题意 给定 \(n\) 个数 \(a_1,a_2,\cdots a_n\),对于每个 \(K\in[1,n]\) ,求出 \(n\) 个数的每个子集的前 \(K\) 大数的和,输出每个值,对 \(998244353\) 取模. \(1\leq n \leq 10^5\) 思路 设 \(K\) 为 \(k\) 时的答案为 \(ans_k\) 有 \[ ans_k=\sum_{i=1}^na_i2^{n-i}\sum_{j=0}^{k-1}{i-1\choose j} \] \(j\) 为在 \(a…

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has n numbers A[1]~A[n] and a number K. For any none empty subset S of the numbers, the value of S is…

http://acm.hdu.edu.cn/showproblem.php?pid=6092 题意: 给出两个数组A和B,A数组一共可以有(1<<n)种不同的集合组合,B中则记录了每个数出现的次数,现在要根据B数组来推出A数组最小的序列. 思路: 如果$B_{i}$是 B 数组中除了$B_{0}$ 以外第一个值不为 0 的位置,那么显然 i 就是 A 中的最小数. 那么我们每次取出$B_{i}$一个数,对于后面的数组来说,满足$B_{j}=B_{j}-B_{j-i}$,为什么? 其实仔细想想就…

Rikka with Subset Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1122    Accepted Submission(s): 541 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation…

题目链接 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has n positive A1−An and their sum is m. Then for each subset S of A, Yuta ca…

题意 \(n\) 局石头剪刀布,设每局的贡献为赢的次数与输的次数之 \(\gcd\) ,求期望贡献乘以 \(3^{2n}\) ,定义若 \(xy=0\) 则,\(\gcd(x,y)=x+y\) 思路 不难得出 \[ ans=3^n\sum_{i=0}^n\sum_{j=0}^{n-i}{n\choose i}{n-i\choose j}\gcd(i,j) \] 对于正整数 \(n\) ,有如下表达式 \[ n=\sum_{d|n}\varphi(d) \] 那么 \[ ans=3^n\sum_{…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6092 #include <cstdio> #include <iostream> #include <cstring> using namespace std; ; int b[maxn] , dp[maxn] , n , m; int main() { int t; scanf("%d", &t); while(t--) { scanf(&qu…

Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: For a tree T, let F(T,i) be the distance between vertice and vertice i.(The length of e…

题意: 输入一棵树,判断这棵树在以节点1为根节点时,是否是一棵特殊的树. 相关定义: 1.  定义f[A, i]为树A上节点i到节点1的距离,父节点与子节点之间的距离为1. 2.  对于树A与树B,如果A与B的节点数相同,且无论i为何值,f[A, i]与f[B, i]都相等,则A与B为两棵相似的树. 3.  对于一棵树A,在以节点1为根节点的情况下,如果不存在与其它树与A相似,则A是一棵特殊的树. 输入: 包含多组输入样例. 每组输入样例中首先输入一个整数n,表示一棵含有n个节点的树. 接下来n…

n个点最少要n-1条边才能连通,可以删除一条边,最多删除2条边,然后枚举删除的1条边或2条边,用并查集判断是否连通,时间复杂度为O(n^3) 这边犯了个错误, for(int i=0;i<N;i++){ fa[i]=i; } 这个将i<=N,导致错误,值得注意 AC代码: #pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio>…