HDU-4059 The Boss on Mars 容斥定理

将与N不互质的数全部找出来，再应用容斥定理求得最后的结果。这题在求 a/b MOD c 的时候使用费马小定理等价于 a*b^c(-2) MOD c.用x/lnx 可以估计出大概有多少素数。

代码如下：

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#define MOD 1000000007LL

using namespace std;



typedef long long int Int64;

typedef long long int LL;



Int64 rec[55000000], res;



int idx;



Int64 N, sum;



Int64 _pow(Int64 a, Int64 b)

{

    a %= MOD;

    Int64 ret = 1;

    while (b) {

        if (b & 1) {

            ret *= a;

            ret %= MOD;

        }

        a *= a;

        a %= MOD;

        b >>= 1;

    }

    return ret;

}



void PowMode(  )

{

    res = 1;

    LL a = 30LL;

    LL b = MOD  - 2;

    while( b )

    {

         if( b&1 ) res = (res*a)%MOD;

         a = ( a * a )%MOD;

         b >>= 1;

    }

}



Int64 Get_sum( Int64 N )

{

   Int64 ans = N;

   ans = ( ans * ( N + 1 ) )%MOD;

   ans = ( ans * ( 2 *N + 1 ) )%MOD;

   ans = ( ans * ( ( 3 * N * N )%MOD + (3*N)%MOD - 1 + MOD)%MOD )%MOD;

   return (ans * res)%MOD;

}



void cut(Int64 x)

{

    idx = 0;

    int LIM = (int)sqrt(double(x));

    if (!(x & 1)) {

        rec[++idx] = 2;

        while (!(x & 1)) {

            x /= 2;

        }

    }

    for (int i = 3; i <= LIM; i += 2) {

        if (x % i == 0) {

            rec[++idx] = i;

            while (x % i == 0) {

                x /= i;

            }

        }

    }

    if (x != 1) {

        rec[++idx] = x;

    }

}



void dfs(int p, Int64 num, int sign)

{

    if (p == idx) {

        if (num > 1) {

            sum += sign * (_pow(num, 4) * Get_sum(N/num)) % MOD;

            sum = ((sum % MOD) + MOD) % MOD;

        }

        return;

    }

    dfs(p+1, num, sign);

    dfs(p+1, num*rec[p+1], -sign);

}



int main()

{

    int T;

    PowMode();

    scanf("%d", &T);

    while (T--) {

        sum = 0;

        scanf("%I64d", &N);

        cut(N);

        dfs(0, 1, -1);

        printf("%I64d\n", (((Get_sum(N)-sum) % MOD) + MOD) % MOD);

    }

    return 0;

}