[参考博客]: https://blog.csdn.net/meopass/article/details/82952087 Farey Sums 题目描述 Given a positive integer, N, the sequence of all fractions a/b with (0 < a ≤ b), (1 < b ≤ N) and a and b relatively prime, listed in increasing order, is called the Farey…

题意:定义 Fn 序列表示一串 <1 的分数,分数为最简分数,且分母 ≤n .问该序列的个数.(2≤N≤10^6) 解法:先暴力找规律(代码见屏蔽处),发现 Fn 序列的个数就是 Φ(1)~Φ(n) 的和.于是用欧拉筛预处理就好了. 注意--求前缀和要用 long long 的类型. 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<iostream> 5 usi…

题意:构造一个数列,使得它们的区间和的种类最少,其中数列中不同的数的数目不少于k. 析:我们考虑0这个特殊的数字,然后0越多,那么总和种类最少,再就是正负交替,那么增加0的数量. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath&g…

就让0出现得尽可能多嘛……大概感受一下就是这样…… 0 0 ... 0 0 0 0 4 -4 3 -3 2 -2 1 -1 #include<cstdio> using namespace std; int n,m,a[510]; int main() { scanf("%d%d",&n,&m); int now=0; for(int i=n,j=1;i>=1;--i,++j) { if(m==1) a[i]=0; else { now=-now; i…

开始想的是O(n2logk)的算法但是显然会tle.看了解题报告然后就打表找起规律来.嘛是组合数嘛.时间复杂度是O(nlogn+n2)的 #include<cstdio> #include<cstring> #include<cctype> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s;i<=t;i++) #define dwn(i,s,t) for(in…

Backward Digit Sums Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5664   Accepted: 3280 Description FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum ad…

1712: [Usaco2007 China]Summing Sums 加密 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 338  Solved: 127[Submit][Status][Discuss] Description     那N只可爱的奶牛刚刚学习了有关密码的许多算法,终于,她们创造出了属于奶牛的加密方法．由于她们并不是经验十足,她们的加密方法非常简单:第i只奶牛掌握着密码的第i个数字,起始的时候是Ci(0≤Ci<90000000).…

Codeforces 1091D New Year and the Permutation Concatenation https://codeforces.com/contest/1091/problem/D 题目: Let n be an integer. Consider all permutations on integers 1 to n in lexicographic order, and concatenate them into one big sequence p. For…

考试时候遇到这种题只会找规律 You've got an array a, consisting of n integers. The array elements are indexed from 1 to n. Let's determine a two step operation like that: First we build by the array a an array s of partial sums, consisting of n elements. Element nu…

n阶的法里数列是0和1之间最简分数的数列,由小至大排列,每个分数的分母不大于n. Stern-Brocot树(SB Tree)可以生成这个序列 {0/1,1/1} {0/1,1/2,1/1} {0/1,1/3,1/2,2/3,1/1} {0/1,1/4,1/3,1/2,2/3,3/4,1/1} {0/1,1/5,1/4,1/3,2/5,1/2,3/5,2/3,3/4,4/5,1/1} {0/1,1/6,1/5,1/4,1/3,2/5,1/2,3/5,2/3,3/4,4/5,5/6,1/1} {0…