3518: 点组计数

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3518: 点组计数

Time Limit: 20 Sec Memory Limit: 128 MB
Submit: 113 Solved: 62
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Description

平面上摆放着一个n*m的点阵（下图所示是一个3*4的点阵）。Curimit想知道有多少三点组(a，b，c)满足以a，b，c三点共线。这里a，b，c是不同的3个点，其顺序无关紧要。(即(a，b，c)和
(b，c，a)被认为是相同的）。由于答案很大，故你只需要输出答案对1，000，000，007的余数就可以了。

Input

有且仅有一行，两个用空格隔开的整数n和m。

Output

有且仅有一行，一个整数，表示三点组的数目对1，000，000，007的余
数。（1，000。000。007是质数）

Sample Input

3 4

Sample Output

2 0

HINT

对于100%的数据，1< =N．m< =50000

Source

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#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#include<vector>#include<queue>#include<set>#include<map>#include<stack>#include<bitset>#define min(a,b) ((a) < (b) ? (a) : (b))#include<ext/pb_ds/priority_queue.hpp>using namespace std; const int maxn = 5E4 + 50;typedef long long LL;const LL mo = 1000000007; int n,m,Ans,tot,mu[maxn],pri[maxn],sum[maxn];bool not_pri[maxn]; inline int Mul(const LL &x,const LL &y) {return x * y % mo;}inline int Add(const int &x,const int &y) {return x + y < mo ? x + y : x + y - mo;}inline int Dec(const int &x,const int &y) {return x - y >= 0 ? x - y : x - y + mo;} int C(int N){    int A = N,B = N - 1,C = N - 2;    if (A % 2 == 0) A /= 2;    else if (B % 2 == 0) B /= 2;    else if (C % 2 == 0) C /= 2;    if (A % 3 == 0) A /= 3;    else if (B % 3 == 0) B /= 3;    else if (C % 3 == 0) C /= 3;    return Mul(A,Mul(B,C));} inline int Sum(int tmp,int N,int M){    int A = Mul(N,M);    int B = Mul(sum[N],tmp);    return Dec(A,B);} inline int Solve(int d,int N,int M){    int ret = 0;    for (int t = 1; t <= N; t++)    {        if (!mu[t]) continue;        int A = Sum(Mul(d,t),N / t,n);        int B = Sum(Mul(d,t),M / t,m);        ret = Add(ret,Mul(mu[t],Mul(A,B)));    }    return ret;} int Calc(){    int ret = 0;    for (int d = 2; d <= n; d++)        ret = Add(ret,Mul(d - 1,Solve(d,n / d,m / d)));    return ret;} int main(){    #ifdef DMC        freopen("DMC.txt","r",stdin);    #endif         cin >> n >> m;    if (n > m) swap(n,m); mu[1] = 1;    for (int i = 2; i < maxn; i++)    {        if (!not_pri[i])            pri[++tot] = i,mu[i] = mo - 1;        for (int j = 1; j <= tot; j++)        {            int Nex = i * pri[j];            if (Nex >= maxn) break;            not_pri[Nex] = 1;            if (i % pri[j] == 0) break;            mu[Nex] = Mul(mu[i],mo - 1);        }    }    for (int i = 1; i < maxn; i++) sum[i] = Add(i,sum[i - 1]);    Ans = Add(Mul(C(n),m),Mul(C(m),n));    Ans = Add(Ans,Mul(2,Calc())); cout << Ans << endl;    return 0;}

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