HDU 4059 The Boss on Mars(容斥原理 + 四次方求和)

时间：2021-01-29 08:29:18

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The Boss on Mars

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2462    Accepted Submission(s): 760

Problem 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
 
Author
ZHANG, Chao
 
Source
 

题目大意：

给定 T 组数据。每组数据一个数 n ，然后让你求 <n且与 n 互素的数的四次方。详见例子。

解题思路：

就是就一个 S = a1^4 + a2^4 + ... + ak^4;

这个我们能够考虑容斥原理：也就是 1^4 + 2^4 + 3^4 + ... + n^4 - 不互素的数的四次方。

然后遇到了一个 非常严重的问题 就是 1^4 + 2^4 + 3^4 + ... + n^4怎么求；

首先 要推一下Sum =  1 + 2 + 3 + ... +n

(n+1)^2 - n^2 = 2*n+1;

n^2 - (n-1)^2 = 2*(n-1)+1;

...

3^2 - 2^2 = 2*2+1;

2^2 - 1^2 = 2*1+1;

将上述等式 左边加左边 右边加右边得到：

(n+1)^2 - 1 = 2*(1+2+3+...+n)+n*1;

又由于 我们所求 Sum =  1 + 2 + 3 + ... +n;

所以：2*Sum + n = (n+1)^2-1 =n^2+2*n+1-1 = n^2+2*n;

Sum = (n^2+n)/2。

所以 我们要求的四次方之和 就得用到 (n+1)^5 - n^5。(三次方的公式 首先得知道 就是通过刚才那个推出来的)

(x+1)^5 = x^5+5*x^4+10*x^3+10*x^2+5*x+1;

推完之后就是以下这个公式 还是挺费事儿的：

(1^4+2^4+……+n^4)=(n*(n+1)*(2n+1)*(3*n*n+3*n-1))/30;在这里 我们还要求一下 30 关于 MOD = 1e9+7的逆元，这个能够通过扩展欧几里得算法实现...

然后 就是 分析一下 代码的实现过程了：

第一步：我们首先进行素数筛 ；

第二步：我们就要进行素因子分解；

第三步：写一个高速幂 计算 X^4对 MOD取模

第四步：二进制枚举 。奇加偶减；

完事儿。

My Code：

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

typedef long long LL;

const LL MOD = 1e9+7;

const LL MAXN = 1e6+5;

LL Scan()///输入外挂

{

    LL res=0,ch,flag=0;

    if((ch=getchar())=='-')

        flag=1;

    else if(ch>='0'&&ch<='9')

        res=ch-'0';

    while((ch=getchar())>='0'&&ch<='9')

        res=res*10+ch-'0';

    return flag?-res:res;

}

inline void Out(LL a)///输出外挂

{

    if(a>9)

        Out(a/10);

    putchar(a%10+'0');

}

LL n;

/**

void Ex_gcd(int a, int b, int &x, int &y)

{

    if(b == 0)

    {

        x = 1;

        y = 0;

        return;

    }

    int x1, y1;

    Ex_gcd(b, a%b, x1, y1);

    x = y1;

    y = x1 - (a/b)*y1;

}**/

bool prime[MAXN];

int p[MAXN];

LL k;

void isprime()

{

    memset(prime, false, sizeof(prime));

    k = 0;

    for(LL i=2; i<MAXN; i++)

    {

        if(!prime[i])

        {

            p[k++] = i;

            for(LL j=i*i; j<MAXN; j+=i)

                prime[j] = true;

        }

    }

}

int fac[MAXN/100], num[MAXN/100];

LL cnt;

void Dec(LL m)

{

    cnt = 0;

    memset(num, 0, sizeof(num));

    for(LL i=0; i<k&&p[i]*p[i]<=m; i++)

    {

        if(m%p[i] == 0)

        {

            fac[cnt] = p[i];

            while(m%p[i]==0)

            {

                m /= p[i];

                num[cnt]++;

            }

            cnt++;

        }

    }

    if(m > 1)

    {

        fac[cnt] = m;

        num[cnt++] = 1;

    }

    /**cout<<cnt<<endl;

    for(int i=0; i<cnt; i++)

        cout<<fac[i]<<" ";*/

}

LL Get_ans(LL x)

{

    LL ans = x%MOD;

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

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

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

    ans = ans*233333335%MOD;///233333335是逆元

    return ans;

}

LL quick_mod(LL a, LL b)

{

    LL ans = 1;

    while(b)

    {

        if(b & 1)

            ans = ans*a%MOD;

        b>>=1;

        a = a*a%MOD;

    }

    return ans%MOD;

}

int main()

{

    isprime();

    ///Dec(6);

    ///int x, y;

    ///Ex_gcd(30,MOD, x, y)

    int T;

    T = Scan();

    while(T--)

    {

        n = Scan();

        Dec(n);

        LL res = 0;

        for(LL i=1; i<(1<<cnt); i++)

        {

            LL ans = 1, sum = 0;

            for(LL j=0; j<cnt; j++)

            {

                if(i & (1<<j))

                {

                    sum++;

                    ans = ans*fac[j]%MOD;

                }

            }

            if(sum&1)

                res = (res+Get_ans(n/ans)*quick_mod(ans,4)+MOD)%MOD;

            else

                res = (res-Get_ans(n/ans)*quick_mod(ans,4)+MOD)%MOD;

        }

        res = (Get_ans(n)-res+MOD)%MOD;

        res=(res%MOD+MOD)%MOD;

        printf("%lld\n",res);

    }

    return 0;

}

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