= MO;
    }
    return mp[x] = (ans + MO) % MO;
}

LL F(LL x) {
    LL temp = (R / x) - (L / x);
    return qpow(temp, n);
}

int main() {
    getp(T);
    scanf("%lld%lld%lld%lld", &n, &k, &l, &r);
    LL ans = 0;
    L = (l - 1) / k, R = r / k;
    for(LL i = 1, j; i <= R; i = j + 1) {
        if(L / i) j = std::min(R / (R / i), L / (L / i));
        else if(R / i) j = R / (R / i);
        else j = R;
        ans += F(i) * (getMiu(j) - getMiu(i - 1)) % MO;
        ans %= MO;
    }
    printf("%lld
", (ans + MO) % MO);

    return 0;
}

整除分块的时候要判断是不是0啊......