hdu4059 The Boss on Mars(差分+容斥原理)

题意：

求小于n (1 ≤ n ≤ 10^8)的数中，与n互质的数的四次方和。

知识点：

差分：

一阶差分： 设 

则    为一阶差分。

二阶差分：

n阶差分：     且可推出   

性质： 1.

2.

差分序列：

给你一列数 a[i][1]，a[i][2]，a[i][3]，a[i][4]，a[i][5]……

那么a[i][j]=a[i-1][j+1]-a[i-1][j]， 即后一行是上一行相邻两项的差（第一行除外）。

如果给你一个多项式， 比如 f（x）=(x+1)*(x+2)*……*（x+p）,即多项式最高项指数为p。

则得到的差分序列有如下性质：

1. f(0),f(1)…f(p)组成多项式的第一行，后面的差分序列可以由上一行推出。第p+1行以后差分序列的值都为0。

2.我们这里要用的差分序列是其第0行对角线的数。 我们设他们为c0、c1、c2、……cp；   则：

第n项的值：f（n）=c0*C(n,0)+c1*C(n,1)+c2*C(n,2)+……+cp*C(n,p)；

前n项的值：Sum（n）=c0*C(n+1,1)+c1*C(n+1,2)+c2*C(n+1,3)+……+cp*C(n+1,p+1)；

把求前n项和组合公式给化简出来Sum（n）=(n^5)/5+(n^4)/2+(n^3)/3-n/30

=(n*(n+1)*(2n+1)*(3*n*n+3*n-1))/30

参考文献

题解：反面考虑，容斥原理，sum(n)=1^4+2^4+…n^4=(n*(n+1)*(2n+1)*(3*n*n+3*n-1))/30,减掉与n不互质的数4次方，将n质因子分解后减掉一个因子的倍数的4次方结果，加上两个因子乘积的倍数的4次方结果，减去……以此类推。

其中还涉及逆元，因为MOD为素数，用费马小定理求。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
const LL MOD=1e9+7;
const LL NN=1e8+5;
const int N=1e4+5;
const LL ni=233333335;  //30 mod MOD 的逆
LL n,syz[15],ans;
int ycnt;

LL mutisum(LL n)
{
    LL ans1=1;
    //long long 范围<18,446,744,073,709,551,616 约10^20    30*MOD> LL范围，此法不可，实力被坑
    //LL mod=30*MOD;
    //ans1=(((((n*(n+1))%mod)*(2*n+1))%mod)*((3*n*n+3*n-1)%mod))%mod;
    ans1=(((((((n*(n+1))%MOD)*(2*n+1))%MOD)*((3*n*n+3*n-1)%MOD))%MOD)*ni)%MOD;
    //ans1/=30;
    return ans1%MOD;
}

int prime[N];
bool vis[N];
int pcnt;
void is_prime()
{
    pcnt=0;
    memset(vis,0,sizeof(vis));
    for(int i=2;i<N;i++)
    {
        if(!vis[i])
        {
            prime[pcnt++]=i;
            for(int j=i+i;j<N;j+=i)
                vis[j]=1;
        }
    }
}

void fenjie(LL n1)
{
    ycnt=0;
    for(int i=0;i<pcnt&&prime[i]<=n1;i++)
    {
        if(n1%prime[i]==0)
            syz[ycnt++]=prime[i];
        while(n1%prime[i]==0)
            n1/=prime[i];
    }
    if(n1>1)
        syz[ycnt++]=n1;
}

void dfs(int c,int cur,int j,LL ans1)      //dfs(c,1,0,1);
{
    if(cur==c+1)
    {
        LL nn=(n-1)/ans1,as1=ans1%MOD;
        if(c&1)
            ans-=(((((((mutisum(nn)*as1)%MOD)*as1)%MOD)*as1)%MOD)*as1)%MOD;
        else
            ans+=(((((((mutisum(nn)*as1)%MOD)*as1)%MOD)*as1)%MOD)*as1)%MOD;
        ans%=MOD;
        return;
    }
    for(;j<ycnt;j++)
    {
        dfs(c,cur+1,j+1,ans1*syz[j]);
    }
}

void test()
{
    for(int i=0;i<ycnt;i++)
        cout<<syz[i]<<' ';
    cout<<endl;

}

int main()
{
    int t;
    scanf("%d",&t);
    is_prime();
    while(t--)
    {
        scanf("%lld",&n);
        if(n==1)
        {
            printf("1\n");
            continue;
        }
        fenjie(n);
        ans=mutisum(n-1);
        for(int c=1;c<=ycnt;c++)
            dfs(c,1,0,1);
        if(ans<0)
            ans=(ans+MOD)%MOD;
        printf("%lld\n",ans);
       // test();
    }
}

The Boss on Mars

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4059

Description

On Mars, there is a huge company called ACM (A huge Company on Mars), and it’s owned by a younger boss.

Due to no moons around Mars, the employees can only get the salaries per-year. There are n employees in ACM, and it’s time for them to get salaries from their boss. All employees are numbered from 1 to n. With the unknown reasons, if the employee’s work number is k, he can get k^4 Mars dollars this year. So the employees working for the ACM are very rich.

Because the number of employees is so large that the boss of ACM must distribute too much money, he wants to fire the people whose work number is co-prime with n next year. Now the boss wants to know how much he will save after the dismissal. 

 

Input

The first line contains an integer T indicating the number of test cases. (1 ≤ T ≤ 1000) Each test case, there is only one integer n, indicating the number of employees in ACM. (1 ≤ n ≤ 10^8) 

 

Output

For each test case, output an integer indicating the money the boss can save. Because the answer is so large, please module the answer with 1,000,000,007. 

 

Sample Input

2
4
5

 

Sample Output

82
354

Hint

 Case1: sum=1+3*3*3*3=82 Case2: sum=1+2*2*2*2+3*3*3*3+4*4*4*4=354