hihocoder #1456 : Rikka with Lattice(杜教筛)

题意

给你一个(n*m)方格图,统计上面有多少个格点三角形,除了三个顶点,不覆盖其他的格点(包括边和内部).

答案对于(998244353)取模... ((n,m le 5 * 10^9))

题解

这个题十分的巧妙... 集训时是大佬ztzshiwo出的..

据他所说,是不那么杜教筛的杜教筛QAQ

考试时候提示了一个皮克定理...

皮克定理:

[S=a+frac{b}{2}-1 ]

(S)为格点多边形面积,(a)为多边形内部点数,(b)为多边形边上点数.

然而我还是只会暴力,正解是真的太神了啊QAQ.

我们考虑一个(a*b)的矩形,以它对顶点为端点的三角形,只当(a ot b)时存在四个解.

这个我是听wearry证明的qwq 十分巧妙

简单的证明:

代入皮克定理,可知

当且仅当格点三角形面积为(S=0+frac{3}{2}-1=frac{1}{2})时才计入答案

我们可以用叉积的形式表示这个面积. (S=frac{1}{2} |AB| |BC| sin heta =frac{1}{2} |vec {AB} imes vec{AC}|).(高中必修内容QAQ)

 我们令之前那个矩形的一条对角线为三角形的一条边,

令左下角为向量出发点,也就是其中一个顶点,然后这条边的向量坐标表达就为((a,b)).

我们令另外一条边为((i,j)),然后三角形面积就是(frac{1}{2}|(a,b) imes (i,j)| = frac{1}{2}|aj - bi|).

这个为(frac{1}{2})所以(|aj-bi|=1). ( herefore aj-bi= pm 1)

我们是要求解((i,j)) 所以不难发现这是一个扩欧的形式,当且仅当(a ot b)时有整数解.

又(ecause 0 < i < a, 0 < j < b).. ( herefore)可以通过扩欧的相邻解确定在这个区域仅一解.

所以(pm 1)各有一解,换个对角线又有对称的一组解.所以最后总共(2*2=4)组解.

所以我们要求的就是原图中每个矩形的贡献就行了...

此处(n),(m)都是要减一的... (至于为啥...手推就知道了QAQ)

[displaystyle mathrm {ans}=sum_{i=1}^{n} sum _{j=1}^{m} 4 cdot (n-i)(m-j) [i ot j] ]

[displaystyle =4sum_{i=1}^{n} sum_{j=1}^{m} (n-i)(m-j) sum _ {x|gcd (i,j)} mu(x) ]

[=displaystyle 4 sum_{x=1}^{min(n,m)} mu(x) sum_{i=1}^{lfloor frac{n}{x} floor} sum_{j=1}^{lfloor frac{m}{x} floor} nm - mix - njx + ijx^2 ]

令$$displaystyle S(i)=sum_{i=1}^{n} i = frac{n(n+1)}{2}$$.

[displaystyle mathrm{sum1}=nm sum_{x=1}^{min(n,m)} mu(x) lfloor frac{nm}{x^2} floor ]

[displaystyle mathrm{sum2}=m sum_{x=1}^{min(n,m)} (mu(x) cdot x) S(lfloor frac{n}{x} floor) lfloor frac{m}{x} floor ]

[displaystyle mathrm{sum3}=n sum_{x=1}^{min(n,m)} (mu(x) cdot x) lfloor frac{n}{x} floor S(lfloor frac{m}{x} floor) ]

[displaystyle mathrm{sum4}=sum_{x=1}^{min(n,m)} (mu(x) cdot x^2) S(lfloor frac{n}{x} floor) S(lfloor frac{m}{x} floor) ]

[mathrm{ans}=mathrm{sum1}-mathrm{sum2}-mathrm{sum3}+mathrm{sum4} ]

这个用根号分块就能做到(Theta (n+sqrt {n}))复杂度啦... 具体推导证明看我的一篇博客线性筛与莫比乌斯反演.

然而这并不能满分...fuck

所以就有杜教筛卡了30分.

(displaystyle sum _{i=1}^{n} mu(i))之前那篇博客杜教筛小结中有介绍.

然后就介绍另外两个套路求的东西吧..

令(displaystyle id(x)=x, mx(x)= mu(i) i). (然后之后默认把第一个字母大写记作前缀和比如(displaystyle Id(x)=sum_{i=1}^{x} id(i) = frac{x(x+1)}{2}))

所以就有

[mx * id (n) ]

[displaystyle = sum _{d|n} mu(d) cdot d cdot frac{n}{d} =displaystyle sum _{d|n} mu(d) cdot n ]

[=displaystyle n sum_{d|n} mu(d) = n cdot [n=1] = epsilon ]

代入之前的套路式子就有

[displaystyle 1 - Mx(n) = sum _{i=2}^{n} i cdot Mx(lfloor frac{n}{i} floor) ]

然后就可以尝试推出(displaystyle sum _{i=1}^{n} mu(i) cdot i cdot i).

这个也不麻烦QAQ...

然后本人比较懒 就直接用c++11 中的unordered_map了(这个基于哈希算法)

有些地方有点细节(5*10^9 * 5*10^9 = 2.5 * 10^{19})会爆long long所以很多地方都要记得先取模!!!

代码

#include <bits/stdc++.h>
#define For(i, l, r) for(register ll i = (l), _end_ = (ll)(r); i <= _end_; ++i)
#define Fordown(i, r, l) for(register ll i = (r), _end_ = (ll)(l); i >= _end_; --i)
#define Set(a, v) memset(a, v, sizeof(a))
using namespace std;

typedef long long ll;
inline ll read() {
    ll x = 0, fh = 1; char ch = getchar();
    for (; !isdigit(ch); ch = getchar() ) if (ch == '-') fh = -1;
    for (; isdigit(ch); ch = getchar() ) x = (x<<1) + (x<<3) + (ch ^ '0');
    return x * fh;
}

void File() {
#ifdef zjp_shadow
    freopen ("1456.in", "r", stdin);
    freopen ("1456.out", "w", stdout);
#endif
}

const ll Mod = 998244353;

ll n, m;

const int N = 1e7 + 1e3;
int prime[N], cnt = 0;
int Limit = N - 1e3;

ll mux[N], muxx[N], mu[N];
bitset<N> is_prime;

void Init(int maxn) {
    int res;
    mu[1] = 1;
    is_prime.set(); is_prime[0] = is_prime[1] = false;
    For (i, 2, maxn) {
        if (is_prime[i]) { prime[++cnt] = i; mu[i] = -1; }
        For (j, 1, cnt) {
            res = prime[j] * i;
            if (res > maxn) break;
            is_prime[res] = false;
            if (i % prime[j]) mu[res] = -mu[i];
            else { mu[res] = 0; break ; }
        }
    }
    For (i, 1, maxn) {
        mux[i] = mux[i - 1] + 1ll * mu[i] * i % Mod;
        mux[i] = (mux[i] % Mod + Mod) % Mod;

        muxx[i] = muxx[i - 1] + 1ll* mu[i] * i % Mod * i % Mod;
        muxx[i] = (muxx[i] % Mod + Mod) % Mod;

        mu[i] += mu[i - 1];
        mu[i] = (mu[i] % Mod + Mod) % Mod;
    }
}

ll fpm(ll x, ll power) { ll res = 1; x %= Mod; for (; power; power >>= 1, (x *= x) %= Mod) if (power & 1) (res *= x) %= Mod; return res; }
const ll inv2 = fpm(2, Mod - 2), inv6 = fpm(6, Mod - 2);
ll Sum(ll x) { x %= Mod; return (x + 1) * x % Mod * inv2 % Mod; }

ll sum1, sum2, sum3, sum4;
ll Nextx;

unordered_map<ll, ll> MU, MUX, MUXX;
ll mu_(ll x) {
    if (x <= Limit) return mu[x];
    if (MU.count(x)) return MU[x];
    ll res = 1, Nextx;
    For (i, 2, x) { Nextx = x / (x / i); (res += Mod - (Nextx - i + 1) * mu_(x / i) % Mod) %= Mod; i = Nextx; }
    return (MU[x] = res);
}

inline ll Sum1(ll x) { x %= Mod; return x * (x + 1) % Mod * inv2 % Mod; }
ll mux_(ll x) {
    if (x <= Limit) return mux[x];
    if (MUX.count(x)) return MUX[x];
    ll res = 1, Nextx;
    For (i, 2, x) { Nextx = x / (x / i); (res += Mod - (Sum1(Nextx) - Sum1(i - 1) + Mod) * mux_(x / i) % Mod) %= Mod; i = Nextx; }
    return (MUX[x] = res);
}

inline ll Sum2(ll x) { x %= Mod ; return x * (x + 1) % Mod * (2 * x + 1) % Mod * inv6 % Mod; }
ll muxx_(ll x) {
    if (x <= Limit) return muxx[x];
    if (MUXX.count(x)) return MUXX[x];
    ll res = 1, Nextx;
    For (i, 2, x) { Nextx = x / (x / i); (res += Mod - (Sum2(Nextx) - Sum2(i - 1) + Mod) * muxx_(x / i) % Mod) %= Mod; i = Nextx; }
    return (MUXX[x] = res);
}

int main () {
    File();
    n = read(); m = read();
    if (n > m) swap(n, m);
    Init(Limit);

    For (x, 1, n) {
        Nextx = min(n / (n / x), m / (m / x));
        (sum1 += (mu_(Nextx) - mu_(x - 1)) * (n / x) % Mod * (m / x) % Mod * n % Mod * m % Mod) %= Mod;
        (sum2 += (mux_(Nextx) - mux_(x - 1)) * Sum(n / x) % Mod * (m / x) % Mod) %= Mod;
        (sum3 += (mux_(Nextx) - mux_(x - 1)) * Sum(m / x) % Mod * (n / x) % Mod) %= Mod;
        (sum4 += (muxx_(Nextx) - muxx_(x - 1)) * Sum(n / x) % Mod * Sum(m / x) % Mod) %= Mod;
        x = Nextx;
    }
    ll ans = sum1 - 1ll * m * sum2 % Mod - 1ll * n * sum3 % Mod + sum4;
    ans = (ans % Mod + Mod) % Mod;
    ans = ans * 4ll % Mod;
    printf ("%lld
", ans);
    return 0;
}