B. Powers of Two time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2 (i.…

Ivan and Powers of Two Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 404C Description Valera had an undirected connected graph without self-loops and multiple edges consisting of n ver…

Powers of Two 题意: 让求ai+aj=2的x次幂的数有几对,且i < j. 题解: 首先要知道,排完序对答案是没有影响的,比如样例7 1一对,和1 7一对是样的,所以就可以排序之后二分找2的x次幂相减的那个数就好了,注意:打表时2的x次幂不能只小于1e9,因为有可能是2个5e8相加,之后就超出了1e9,但是你打表的时候又没有超出1e9的那个2的x次幂,所以答案总是会少几个.所以尽量就开ll 能多打表就多打. 代码: #include <bits/stdc++.h> usin…

The Sum of the k-th Powers There are well-known formulas: , , . Also mathematicians found similar formulas for higher degrees. Find the value of the sum modulo 109 + 7 (so you should find the remainder after dividing the answer by the value 109 + 7).…

Consider all integer combinations ofabfor 2a5 and 2b5: 22=4, 23=8, 24=16, 25=32 32=9, 33=27, 34=81, 35=243 42=16, 43=64, 44=256, 45=1024 52=25, 53=125, 54=625, 55=3125 If they are then placed in numerical order, with any repeats removed, we get the f…

UVA 10622 - Perfect P-th Powers 题目链接 题意:求n转化为b^p最大的p值 思路:对n分解质因子,然后取全部质因子个数的gcd就是答案,可是这题有个坑啊.就是输入的能够是负数,负数的情况比較特殊,p仅仅能为奇数.这时候是要把答案不断除2除到为奇数就可以. 代码: #include <stdio.h> #include <string.h> #include <math.h> long long n; int prime[333333],…

标题手段: 他给了一个无限Pascal阵,定义了powers,然后询问power为P的pascal矩阵里面的第R行C列的元素是多少. 最開始读错题意了...然后 就成了一个神得不得了的题了.后来请教的别人. 感觉能够用矩阵高速幂做. 然后,不用高速幂的话,你会惊奇的发现,变成了找规律的题了... 答案成了 comb(i,j) * P^(i-j) 对于comb(i,j),利用组合数性质,能够得到,当j>i-j 的时候 j=i-j; 由于题目说答案不会爆long long  所以能够预处理,大概56…

B. Powers of Two time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2 (i.…

The Super Powers Time Limit: 1000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status] Description   A The Super Powers   We all know the Super Powers of this world and how they manage to get advantages in politica…

Perfect Pth Powers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16383   Accepted: 3712 Description We say that x is a perfect square if, for some integer b, x = b2. Similarly, x is a perfect cube if, for some integer b, x = b3. More g…

[CSAcademy]Sum of Powers 题目大意: 给定\(n,m,k(n,m,k\le4096)\).一个无序可重集\(A\)为合法的,当且仅当\(|A|=m\)且\(\sum A_i=n\).定义一个集合的贡献为\(\sum A_i^k\),求所有满足条件的集合的贡献之和. 思路: \(f[i][j]\)表示将\(j\)个数之和为\(i\)的方案数,有如下两种转移: \(f[i][j]+=f[i-1][j-1]\),表示新加入一个元素\(1\): \(f[i][j]+=f[i-j]…

Codeforces Round #486 (Div. 3) D. Points and Powers of Two 题目连接: http://codeforces.com/group/T0ITBvoeEx/contest/988/problem/D Description There are n distinct points on a coordinate line, the coordinate of i-th point equals to xi. Choose a subset of…

题目链接:Sad powers 题意:给出n个l和r,求出每个给出的[l,r]之间的可以使是另外一个数的k次方的数.(k>=2) 题解:题目给出的数据范围最大是1E18所以如果要直接把所有的从1-1E18的满足条件的数存下来的话一定会超时,但是我们可以注意到1e6^3 = 1e18,所以我们可以先把指数大于3的满足条件的数存下来,指数为2的分开来算(这里要把是别的数平方的数去掉,为了在后面算指数为2的数的时候更方便).在算指数为2的数的时候要注意不可以用floor(sqrt(x))这种方法来取整…

一段时间不写线段树标记,有些生疏了 codeforces 679e Bear and Bad Powers of 42 - CHADLZX - 博客园 关键点是:42的次幂,在long long范围内只有11个 考虑暴力修改 记录每个点距离下一个42次幂的距离,一般是负数 再记录每个点的等级,则有num=mi[lev+1]+val 特别地,当val=0的时候,num就是第lev个42的次幂 假如只有3操作,区间加,如果当前区间最大值大于等于0, 那么暴力下去升级:如果区间最大值大于等于0,就接着…

A positive integer xx is called a power of two if it can be represented as x=2y, where y is a non-negative integer. So, the powers of two are 1,2,4,8,16,… You are given two positive integers nn and k. Your task is to represent nn as the sumof exactly…

\(>Codeforces \space 622\ F. The\ Sum\ of\ the\ k-th\ Powers<\) 题目大意 : 给出 \(n, k\),求 \(\sum_{i=1}^{n} i^k\) 对 \(10^9 +7\) 取模的值 \(1 \leq n \leq 10^9, 0 \leq k \leq 10^6\) 解题思路 : 考虑 \(k\) 比较大,且模数不太好 \(NTT\),于是考虑插值做法, 学习了一波插值之后,发现就是个板子题 (雾) 拉格朗日插值: 对于…

A. Da Vinci Powers 题目连接: http://www.codeforces.com/contest/656/problem/A Description The input contains a single integer a (0 ≤ a ≤ 35). Input Output a single integer. Output In the first line print "YES" (without quotes), if Polycarpus can perf…

F. The Sum of the k-th Powers 题目连接: http://www.codeforces.com/contest/622/problem/F Description There are well-known formulas: , , . Also mathematicians found similar formulas for higher degrees. Find the value of the sum modulo 109 + 7 (so you shoul…

Ivan and Powers of Two time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Ivan has got an array of n non-negative integers a1, a2, ..., an. Ivan knows that the array is sorted in the non-decre…

/** 题目:The Super Powers UVA 11752 链接:https://vjudge.net/contest/154246#problem/Y 题意:求无符号长整形以内的数满足至少可以用两种不同的次方来表示.比如64 = 2^6 = 8^2: 一个数的1次方不算数. 思路: 分析过程如下: 1 = 1^1 1^2 1^3 16 = 2^4 4^2 64 = 2^6 8^2 81 = 3^4 9^2 256 = 2^8 16^2 512 = 2^9 8^3 设Max为最大的可能获…

题目链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2852 题意:找到在[1,2^64-1]区间范围内的所有Super Powers数,Super Powers数指的是可以写成另外两个正数的次幂: 例如:1=1^1,1=1^20;   64=8^2,64=4^4; 思路:1另外算,从2开始,他的指数如果不是素数,由于算数基本定理,…

B. Powers of Two   You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2 (i. e. some integer xexists so that ai + aj = 2x). Input The first line contains the single positive integer n …

题目链接 The Sum of the k-th Powers 其实我也不懂为什么这么做的……看了无数题解觉得好厉害哇…… #include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >= (b); --i) ; ; int v[N], p[N], f[N], r[N], b…

                                                         id=1730">Perfect Pth Powers Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16746   Accepted: 3799 Description We say that x is a perfect square if, for some integer b, x = b2.…

题目链接: B. Powers of Two time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of…

/* UVA11752 The Super Powers https://vjudge.net/contest/153365#problem/Y 数论 注意a^n=b要用除法求,并且求得的n比实际大1 */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <vector> #include <queue> #incl…

UVA 10515 - Powers Et Al. 题目链接 题意:求出m^n最后一位数 思路:因为m和n都非常大,直接算肯定是不行的,非常easy想到取最后一位来算,然后又非常easy想到最后一位不断乘自身肯定会形成周期.而且这个周期还挺短的,于是先求出周期,然后用n去取模该周期求出答案是第几个就可以 代码: #include <stdio.h> #include <string.h> int t[10]; int save[10][10]; char n[105], m[105…

Points and Powers of Two time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output There are nn distinct points on a coordinate line, the coordinate of ii-th point equals to xixi. Choose a subset of…

题目描述: The Sum of the k-th Powers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There are well-known formulas: , , . Also mathematicians found similar formulas for higher degrees. Find th…

转帖:Euler's Sum of Powers Conjecture 存不存在四个大于1的整数的五次幂恰好是另一个整数的五次幂? 暴搜:O(n^4) 用dictionary:O(n^3) import itertools def euler(m): """Yield tuples (a, b, c, d, e) such that a^5 + b^5 + c^5 + d^5 = e^5, where all are integers, and 1 < a ≤ b ≤…