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;
}