分析:我们可以看出这道题目的描述并不是很复杂,就是说对于一个给定的整数n,我们能否把他拆成k个powerful的数,也就是说这k个数要么是2的幂次,要么是某个数的阶乘,并且我们要让当前的k越小越好:然后如果不能被拆的话输出-1: 我们这样来看,先看会不会输出-1,我们如果把这个整数n用二进制的方法写出来,每个1都表明可以写成某个powerful的数,所以不可能输出-1: 那么我们就可以发现了k的个数就是这里二进制表示中1的个数,但是我们考虑到还有阶乘,我们令阶乘的和为s,个数为cnt,则k =…

传送门 今天在HackerRank上翻到一道高精度题,于是乎就写了个高精度的模板,说是模板其实就只有乘法而已. Extra long factorials Authored by vatsalchanana on Jun 16 2015 Problem Statement You are given an integer N. Print the factorial of this number. N!=N×(N−1)×(N−2)×⋯×3×2×1 Note: Factorials of N>20…

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…

时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:2109 解决:901 题目描述: John von Neumann, b. Dec. 28, 1903, d. Feb. 8, 1957, was a Hungarian-American mathematician who made important contributions to the foundations of mathematics, logic, quantum physics, meteorology, scienc…

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).…

Description John von Neumann, b. Dec. 28, 1903, d. Feb. 8, 1957, was a Hungarian-American mathematician who made important contributions to the foundations of mathematics, logic, quantum physics,meteorology, science, computers, and game theory. He wa…

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.…

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. Find the sum of all numbers which are equal to the sum of the factorial of their digits. Note: as 1! = 1 and 2! = 2 are not sums they are not included. 题目大意: 145 是一个奇怪的数字, 因为 1! + 4! + 5!…

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…