General Formula for the principle of hdu 4059 refresh

In fact, the difficulty of this question lies in the details of Large Numbers crossing the border, taking the remainder, and so on.

First, find the generic formula

n(2n+1)(n+1)(3n² +3n-1)/30

There are two ways to evaluate (a/B) % mod

1. Original formula = a % (B * n)/B

2. Original formula = a * B ^ (phi (mod)-1) % mod; (in which B interacts with mod)

In this question, because mod is too large, the first method will cause an error, so the second method is used.

For example, four times of 2, four times of 4, and four times of 6 are added in the Process of rejection .... It can be extracted four times of 2, and the other item is changed to four vertices.

View Code

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
const __int64 mod = 1000000007;
__int64 n;
__int64 power(__int64 a,__int64 b,__int64 c)
{
    __int64 ret=1;
while(b>0)
    {
if(b&1)
            ret=ret*a%c;
        a=a*a%c;
        b>>=1;
    }
return ret;
}
__int64  calc(__int64 n)
{
       __int64 a=n,b=(2*n+1),c=(n+1),d=(3*n*n+3*n-1)%mod;
       __int64 sum=1;
       sum*=a;sum%=mod;
       sum*=b;sum%=mod;
       sum*=c;sum%=mod;
       sum*=d;sum%=mod;
       __int64 t=power(30,mod-2,mod);
return sum*t%mod;
}
__int64 sici(__int64 n)
{
    __int64 sum=1;
for(int i=1;i<=4;i++)
    {
        sum*=n;
if(sum>=mod) sum%=mod;
    }
return sum;
}
__int64 solve(__int64 r,__int64 n)
{
    vector<__int64> p;
    __int64 i;
for(i=2;i*i<=r;i++)
    {
if(r%i==0)
        {
            p.push_back(i);
while(r%i==0) r/=i;
        }
    }
if(r>1) p.push_back(r);
    __int64 sum=0;
for(__int64 num=1;num<(1<<p.size());num++)
    {
        __int64 mult=1,ones=0;
for(i=0;i<p.size();i++)
        {
if(num&(1<<i))
            {
                ones++;
                mult*=p[i];
if(mult>=mod) mult%=mod;
            }
        }
if(ones%2) sum+=sici(mult)*calc(n/mult);
else sum-=sici(mult)*calc(n/mult);
        sum%=mod;
    }
return sum;
}
int main()
 {
     __int64 t,i,j;
     scanf("%I64d",&t);
while(t--)
     {
         scanf("%I64d",&n);
         printf("%I64d\n",(calc(n)-solve(n,n)%mod+mod)%mod);
     }
return 0;
 }