(权限题)NFLSoj #35: Large Banner 题解

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#莫比乌斯反演

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这篇博客详细解说了USACO2010竞赛中的一道题目——NFLSoj #35，探讨了如何运用莫比乌斯反演这一数学工具进行求解，为读者提供了清晰的思路和步骤。

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这好像是USACO2010的原题

A n s = \sum i = 1 m \sum j = 1 n [g c d (i, j) = 1 且 l 2 \leq i 2 + j 2 \leq h 2] * 2 (m - i + 1) (n - j + 1)

$Ans=\sum_{i=1}^{m}\sum_{j=1}^{n}[gcd(i,j)=1 且 l^{2}\leq i^{2}+j^{2}\leq h^{2}]*2(m-i+1)(n-j+1)$

令 f (k) = \sum i = 1 m \sum j = ⌈ l 2 - i 2 \sqrt ⌉ ⌊ h 2 - i 2 \sqrt ⌋ [g c d (i, j) = k] * 2 (m - i + 1) (n - j + 1) ， A n s = f (1)

$令f(k)=\sum_{i=1}^{m}\sum_{j=\left\lceil\ \sqrt{l^{2}-i^{2}} \right\rceil}^{\left\lfloor\ \sqrt{h^{2}-i^{2}} \right\rfloor}[gcd(i,j)=k]*2(m-i+1)(n-j+1)，Ans=f(1)$

F (k) = \sum k | d f (d) = \sum i = 1 ⌊ m k ⌋ 2 (m - i k + 1) \sum j = ⌈ l 2 - ( i k ) 2 \sqrt k ⌉ ⌊ h 2 - ( i k ) 2 \sqrt k ⌋ (n - j k + 1)

$F(k)=\sum_{k\left.\right|d}f(d)=\sum_{i=1}^{\left\lfloor\frac{m}{k}\right\rfloor}2(m-ik+1)\sum_{j=\left\lceil\ \frac{\sqrt{l^{2}-(ik)^{2}}}{k} \right\rceil}^{\left\lfloor \frac{\sqrt{h^{2}-(ik)^{2}}}{k} \right\rfloor}(n-jk+1)$

令 s = ⌈ l 2 - ( i k ) 2 - - - - - - - - \sqrt k ⌉, t = ⌊ h 2 - ( i k ) 2 - - - - - - - - \sqrt k ⌋

$令s=\left\lceil\ \frac{\sqrt{l^{2}-(ik)^{2}}}{k} \right\rceil,t=\left\lfloor\ \frac{\sqrt{h^{2}-(ik)^{2}}}{k} \right\rfloor$

F (k) = \sum i = 1 ⌊ m k ⌋ 2 (m - i k + 1) ( n - s k + 1 ) ( n - t k + 1 ) ( t - s + 1 ) 2

$F(k)=\sum_{i=1}^{\left\lfloor\frac{m}{k}\right\rfloor}2(m-ik+1)\frac{(n-sk+1)(n-tk+1)(t-s+1)}{2}$
*Mobius Inversion

f (k) = \sum k | d F (d) μ (d k)

$f(k)=\sum_{k\left.\right|d}F(d)\mu(\frac{d}{k})$

f (1) = \sum d = 1 m F (d) μ (d)

$f(1)=\sum_{d=1}^{m}F(d)\mu(d)$
Complexity:

O(m)O(m) $O(m)$

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <utility> 
#include <cctype>
#include <algorithm>
#include <bitset>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <cmath>
#define LL long long
#define LB long double
#define x first
#define y second
#define Pair pair<int,int>
#define pb push_back
#define pf push_front
#define mp make_pair
#define LOWBIT(x) x & (-x)
using namespace std;

const int MOD=1e9+7;
const LL LINF=2e16;
const int INF=2e9;
const int magic=348;
const double eps=1e-10;

inline int getint()
{
    char ch;int res;bool f;
    while (!isdigit(ch=getchar()) && ch!='-') {}
    if (ch=='-') f=false,res=0; else f=true,res=ch-'0';
    while (isdigit(ch=getchar())) res=res*10+ch-'0';
    return f?res:-res;
}

LL m,n,l,h,b;
LL mu[2000048];
bool isprime[200048];int prime[200048],tot=0;

void sieve()
{
    int i,j;
    mu[1]=1;
    memset(isprime,true,sizeof(isprime));
    for (i=2;i<=200000;i++)
    {
        if (isprime[i]) prime[++tot]=i,mu[i]=-1;
        for (j=1;j<=tot && (long long)i*prime[j]<=200000;j++)
        {
            isprime[i*prime[j]]=false;
            if (i%prime[j]==0) mu[i*prime[j]]=0; else mu[i*prime[j]]=-mu[i];
            if (i%prime[j]==0) break;
        }
    }
}

int main ()
{
    m=getint();n=getint();l=getint();h=getint();b=getint();
    sieve();
    LL d,i,j,ss,tt;
    LL tmp,ans=0;
    for (d=1;d<=m;d++)
    {
        LL cur_ans=0; 
        for (i=1;i<=m/d;i++)
        {
            if (h*h-i*i*d*d<=0) continue;
            if (l*l-i*i*d*d<=0) ss=1; else ss=ceil(sqrt(l*l-i*i*d*d)/d);
            tt=min((long long)(sqrt(h*h-i*i*d*d)/d),n/d);
            if (ss>tt) continue;
            tmp=(n-d*ss+1+n-d*tt+1);tmp*=(tt-ss+1);tmp%=b;
            tmp*=(m-i*d+1);tmp%=b;
            cur_ans=(cur_ans+tmp)%b;
        }
        cur_ans=((cur_ans*mu[d])%b+b)%b;ans=(ans+cur_ans)%b;
    }
    if (l==1ll) 
    {
        ans=(ans+m*(n+1))%b;
        ans=(ans+n*(m+1))%b;
    }
    ans%=b;
    cout<<ans<<endl;
    return 0;
}