先放知识点: 莫比乌斯反演 卢卡斯定理求组合数 乘法逆元 快速幂取模 GCD of Sequence Alice is playing a game with Bob. Alice shows N integers a 1, a 2, -, a N, and M, K. She says each integers 1 ≤ a i ≤ M. And now Alice wants to ask for each d = 1 to M, how many different sequences b…

题意:给出序列[a1..aN],整数M和k,求对1-M中的每个整数d,构建新的序列[b1...bN],使其满足: 1. \(1 \le bi \le M\) 2. \(gcd(b 1, b 2, -, b N) = d\) 3. 恰好有k个位置 \(bi!=ai\) 求对每个d,有多少种满足条件的序列 分析:对于前两个条件,就是单纯的莫比乌斯反演. 令\(F(d) = [d|gcd(b1...bN)]\) \(f(d) = [gcd(b1...bN)]=d]\) 则$f(n) = \sum_{x…

HDU 1061 题目大意:给定数字n(1<=n<=1,000,000,000),求n^n%10的结果 解题思路:首先n可以很大,直接累积n^n再求模肯定是不可取的, 因为会超出数据范围,即使是long long也无法存储. 因此需要利用 (a*b)%c = (a%c)*(b%c)%c,一直乘下去,即 (a^n)%c = ((a%c)^n)%c; 即每次都对结果取模一次 此外,此题直接使用朴素的O(n)算法会超时,因此需要优化时间复杂度: 一是利用分治法的思想,先算出t = a^(n/2),若…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1097 分析:简单题,快速幂取模, 由于只要求输出最后一位,所以开始就可以直接mod10. /*A hard puzzle Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 33036 Accepted Submission(s): 11821 Pr…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2817 解题思路:arithmetic or geometric sequences 是等差数列和等比数列的意思, 即令输入的第一个数为a(1),那么对于等差数列 a(k)=a(1)+(k-1)*d,即只需要求出 a(k)%mod   又因为考虑到k和a的范围, 所以对上式通过同余作一个变形:即求出 (a(1)%mod+(k-1)%mod*(d%mod))%mod 对于等比数列 a(k)=a(1)*q…

GCD of Sequence Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 46    Accepted Submission(s): 14 Problem Description Alice is playing a game with Bob.Alice shows N integers a1, a2, …, aN, and M,…

Describtion First we define: (1) lcm(a,b), the least common multiple of two integers a and b, is the smallest positive integer that is divisible by both a and b. for example, lcm(2,3)=6 and lcm(4,6)=12. (2) gcd(a,b), the greatest common divisor of tw…

http://www.cnblogs.com/BLADEVIL/p/3490321.html http://www.cnblogs.com/zyfzyf/p/3997986.html 翻了翻题解,这两个合起来比较明白…… 题意:求1~n!中与m!互质的数的数量(mod R). ∵由欧几里得算法得gcd(a,b)=gcd(b,a%b) ∴gcd(a+b,b)=gcd(b,(a+b)%b)=gcd(b,a) 即 gcd(a,b)=gcd(a+b,b) 推广:gcd(a,b)=gcd(a+k*b,b)…

题意: 给出\(M\)和\(a数组\),询问每一个\(d\in[1,M]\),有多少组数组满足:正好修改\(k\)个\(a\)数组里的数使得和原来不同,并且要\(\leq M\),并且\(gcd(a_1,a_2,\dots,a_n)=d\). 思路: 对于每一个\(d\),即求\(f(d)\):修改\(k\)个后\(gcd(a_1,a_2,\dots,a_n)=d\)的对数. 那么假设\(F(d)\):修改\(k\)个后\(gcd(a_1,a_2,\dots,a_n)\)是\(d\)倍数的对数.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4675 题意:给出n,m,K,一个长度为n的数列A(1<=A[i]<=m).对于d(1<=d<=m),有多少个长度为n的数列B满足: (1)1<=B[i]<=m; (2)Gcd(B[1],B[2],……,B[n])=d: (3)恰有K个位置满足A[i]!=B[i]. 思路: i64 p[N]; void init(){    p[0]=1;    int i;    FOR1…