题目The sum of consecutive integers could be( ). A: B: C: D: 相关知识点: 试题来源: 解析 C The sum of the consecutive integers from to is . 个连续整数的和可以是( ). . . . . 从到的个连续整数的和是. 故选.反馈 收藏
Sum of Consecutive Integers 题目链接 题意 问N能够分解成多少种不同的连续数的和. 思路 连续数是一个等差数列:$$ \frac{(2a1 + n -1)n}{2} = T$$ 那么2∗Tn−n=2∗a1−12∗Tn−n=2∗a1−1,所以当nn为TT的奇因子的时候符合要求. 那么当nn为偶数的时候2∗Tn−(n−1)=2...
sum = m*(a1+a1+m-1)/2 ---> a1 = sum/m - (m-1)/2 a1 和 m 一定是整数,所以sum%m = 0 并且(m-1)%2=0, 所以m是sum的因子,并且要是奇数; 所以我们只要求n的奇数因子的个数即可,求一个数的因子个数是所有素数因子的幂+1,相乘起来就是,那么素数只有2是偶数, 所以奇数因子的个数就是所...
Given an integer NN, you have to find the number of ways you can express NN as sum of consecutive integers. You have to use at least two integers. For example, N=15N=15has three solutions, (1+2+3+4+5),(4+5+6),(7+8).(1+2+3+4+5),(4+5+6),(7+8). Input Input ...
lightOJ 1278 Sum of Consecutive Integers(数论,数学推导),题目:给出一个数n,求出有多少种连续整数和等于n(连续整数至少两个)。“Forexample,N=15hasthreesolutions,(1+2+3+
Sum of Consecutive Integers LightOJ - 1278———(算术基本定理+容斥),GivenanintegerNNN,youhavetofindthenumberofwaysyoucanexpressNN
解析 A If the sum of 7 consecutive integers is 63, the middle one is 63÷7=9, The desired sum is 6+12=18. 如果7个连续整数的和是63,那么7个整数中最大和最小的和是( ). A.18 B.24 C.42 D.64 如果7个连续整数的和是63,中间的数是63÷7=9,期待的和是6+12=18. 故选A....
integersPartitionprimegapRiemannHypothesisLet A be a set of natural numbers, rA(n) be the number of ways to represent n as a sum of consecutive elements in A, and MA(x) := ∑n≤ x rA(n). Under various circumstances, we show that Mℕ(x) ~ ∑ MAi(x), where ℕ = ⊔ Ai ...
The sum of six consecutive positive integers is . What is the largest of these six integers?( ). A: B: C: D: E: 相关知识点: 试题来源: 解析 B 连续六个正整数的和是.则其中最大的数是? 这些数字的平均值是.因此数字是,,,所以答案是. 故选. 设第四个数是.我们想求的数是. 我们的整数...
所以对于唯一分解定理,其答案就是奇数质因子的幂次+1的乘积,最后去掉1。 #include<stdio.h>#include<math.h>#include<string.h>#include#include<vector>#include<algorithm>#defineN 2000006#definell long long#defineull unsigned long longusingnamespacestd;boolvis[10000010];intprime[N];intmain() {intk...