//求1到n之间与n互质的数的四次方的和 //segma(n^4) = (6n^5+15n^4+10n^3-n)/30 //对于(a/b)%mod能够转化为(a*inv(b))%mod //inv(b)为b的逆元 //由费马小定理a^(p-1) = 1(modp) a , p 互质可得 //30的逆元为30^(mod-2) //由容斥原理非常easy得到与n不互质的数之和为 //对于全部的n的素数因子 //一个素数因子的全部数的四次方之和-有两个素数因子的全部数的四次方之和+有三个. ... /…

The Boss on Mars Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1528    Accepted Submission(s): 452 Problem Description On Mars, there is a huge company called ACM (A huge Company on Mars), and…

传送门 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)…

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 salarie…

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…

The Boss on Mars Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2494    Accepted Submission(s): 775 Problem Description On Mars, there is a huge company called ACM (A huge Company on Mars), an…

The Boss on Mars Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2327    Accepted Submission(s): 718 Problem Description On Mars, there is a huge company called ACM (A huge Company on Mars), and…

The Boss on Mars Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1934    Accepted Submission(s): 580 Problem Description On Mars, there is a huge company called ACM (A huge Company on Mars), an…

The Boss on Mars Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1335    Accepted Submission(s): 401 Problem Description On Mars, there is a huge company called ACM (A huge Company on Mars), an…

The Boss on Mars Problem's Link Mean: 给定一个整数n,求1~n中所有与n互质的数的四次方的和.(1<=n<=1e8) analyse: 看似简单,倘若自己手动推公式的话,还是需要一定的数学基础. 总的思路:先求出sum1=(1^4)+(2^4)+...(n^4),再求出sum2=(1~n中与n不互质的数的四次方的和),answer=sum1-sum2. 如何求sum1呢? 有两种方法: 1.数列差分.由于A={Sn}={a1^4+a2^4+...an^4}…