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): 1335    Accepted Submission(s): 401 Problem Description On Mars, there is a huge company called ACM (A huge Company on Mars), an…

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…

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)…

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}…

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…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4059 题目大意: 给一个n,求1~n中与n互质的数的4次方的总和. 解题思路: 容斥原理.逆元.公式. 其实是蛮简单的一道题.囧囧. 首先必须知道1^4+2^4+...+n^4=n*(n+1)*(2n+1)(3n^2+3n-1)/30. 除以30可以转化成乘以30的逆元,30^(M-2) 由费马小定理很快可以得到. 先求出1^4+2^4+...+n^4然后减去与n不互质的.  假设n=p1^a1*…